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Number Theory
Theory and questions for topic based enrichment
activities/teaching
Contents
Prerequisites ................................................................................................................................................................................................................ 1
Questions ..................................................................................................................................................................................................................... 2
Investigative ............................................................................................................................................................................................................. 2
General Questions ................................................................................................................................................................................................... 2
Senior Maths Challenge Questions .......................................................................................................................................................................... 3
British Maths Olympiad Questions .......................................................................................................................................................................... 8
Key Theory ................................................................................................................................................................................................................... 9
Solutions..................................................................................................................................................................................................................... 13
Theory Questions ................................................................................................................................................................................................... 13
General Questions ................................................................................................................................................................................................. 15
SMC Questions ....................................................................................................................................................................................................... 20
British Maths Olympiad ......................................................................................................................................................................................... 33
Prerequisites




For some questions, appreciation of the factorial function n!
For one of the questions, knowledge of the difference between a convergent and divergent infinite series.
For one of the questions, knowledge of logarithms and determining a turning point using calculus.
Appreciation of big-sigma  to indicate a summation over a range.
Questions
Investigative
These questions are intended to highlight key theory in number theory or proofs you should be aware of (the latter of which you should
research, as you would not be expected to prove them). For more general questions, see those in the next section or the SMC/BMO questions.
1. If x is divisible by a, then is x2 divisible by a?
2. If x2 is divisible by a, then is x divisible by a? More generally, if xn is divisible by a for some integer n, then is x divisible by a?
3. Prove that √2 is irrational.
4. Prove that
is divergent.
5. Show that there are an infinite number of prime numbers.
6. Why is 1 not considered to be prime?
General Questions
1.
2.
3.
4.
5.
6.
Show that the sequence 2, 5, 8, 11, 14, 17, … can never be a square number.
Show that 2n = n3 has no integer solution for n.
Does n have an integer solution to
Find integer solutions to xy + x + y = 2004 where x>0, y>0.
How many positive integer values of n is
also an integer?
Suppose an integer has d distinct prime factors, where each is repeated a i times. How many divisors does the number have? How many
factors does 1010 have?
7. Find all integer solution to the equation ab = ba that are non-trivial (ab), and show that these are the only solutions.
8. (a) Suppose n > 2 is a positive integer and that 2k is the highest power of 2 that divides any term in the sequence 2, 3, 4, ... , n. Show
that only one term of the sequence is divisible by 2k. Deduce that lcm(2, 3, 4, ... , n) = 2kc where c is an odd integer.
(b) By expressing the number
in the form a/b, show that x cannot be an integer.
9. Show that if a number is divisible by 3, its digits add up to a multiple of 3.
10. Determine the largest power of 10 that is a factor of the following:
(a)
50! (b)
1000! (c)
In general, n!
Senior Maths Challenge Questions
Level 1
1. How many prime numbers are there less than 20?
A
6
B
7
C
8
D
9
E
10
Level 3
1. Which of these five expressions represents the largest
number?
A
9 (9
9
)
B 999
C
999
D (99)9
E
999
2. What is the remainder when the number 743589 × 301647
is divided by 5?
A
0
B
1
C
2
D
3
E
4
2. What is the value of
1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 +
2))))))))))?
A 210 +1 B 211 −1 C 211 +1 D 212 −1 E
212 + 1
3. Exactly one of the following numbers is divisible by 11.
Which is it?
A
107 − 11
B
107 −1
C
107
D
107 +1
E
107 + 11
3. I am trying to do a rectangular jigsaw puzzle. The puzzle was
made by starting with a rectangular picture and then cutting
it into 1000 pieces by sawing along the lines of a (wiggly!)
rectangular grid. I start by separating out all the edge and
corner pieces. Which of the following could not possibly be
the number of corner and edge pieces of such a jigsaw?
A
126 B
136 C
216
D
316 E
A-D are all possible
Level 2
1. The mean of seven consecutive odd numbers is 21. What is
the sum of the first, third, fifth and seventh of these
numbers?
A
16
B
21
C
84
D
147 E
more information needed
2. The value of the product wxyz is 2002 and w, x, y and z are
prime numbers. What is the value of w2 + x2 + y2 + z2?
A
66
B
203 C
260
D
285 E
343
4. Observe that 18 = 42 + 12 + 12 + 02. How many of the first
fifteen positive integers can be written as the sum of the
squares of four integers?
A
11
B
12
C
13
D
14
E
15
5. The probability of a single ticket winning the jackpot in the
National Lottery is
6
5
4
3
2
1





.
49 48 47 46 45 44
If I buy one ticket every week, approximately how often
might I expect to win the jackpot?
A once every hundred years
B once every twenty thousand years
C once every hundred thousand years
D once every quarter of a million years
E once every million years
6. The factorial of n, written n!, is defined by n! = 1 × 2 ×
3×…×(n − 2) × (n − 1) × n
e.g. 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720. What is the smallest
positive integer which is not a factor of 50! ?
A
51
B
52
C
53
D
54
E
55
7. What is the value of (612 − 392) ÷ (512 − 492)?
A
10·5 B
11
C
12
D
21
E
22
8. Given that a and b are integers greater than zero, which of
the following equations could be true?
A
a–b=a b
B
a+b=a b
C
a–b=a  b
D a+b=a–b
a+ b = a + b
E
9. Observe that 2000 = 24 × 53. What is the number of the next
year after the year 2000 which can be written ab × cd where
a, b, c, d are 2, 3, 4, 5 in some order?
A
D
2016 B
2048 E
2025 C
2050
2040
10. Which of the following numbers n gives a counter-example
for the statement: ‘If n is a prime number then n2 + 2 is also
a prime number’?
A
3
B
5
C
6
D
9
E
none of them
n 20
n
11. When n = 81, what is the value of 3 ?
1
1
100
3
A less than
B
C
1
D
3
E
more than 100
12. The number of the year 2003 is prime.
How many square numbers are factors of 20032003?
A
0
B
1
C
44
D
1002 E
2003
13. A square number is divided by 6. Which of the following
could not be the remainder?
A
0
B
1
C
2
D
3
E
4
Level 4
1. What is the maximum possible value of the product of a list
of positive integers, which are not necessarily all different,
given that the sum of these numbers is 100?
A
1010
B
250
C
220 × 320
D
520
E
22 × 332
2. Sam correctly calculates the value of 58 × 85. How many
digits does her answer contain?
A
11
B
12
C
13
D
14
E
15
3. Note that 2001 = 3 × 23 × 29. What is the number of the
next year which can be written as the product of three
distinct primes?
A
2002 B
2004 C
2006
D
2007 E
none of these
4. For how many integer values of n does the equation x2 + nx
− 16 = 0 have integer solutions?
A
2
B
3
C
4
D
5
E
6
5. Which of the following is divisible by 3 for every whole
number x?
A
x3 − x
B
x3 − 1
C
x3
D
x3 + 1
E
x3 + x
6. What is the value of 22003 − 22002 − 22001 − 22000?
A
−22002 B
0
C
2−4000
D
64
E
22000
7. The value of 12004 + 32004 + 52004 + 72004 + 92004 is calculated
using a powerful computer. What is the units digit of the
correct answer?
A
9
B
7
C
5
D
3
E
1
8. Which is the lowest positive integer by which 396 must be
multiplied to make a perfect cube?
A
11
B
66
C
99
D
121 E
726
9. How many two-digit numbers N have the property that the
sum of N and the number formed by reversing the digits of
N is a square?
A
2
B
5
C
6
D
7
E
8
10. Damien wishes to find out if 457 is a prime number. In order
to do this he needs to check whether it is exactly divisible by
some prime numbers. What is the smallest number of
possible prime number divisors that Damien needs to check
before he can be sure that 457 is a prime number?
A
8
B
9
C
10
D
11
E
12
1. Three people each think of a number which is the product
of two different primes. Which of the following could be the
product of the three numbers which are thought of?
A
120 B
144 C
240
D
3000 E
12100
6. Let N be a positive integer less than 102002. When the digit 1
is placed after the last digit of N, the number formed is
three times the number formed when the digit 1 is placed in
front of the first digit of N. How many different values of N
are there?
A
1
B
42
C
333
D
667 E
2002
2. Just one of the following is a prime number. Which one is it?
A
10002 + 1112 B
5552 + 6662
C
20002 – 9992 D
10012 + 10022
E
10012 + 10032
7. Given an unlimited supply of 50p, £1 and £2 coins, in how
many different ways is it possible to make a sum of £100?
A
1326 B
2500 C
2601
D
5050 E
10 000
3. The statement “There are exactly four integer values of n
for which (2n + y)/ (n − 2) is itself an integer” is true for
certain values of y only. For how many values of y in the
range 1 ≤ y ≤ 20 is the statement true?
A
0
B
7
C
8
D
10
E
20
8. How many pairs of positive integers (x, y) are solutions of
Level 5
4. How many pairs of positive integers (x, y) satisfy the
equation x – 17  y ?
A
0
B
1
C
2
D
17
E
infinitely many
5. For how many values of n are both n and 4
A
1
B
2
C
3
D
4
E
5
 n 1 


 n 1 
integers?
the equation
A
D
0
3
1 2
3
+ =
?
x y 19
B
E
1
C
2
more than 3
9. The number N is exactly divisible by 7. It has 4008 digits.
Reading from left to right, the first 2003 digits are all 2s, the
next digit is n and the last 2004 digits are all 8s. What is the
value of n?
A
4
B
5
C
0 or 3
D
2 or 9
E
1 or 8
10. Positive integers x and y satisfy the equation:
x + 1 y – x – 1 y = 1.
2
2
Which of the following is a possible value of y?
A
5
B
6
C
7
D
8
E
9
11. What is the sum of the values of n for which both n and
n2 – 9
are integers?
n –1
A
D
−8
4
B
E
−4
8
C
0
12. Given that S = (x + 20) + (x + 21) + (x + 22) +…+ (x + 100),
where x is a positive integer, what is the smallest value of x
such that S is a perfect square?
A
1
B
2
C
4
D
8
E
64
13. The factorial of n, written n!, is defined by n! = 1 × 2 × 3 ×…×
(n − 2) × (n − 1) × n. For how many positive integer values of
k less than 50 is it impossible to find a value of n such that n!
ends in exactly k zeros?
A
0
B
5
C
8
D
9
E
10
British Maths Olympiad Questions
1. One number is removed from the set of integers from 1 to
n. The average of the remaining numbers is 40.75. Which
integer was removed?
2. Determine the sequences a0, a1, a2, … which satisfy all of the
following conditions:

for every integer
,

is a rational number and

for some I, j with
3. Find all positive integers n such that both n+8000 divides n2
+ 2008 and n+2009 divides n2 + 2009.
4. Find four prime numbers less than 100 which are factors of
332 – 232.
5. Let n be an integer. Show that, if
is an
integer, then it is a perfect square.
6. Each of Paul and Jenny has a whole number of pounds. He
says to her: “If you give me £3, I will have n times as much
as you.” She says to him: “If you give me £n, I will have 3
times as much as you”. Given that all these statements are
true and that n is a positive integer, what are the possible
values for n?
7. Determine the least possible value of the largest term in an
arithmetic progression of seven distinct primes.
8. Let s be an integer greater than 6. A solid cube of side s has
a square hole of side x<6 drilled directly through from one
face to the opposite face (so the drill removes a cuboid).
The volume of the remaining solid is numerically equal to
the total surface area of the remaining solid. Determine all
possible integer values of x.
9. Determine the least natural number n for which the
following result holds: No matter how the elements of the
set {1, 2, … , n} are coloured red or blue, there are integers,
x, y, z, w in the set (not necessarily distinct) of the same
colour such that x + y + z = w.
10. Find, showing your method, a six-digit integer n with the
following properties: (i) n is a perfect square, (ii) the number
formed by the last three digits of n is exactly one greater
than the number formed by the first three digits of n. (Thus
n might look like 123124, although this is not a square.)
Key Theory
What kind of problems do we find in
number theory?
If a and b are integers, then ab is divisible
by both a and b.
Key term: Modulo Arithmetic
Explanation
Number theory is concerned solely with the study of ‘integers’, i.e. whole numbers, and various properties
concerning them. It’s interested in problems like the following:
 Integer solutions to equations.
 Various properties of prime numbers: Are there an infinite number of them? Is there any way to
enumerate all the prime numbers? What’s the probability that a given number is prime? Can all
integers be expressed as the sum of two primes?
 Divisibility problems: Properties involving factors of numbers and expressions and remainders.
 Properties of rational numbers: Rational numbers are simply those that can be expressed as a
fraction involving integers, so number theory typically asks questions about rational numbers as
well. Is 2 rational? Is e rational?
 Types of numbers and associated questions. e.g. Perfect numbers (where the factors of the number
add up to itself, e.g. 6): do they all end in 6 or 8? Are there an infinite number of them?
This might seem obvious, but is important to keep in mind. If we have an expression like 3x2 say and we
know that x is an integer, then 3x2 is divisible by 3.
Modulo arithmetic is when we do arithmetic on numbers, but where the numbers ‘wrap around’ when
reaching a certain number known as the modulus. We do modulo arithmetic with angles and time on a clock
for example.
Formally, we write:
When we find the remainder of a sum of
numbers when dividing by a, we can find
the remainders of each term in the sum,
to mean that the remainder when we divide both a and b by c is the same. So for example
and
. Sometimes you might see the mod used more in a functional way like say addition
or subtraction, for example 11 mod 2 = 1, which means “the remainder when we divide 11 by 2”. But usually
the ‘mod’ comes at the end and in brackets.
I use the equality symbol = from here on for convenience.
This is one of the features of modulo arithmetic. For example, suppose we were to find the remainder of 7 +
4 when dividing by 3. We could find the remainder when dividing 7 by 3 (i.e. 1), and similarly with 4 (i.e. 1
again), then add these remainders together (i.e. 2). This gives us the same result than if we were to directly
add them together, and then find the
remainder.
divide 11 by 3.
An implication of this is that we can tell 3n + 2 gives us a remainder of 2 when dividing by 3. Because 3n
gives us a remainder of 0, and 2 gives us a remainder of 2. So the overall remainder is 2.
Modulo arithmetic also works with
multiplication.
If a = b (mod c), then an = bn (mod c) for
any positive integer n.
If 5 = 2 (mod 3) and 8 = 2 (mod 3), then we can work out the remainder when we divide 5 x 8 = 40 by 3. We
multiply the remainders, giving 2 x 2 = 4. But since this is modulo arithmetic, we actually get 1 (since the 4
wraps around).
This follows from the above. If a gives a remainder of b when dividing by c, then
gives a remainder of
since we earlier said that we could multiply the remainders. Repeating this n times results in the
stated law.
Examples:
4 = 1 (mod 3), so 16 = 1 (mod 3), and 64 = 1 (mod 3).
7 = 2 (mod 5), so 49 = 4 (mod 5)
5 = 2 (mod 3), so 25 = 1 (mod 3), because the 22 = 4 ‘wraps’ to 1 in modulo 3.
If a = b (mod c), then ka = kb (mod kc)
If a = b (mod c), then a-b is divisible by c.
The factorial function n!
Diophantine Equations
We’ll see how this law can be used in solving various problems.
For example 24 = 3 (mod 7). Then 48 = 6 (mod 14).
If for example 103 gives a remainder of 3 when divided by 10, then clearly 103-3 = 100 is divisible by 10. This
gives an effective means of dealing with many divisibility-themed questions, as will be seen.
Number theory type problems frequently involve the factorial function. It is defined as such:
0! = 1
n! = n x (n-1) x (n-2) x …. x 2 x 1
So 4! = 24 for example. See the Combinatorics set of problems for lots of examples of its use.
Diophantine equations are polynomial equations where the variables are restricted to integers. For example
for constants a, b and c we might be interested in integer solutions for ax + by = c (known as a Linear
Diophantine Equations), or solutions to x2 + y2 = z2 (known as Pythagorean triples).
A famous historical problem known as Hilbert’s 10th Problem (posed in 1900) asks whether there is some
standard procedure by which we can either find all solutions to a Diophantine Equations or show that none
exist (for example in the case x3 + y3 = z3, due to Fermat’s Last Theorem). It’s since been proven that no such
procedure exists.
There are 4 well-known examples of Diophantine Equation:
Form
ax + by = c
If we find the factors of some number x,
then the product of any subset of factors
of x divides x.
Key Theorem: Fundamental Theorem of
Arithmetic (also known as the Unique
Factorisation Theorem)
Key Term: Lowest Common Multiple
(LCM)
Fermat’s Little Theorem
Info
Linear Diophantine Equation: This has infinitely many solutions if c is a
multiple of the greatest common divisor of a and b, and no solutions
otherwise. It may be worth learning how to solve such equations.
The Erdos-Staus Conjecture states that for any integer n>1, 4/n can be
expressed as the sum of 3 positive unit fractions. It’s been verified for values
up to very large n, but not yet proven.
See ‘Famous Theorems’ below.
Known as Pell’s Equation. We’re interested in integer solutions for x and y
(apart from the trivial x=1 and y=0) when n is non-square. Historical interest
in the equation was for its use in finding approximations to square roots of
numbers. For example, in finding solutions for x2 – 2y2 = 1, x/y gives an
approximation for 2. Lagrange proved that there are infinitely many
solutions for any Pell Equation.
For example, 30 = 5 x 3 x 2. Then 3 x 2 divides 30, 5 x 2 divides 30, 2 divides 30, etc.
This might seem obvious, but is a principle used in some SMC questions.
All integers greater than 1 can be uniquely factorised as a product of primes. If 12 = 2 x 2 x 3, then there is
no other product of primes which yields 12 (except by reordering the factors).
When we find the prime factorisation of a number, we tend to group identical factors using powers, i.e. 300
= 22 x 3 x 52. This makes it easier to both count the total prime factors and the distinct factors.
The smallest number which divides a given set of integers.
Not to be confused with ‘Fermat’s Last Theorem.
If p is a prime number, then for any integer a:
or equivalently:
Famous Theorems in Number Theory
So 612 = 1 (mod 13) for example, and 612 – 1 is divisible by 13.
The guidance document for the British Maths Olympiad recommends that you know this for Round 2.
Reading about some of these problems will broaden your general mathematical knowledge (and if
applicable, hopefully provide some good fodder for your university personal statement!)
Name
Goldbach Conjecture
Fermat’s Last
Theorem
Prime Number
Theorem
Euclid’s Theorem
Explanation
Are all positive integers are the sum of two primes?
an + bn = cn has no integer solutions for a, b and c when n>2.
The probability that a randomly chosen integer n is prime is
approximately 1/ln(n) (the reciprocal of the natural log of n).
There are an infinite number of prime numbers (see Theory
Questions).
Euler Product
Formula
Solved?
No.
Yes, by Andrew
Wiles in 1995.
NA
Yes (by Euclid!)
Euler in 1737.
i.e. The formula relates a summations involving the natural
numbers with a product involving all the prime numbers.
e.g.
Riemann Hypothesis
The summation on the LHS of the equation above is known as
the Riemann Zeta Function.
The domain of the function can be extended to complex
numbers. The Riemann Hypothesis states that all roots of the
Riemann Zeta function, i.e. s for which
, have the
form
for any a (except for ‘obvious’ roots known as
‘trivial roots’). Recommended book: “Prime Obsession:
Bernhard Riemann and the Greatest Unsolved Problem in
Mathematics” by John Derbyshire.
No.
Solutions
Investigative Questions
Question
If x is divisible by a, then is x2
divisible by a?
If x2 is divisible by a, then is x
divisible by a? More generally, if xn is
divisible by a for some integer n,
then is x divisible by a?
Solution
To formally prove: If a is a factor of x, then x = ka for some k. Then x2 = k2 a2, where the RHS is clearly divisible by a.
But the answer is clear just from observation. If x is divisible by a, then any whole number times x is still visible by
a.
Teaching points: The idea here is to get students to appreciate that multiplying an integer by another doesn’t
‘lose’ any factors from the original number, only potentially adds them.
Only if a is prime. The key here is prime factorisation. Whatever prime factors we have in x will appear twice in x2,
but ultimately we don’t add any new prime factors when squaring (or taking to any integer power) other than
repeating them. If we have the same unique prime factors in x2 and x, then clearly a will be a factor of x. Exactly
the same argument applies for xn.
There’s simple counter examples if a is not prime. 32 is divisible by 9, but 3 is obviously not divisible by 9. This is
because 9 is not prime.
Prove that √2 is irrational
Teaching points: It’s a fact needed in the proof of root 2’s irrationality amongst others.
We do a ‘proof by contradiction’, i.e. asserting the opposite of the theorem, and showing this leads to a
contradiction.
If √2 is rational, then suppose there are some a and b such that √2 = a/b and a/b is a fraction in its simplest form.
By multiplying by sides by b and squaring both sides, we get 2b2 = a2. We can see 2b2 is divisible by 2. So a2 must
also be divisible by 2. By the theorem above (i.e. if xn is divisible by a and a is prime, then is x divisible by a), a must
be divisible by 2 since 2 is prime.
Since a is divisible by 2 (i.e. is even), then our fraction a/b is not in the simplest form. This contradicts what we
initially asserted, so √2 must be irrational.
This reasoning applies for the root of any prime number.
Teaching points: Appreciating how to solve this problem is much more important than this particular proof itself,
Prove that
divergent.
is
otherwise students are likely to quickly forget the proof. The emphasis in presenting the solution should probably
be the strategy in getting to the each subsequent step of the proof. The hardest part of the proof perhaps (in
terms of knowing what step you would take next) is making the assertion that you need the fraction in its simplest
form – it’s not obvious at that particular point why it will be useful to us later in the proof.
This particular series is known as the harmonic series. A series is divergent when the sum of an infinite number of
terms is infinity. This contrasts to a convergent series such as
, which gradually approaches 2.
There’s two proofs for this on http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29
Show that there are an infinite
number of prime numbers.
Why is 1 not considered to be prime?
Teaching points: Note that the harmonic series is not a geometric series.
This is known as Euclid’s Theorem, published in the book ‘Elements’ in 300BC.
The proof goes as such:
 Suppose we have a list of all the known primes. Multiply them together, and add 1. i.e. n = (p1p2....pk) + 1.
 We know that if we divide n by any of our known primes pi, then the remainder will be 1. Thus none of
these primes can be divisors of n.
 Therefore either n is prime not in our original list, or n is composite, but its prime factors are not in our
original list.
Teaching points: This proof is important as part of a broad number theory knowledge. To teach it, you could
potentially give different sets of cards, just give them the first bullet point of the proof, and see what they notice
for different sets, so as to distinguish between the cases when n is prime and composite. One point to emphasise
is that the method doesn’t enumerate all prime factors; it only shows we can always find a new prime given a
finite list. If we start with just 2 for example, 2 + 1 = 3, then (2 x 3) + 1 = 7, then (2 x 3 x 7) + 1 = 43, (2 x 3 x 7 x 43) +
1 = 1807 = 13 x 39. We so far have generated the primes {2, 3, 7, 13, 43, 139}, which is clearly not a complete list of
the first few prime numbers.
Here’s a nice explanation of the various reasons 1 is not considered a prime:
http://primes.utm.edu/notes/faq/one.html
The most convincing reason is due to the definition of the Fundamental Law of Arithmetic. If 1 is prime, then a
number 10 say can be prime factorised as 5 x 2, or 5 x 2 x 1, or 5 x 2 x 1 x 1 and so on. This violates the condition
that there must be a unique prime factorisation of any integer.
General Questions
1
Question
Show that the sequence 2, 5, 8, 11, 14,
17, … can never be a square number.
Solution
This is equivalent to saying “Show that the square numbers are always a multiple of 3 or one
more than a multiple of 3.
Source
?
JAF’s solution: We can use modulo arithmetic to get the solution. For any number x, then x =
0, 1 or 2 in modulo 3 arithmetic (cycling through these as x increases).
So then x2 = 02, 12 or 22 (mod 3).
Thus x2 = 0, 1 or 1 (mod 3), since 4 is 1 in modulo 3 arithmetic. This tells us that the remainder
of the squares when dividing by 3 oscillates between 0, 1 and 1, i.e. being a multiple of 3,
followed by two square numbers one more than a multiple of 3. But the numbers in the
sequence are all one less than a multiple of 3 (i.e. are all 2 in modulo 3).
2
3
4
Show that 2n = n3 has no integer
solution. for n.
Find integer solutions to xy + x + y =
2004 where x>0, y>0.
Does n have an integer solution to
Teaching points: This law is really incredibly useful in a number of problems involving
divisibility. Students ought to learn it and appreciates its usefulness.
This is one of many questions where we wish to appreciate the prime factors on each side of
the equation. On the LHS of the equation, we only have prime factors of 2. Then n3 can only
have factors of 2, and thus by our usual theorem n can only have factors of 2. If this is the
case, we can put n in the form 2k for some integer k. Substituting this in, we get:
Then 2k = 3k by comparing the powers. The RHS of the equation is divisible by 3, but the LHS is
not. So k cannot be an integer, which contradicts what we asserted about k.
This problem is in the British Maths Olympiad guidance document under ‘key things you
should know about Number Theory’. So take heed!
The key here is to spot potential factorisations. xy + x + y + 1 = (x+1)(y+1). Therefore (x+1)(y+1)
= 2005. The only factors pairs of 2005 are 1 and 2005 (giving 0 and 2004, which is not
allowed), and 5 and 401 (giving 4 and 400 as solutions).
We can use the same method as above. Firstly note we can factorise the RHS as n2(1 + n). If
again the RHS has to only contain prime factors of 2, then each of its factors must also only
BMO
guidance
doc
contain prime factors of two. So the n term we factored out must again only contain prime
factors of two. By again substituting n = 2k, we obtain:
5
How many positive integer values of n
is
also an integer?
6
Suppose an integer has d distinct
prime factors, where each is repeated
ai times. How many divisors does the
number have? How many factors does
1010 have?
7
Find all integer solution to the
equation ab = ba that are non-trivial
(ab), and show that these are the
only solutions.
We can see the RHS must be odd unless k = 0 (which yields n = 1 as a solution). But otherwise,
the RHS is odd and LHS even, so there’s no other solution.
JAF’s solution: The first thing that came to me was “Wouldn’t it be easier if the subtraction
was in the numerator instead of the denominator – then we could simplify”. So I created a
variable n = 100 – x. Substituting this in, you get (100 – x) / x, which conveniently simplifies to
(100/x) – 1. If you choose factors of 100 for x, then the whole expression is clearly an integer.
So x = 1, 2, 4, 5, 10, 20, 25, 50, 100. Since n = 100 – x, we obtain n = 99, 98, 96, 95, 90, 80, 75,
50, 0.
This is more of a combinatorics problem, since we’re concerned with the number of ways of
combining the prime factors to form different divisors.
If we look at 20 for example (2 x 2 x 5), our divisors are obtained using 2, (2 x 2), 5, (2 x 5) and
(2 x 2 x 5), as well as using none of the prime factors (yielding the divisor 1). The factor of 2
can be seen zero, one or two times, and the factor of 5 zero or one time. So the number of
divisors is just:
where the +1 is because we need to include the ‘zero’ number of times a prime factor can
appear in a factor.
For the second part of the question, 1010 = (2 x 5)10 = 210 x 510. Using the formula, this gives 11
x 11 = 121 factors.
This is an incredibly interesting problem! Older students might initially be inclined to take logs
of both sides, giving b log(a) = a log(b). Rearranging, we get log(a) / a = log(b) / b. We can’t
simplify further (i.e. we can’t make a the subject of the formula in a = log(a)). However, if we
define a function f(x) = log(x) / x, then we have a solution if f(a) = f(b).
SMC
JAF
Classic
problem
At this point, we need to sketch the function f(x) = log(x) / x. (The key to sketching it is
identifying the domain, the range, roots and the turning points)
8
(a) Suppose n > 2 is a positive integer
and that 2k is the highest power of 2
that divides any term in the sequence
2, 3, 4, ... , n. Show that only one term
of the sequence is divisible by 2k.
Deduce that
lcm(2, 3, 4, ... , n) = 2kc
where c is an odd integer.
The turning point (using calculus) is at (e, 1/e), and the root is x = 1. By observation of the
shape of the graph, we can see that if f(a) = f(b), then if
, then
. If however a <
1, then the only b for which f(a) = f(b) is when a = b (i.e. our trivial solution). So one of a or b
must be in the range (1,e). 2 is the only integer in this range, and thus is the only integer
solution. So 24 = 42 is the only integer solution.
We had an earlier question on the proof of the harmonic series up to infinity being divergent.
Now we’re trying to show (in part (b)) that it’s not an integer either for any n.
The first part of (a) is rather confusing until we use a concrete example. Suppose we have the
number 2, 3, 4, 5, 6, 7, 8, 9, 10. Then 8 is the highest power of 2. Clearly nothing less than 8 is
divisible by 8, and the lowest number greater than 8 that is divisible by 8 is twice it, i.e. the
next power of 2. But we know this must be greater than n because we defined 2k to be the
highest power of 2 less than n. So 2k itself is the only term in the sequence divisible by 2k.
Oxford
Interview
(b) By expressing the number
in the form a/b,
show that x cannot be an integer.
The next part of (a) is a little tricky to understand. All integers can be expressed in the form 2r
c, where c is odd. So the key here is showing that r = k when we have the lowest common
multiple of 2 to n. We previously showed that 2k is the highest power of 2 that will divide one
of the terms in the sequence 2 to n. By finding the LCM, we’ve found a number which divides
all of 2 to n.
Now,
because we need to ensure this LCM can be divided by the number 2k in the
sequence. But we also don’t need r to be any greater than k, because no term in the sequence
2 to n has more the k twos as factors, and thus the LCM need not have more than k twos to be
divisible by all the numbers.
(b) is a bit more fun. In order to add the numerators, we need to make the denominators the
same, and finding the LCM of 2 to n is sufficient. Then:
for some natural number a. To make the sum of fractions a sum of integers, we can multiply
both sides by 2kc, since we know 2kc / 2 is an integer, 2kc / 3 is an integer, etc. (because of the
fact 2kc is our LCM of 2 to n). So:
9
Show that if a number is divisible by 3,
its digits add up to a multiple of 3.
The LHS is even (because it has a factor of 2). And all of the RHS terms are even, except for the
one where the denominator is 2k, because we’re dividing by a number that contains less than
k twos (by definition of k). However when the denominator is 2k, we end up with c, which by
definition is odd. Therefore the LHS is even but the RHS is odd. This is a contradiction, and
therefore x cannot be an integer.
Suppose the digits of n are nmnm-1...n1 (where there’s m digits). Then
in
general. If its digits add up to a multiple of 3, then
for some integer k. We have
two simultaneous equations! By subtracting the second from the first, we get:
JAF
So we just need to show the RHS is divisible by 3. Certainly 3k is divisible by 3. And
is divisible by 3 because for i=1, 2, 3, etc. we get 0, 9, 99, 999, ... (or we could more formally
show
is divisible by 3, since 10 = 1 (mod 3), and thus 10i-1 = 1i-1 = 1 (mod 3), and
thus
= 0 (mod 3)).
10
Determine the largest power of 10
that is a factor of the following:
(a)
50!
(b)
1000!
(c)
In general, n!
Teaching points: For younger students, get them to think about 2 digit numbers (say using a
and b for the two digits to avoid confusing subscript notation), then 3 digit numbers, and so
on. Hopefully they’ll identify the pattern that emerges in order to give an informal argument
for why the trick works for any number of digits.
A related problem is showing that if the digits add up to a multiple of 9, the number is divisible
by 9.
Once again, consider the prime factors of 50! and 1000!
We note that 10 = 2 x 5. So each time we find a pair of 2 and 5 in the prime factorisation of
some number, we obtain a factor of 10. In both 50! and 1000!, the factor of 2 is going to
appear and awful lot more than 5, so it’s sufficient to count the number of factors of 5,
knowing that we’ll definitely have enough 2s floating about to pair them up with.
a) 50! = 50 x 49 x 48 x ... x 1. How many factors of 5 appear in these 50 numbers? Well,
each multiple of 5 has one. But we also get additional factors of 5 from 25 (= 52) and
50 (= 52 x 2). So that’s 10 plus the additional 2, giving us 12.
b) We spotted from the first part that multiples of 5 give us one factor of 5 each, and
that multiples of 52 give us a bonus 5. Similarly multiples of 53 = 125 will give us a
further 5, as will multiples of 54 = 625. That gives us 200 + 40 + 8 + 1 = 249.
c) Using the pattern from above, we get:
, where
is a function which
rounds down its value (we’re looking for the whole number of times 5k goes into n).
Although I’ve summed to infinity, the floor function is going to give 0 once 5k exceed n
(e.g. in part (ii), none of the numbers between 1 and 1000 have a factor of 55, and so
gives us 0 when rounded down, and is not contributing to the summation). We
could use logs to limit the summation (by finding the maximum power of 5 that
divides k), so a better answer would be
Classic
Problem
SMC Questions
Level 1
1
2
3
Question
How many prime numbers are there less
than 20?
A
6
B
7
C
8
D
9
E
10
Solution
Answer C
The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19.
Ref
1999
Q1
What is the remainder when the number
743589 × 301647 is divided by 5?
A
0
B
1
C
2
D
3
E
4
Exactly one of the following numbers is
divisible by 11.
Which is it?
A
107 − 11
B
107 −1
C
107
D
107 +1
E
107 + 11
Answer D
The units digit of the product is 3; therefore the remainder when that number is divided by 5 is
also 3.
2000
Q1
Answer D
Since 107 is not divisible by 11, then neither is 107 − 11 nor 107 + 11. Therefore, the correct answer
is either B or D.
107 − 1 = 9 999 999 = 11 × 909090 + 9. Therefore
107 + 1 = 11 × 909090 + 11. Hence D is the correct answer.
2001
Q2
Solution
Answer C
Since the four numbers are symmetrical about the mean of the group, they must have the same
mean, 21, giving their total as 84. Alternatively, it is possible to state that the original numbers
must be 15, 17, 19, 21, 23, 25, 27 (since the numbers are given as consecutive and odd). Hence
the required sum is 15 + 19 + 23 + 27 = 84.
Ref
2001
Q6
Level 2
1
Question
The mean of seven consecutive odd
numbers is 21. What is the sum of the first,
third, fifth and seventh of these numbers?
A
16
B
21
C
84
D
147
E
more information needed
2
The value of the product wxyz is 2002 and
w, x, y and z are prime numbers.
What is the value of w2 + x2 + y2 + z2?
A
66
B
203
C
260
D
285
E
343
Answer E
2002 = 2 × 7 × 11 × 13; 22 + 72 + 112 + 132 = 343
2002
Q5
Level 3
1
Question
Which of these five expressions represents
the largest number?
9
(9 )
2
3
A 9
B 999
99
C 9
D (99)9
E 999
What is the value of
1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 +
2(1 + 2(1 + 2(1 + 2))))))))))?
A 210 +1 B 211 −1
C 211 +1 D 212 −1
E
212 + 1
I am trying to do a rectangular jigsaw
puzzle. The puzzle was made by starting
with a rectangular picture and then cutting
it into 1000 pieces by sawing along the
lines of a (wiggly!) rectangular grid. I start
by separating out all the edge and corner
pieces.
Which of the following could not possibly
be the number of corner and edge pieces
Solution
Answer: E
9
999 = 9 × 111 < 9 × 93 = 94 < 999 < (93)9 = 927 < 981 = (99)9 < 999 < 9
9
Answer: D
= 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211.
Now (xn − 1 + xn − 2 + … + x2 + x + 1) (x − 1) = (xn − 1)
 211 + 210 + 29 + … + 22 + 2 + 1) = (212 − 1)/ (2 − 1) = 212 – 1
Ref
1997
Q11
1997
Q12
(JAF: Know your formula for the sum of a geometric series!)
Answer: D
Suppose there are h pieces along each horizontal edge and v pieces along each vertical edge.
Then h.v = 1000 = 23.53.
Thus the only possibilities for the pair {h, v} are {1, 1000}, {2, 500}, {4, 250}, {5, 200},
{8, 125}, {10, 100}, {20, 50}, {25, 40}.
The total number of edge pieces is 2h + 2v − 4 = 2 (h + v − 2) (since 2h + 2v counts each
of the four corners twice).
 126 (= 2 (25 + 40 − 2)), 136 (= 2 (20 + 50 − 2)),
216 (= 2 (10 + 100 − 2)) are all possible; but 316 is not.
1997
Q14
4
5
of such a jigsaw?
A
126
B
136
C
216
D
316
E
A-D are all possible
Observe that 18 = 42 + 12 + 12 + 02.
How many of the first fifteen positive
integers can be written as the sum of the
squares of four integers?
A
11
B
12
C
13
D
14
E
15
The probability of a single ticket winning
the jackpot in the National Lottery is
6
5
4
3
2
1





.
49 48 47 46 45 44
6
If I buy one ticket every week,
approximately how often might I expect to
win the jackpot?
A once every hundred years
B once every twenty thousand years
C once every hundred thousand years
D once every quarter of a million years
E once every million years
The factorial of n, written n!, is defined by
n! = 1 × 2 × 3×…×(n − 2) × (n − 1) × n
e.g. 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720.
What is the smallest positive integer which
is not a factor of 50! ?
A
51
B
52
C
53
D
54
E
55
Answer E
The question really asks how many of the first 15 positive integers can be expressed as the sum
of 4 or fewer perfect squares. These can be listed as follows:−
1 = 12; 2 = 12 + 12; 3 = 12 + 12 + 12; 4 = 22. Expressions for 5, 6, 7 and 8 can be obtained by
adding 22 to each of the ones already given. 9 = 32 and expressions for 10, 11, 12, 13, 14, and
15 can be obtained by adding to the ones for 1, 2, 3, 4, 5 and 6.
Note: In fact, the French mathematician Lagrange proved in 1770 that every positive integer
can be expressed as the sum of four squares.
Answer D
49 48 47 46 45 44
      52
Number of years =
6 5 4 3 2 1
= 49 × 1 × 47 × 46 × 3 × 44 ÷ 52
 503  3  375000.
Note: 375000 is clearly an overestimate since 49 × 1 × 47 × 46 is less than 503 and 3 × 44 ÷ 52
is less than 3.
1998
Q6
Answer C
All of the positive integers from 1 to 50 inclusive must be factors of 50!
51 = 3 × 17 and 52 = 2 × 26 which means that these two numbers are also factors of 50!. 53 is
prime and is not a factor of 50!.
1999
Q9
1998
Q16
7
8
What is the value of (612 − 392) ÷ (512 −
492)?
A
10·5 B
11
C
12
D
21
E
22
Answer B
Given that a and b are integers greater
than zero, which of the following equations
could be true?
A
a–b=a b
B
a+b=a b
C
a–b=a  b
D a+b=a–b
Answer A
B cannot be true since a + b > a but a ÷ b ≤ a; C cannot be true since a − b < a but a × b ≥ a;
E
61
51


– 392 100  22

 11.
2
100  2
– 492
JAF note: Questions that require you to spot the difference of two squares are quite common.
2
a2 − b2 = (a + b) (a − b). Thus
similarly, D cannot be true. Furthermore, E cannot be true since
is true if a = 4 and b = 2.


2000
Q9
2000
Q11
2
a  b  a +b. However, A
a+ b = a + b
9
Observe that 2000 = 24 × 53. What is the
number of the next year after the year
2000
which can be written ab × cd where a, b, c,
d are 2, 3, 4, 5 in some order?
A
2016 B
2025
C
2040 D
2048
E
2050
Answer B
If the number of the year is even then it is either a power of 2, namely 25 × 43 or 23 × 45, both of
which are greater than 2025, or it is 2a × 3b or 2a × 5b , but no number between 2001 and 2024
is of this form. If the number is odd then it is either 34 × 52, which is 2025, or 32 × 54, which is
much greater. Therefore 2025 is the next number of the given form.
2000
Q12
10
Which of the following numbers n gives a
counter-example for the statement:
‘If n is a prime number then n2 + 2 is also a
prime number’?
A
3
B
5
C
6
D
9
E
none of them
Answer B
Since 6 and 9 are not prime, C and D cannot provide counter-examples. For 3, both parts are
true (since 11 is prime) but since 52 + 2 = 27 is not prime, B does provide a counter-example.
2001
Q9
11
n 20
n
When n = 81, what is the value of 3 ?
1
1
100
3
A less than
B
C
1
D
3
E
more than 100
Answer B
12
The number of the year 2003 is prime.
How many square numbers are factors of
20032003?
A
0
B
1
C
44
D
1002
E
2003
Answer D
The only square numbers which are factors of 20032003 have the form 20032n for a nonnegative integer n. But 2n ≤ 2003 so n = 0, 1, … , 1001.
2003
Q15
13
A square number is divided by 6.
Which of the following could not be the
remainder?
A
0
B
1
C
2
D
3
E
4
When divided by 6, a whole number leaves remainder 0, 1, 2, 3, 4 or 5. So the possible
remainders when a square number is divided by 6 are the remainders when 0, 1, 4, 9, 16 and 25
are divided by 6. These are 0, 1, 4, 3, 4 and 1 respectively, so a square number cannot leave
remainder 2 (or remainder 5) when divided by 6.
2005
Q14
20
Note that 81 = 34. Therefore
80
81
3
1
 81  .
81
3
3
3
2002
Q13
Level 4
1
2
3
Question
What is the maximum possible value of the
product of a list of positive integers, which
are not necessarily all different, given that
the sum of these numbers is 100?
A
1010
B
250
C
220 × 320
D
520
E
22 × 332
Sam correctly calculates the value of 58 ×
8 5.
How many digits does her answer contain?
A
11
B
12
C
13
D
14
E
15
Note that 2001 = 3 × 23 × 29.
What is the number of the next year which
can be written as the product of three
distinct primes?
A
2002 B
2004
C
2006 D
2007
E
none of these
Solution
Answer E
Any list whose product is a maximum does not contain any 1s since one 1 and one n could be
replaced by one n+ 1 and this would contribute n + 1 to the product rather than n. We can also
say that any list of maximum product would contain a maximum of two 2s, since any three 2s
could be replaced by two 3s which would contribute 9 to the product rather than 8.
Furthermore, it would contain a maximum of one 4 since two 4s could be replaced by one 2
and two 3s thereby contributing 18 to the product rather than 16. It would not contain any
number greater than 4 since 2 (n − 2) > n when n > 4 and therefore the product would be
increased by replacing one n with one 2 and one n − 2. Any list of numbers whose product is a
maximum, therefore, consists of two 2s and thirty-two 3s or one 4 and thirty-two 3s and the
maximum product is 22 × 332.
58 × 85 = 58 × 215 = (5 × 2)8 × 27 = 128 × 100 000 000 = 12 800 000 000 which has 11 digits.
Ref
2000
Q20
2001
Q15
JAF: Whenever you see a question involving numbers of 0s or number of digits, then always
remember the prime factors of 10 are 2 and 5.
Answer C
2002 = 2 × 1001 = 2 × 7 × 11 × 13 which is the product of four primes.
Since 2003 < 13 × 13 × 13, for 2003 to be a product of three distinct primes,
at least one of them must be less than 13. However, none of 2, 3, 5, 7 or 11 is
a factor of 2003. 2004 = 2 × 2 × 3 × 167 which has a repeated prime factor.
2005 = 5 × 401 which is the product of two primes.
2006 = 2 × 1003 = 2 × 17 × 59 and is therefore the next year which is the
product of three distinct primes.
2001
Q19
4
For how many integer values of n does the
equation x2 + nx − 16 = 0 have integer
solutions?
A
2
B
3
C
4
D
5
E
6
Answer D
For the equation to have integer solutions, it must be possible to write x2 + nx − 16 in the form
(x − α) (x − β), where α and β are integers. Therefore x2 + nx − 16 = x2 − (α + β)x + αβ and we
require that αβ = −16. The possible integer values of α, β are 1,−16; −1, 16; 2, −8; −2, 8; 4, –4
(we do not count −16, 1 as being distinct from 1, −16, for instance).
As n = −(α + β), the possible values of n are 15, −15, 6, −6 and 0.
2002
Q15
5
Which of the following is divisible by 3 for
every whole number x?
A
x3 − x B
x3 − 1
C
x3
D
x3 + 1
E
x3 + x
Answer A
Since x3 − x = x (x2 − 1) = (x − 1) × x × (x + 1), x3 − x is always the product of three consecutive
whole numbers when x is a whole number. As one of these must be a multiple of 3, x3 − x will
be divisible by 3. Substituting 2 for x in the expressions in B,C and E and substituting 3 for x in
the expression in results in D numbers which are not divisible by 3.
2003
Q13
JAF’s alternative answer: Use modulo arithmetic. If x = 0, 1, 2 (mod 3) oscillating between these
as x increases, then:
x3 = 0, 1, 8 = 0, 1, 2 (mod 3)
Then x3 – x = 0-0, 1-1, 2-2 = 0, 0, 0 (mod 3)
i.e. For numbers of x, x3 – x gives us a remainder of 0 when dividing by 3.
6
7
What is the value of 22003 − 22002 − 22001 −
22000?
A
−22002 B
0
−4000
C
2
D
64
2000
E
2
The value of 12004 + 32004 + 52004 + 72004 +
92004 is calculated using a powerful
computer.
What is the units digit of the correct
answer?
A
9
B
7
C
5
D
3
E
1
22003 − 22002 − 22001 − 22000
2003
Q18
= 22000 (23 − 22 − 2 − 1)
= 22000 (8 − 4 − 2 − 1) = 22000.
Answer A
The last digit of 34 is 1, as is the last digit of 74 and the last digit of 92. So the last digit of (34)501,
that is of 32004, is 1. Similarly, the last digit of (74)501, that is of 72004, is 1 and the last digit of
(92)1002, that is of 92004, is 1. Furthermore, 12004 = 1 and the last digit of 52004 is 5. So the units
digit of the expression is 1 + 1 + 5 + 1 + 1, that is 9.
2004
Q13
8
9
10
Which is the lowest positive integer by
which 396 must be multiplied to make a
perfect cube?
A
11
B
66
C
99
D
121
E
726
How many two-digit numbers N have the
property that the sum of N and the
number formed by reversing the digits of N
is a square?
A
2
B
5
C
6
D
7
E
8
Answer E
Expressed as the product of prime factors, 396 = 22 × 32 × 11. Therefore the lowest positive
integer by which it must be multiplied to make a perfect cube is 2 × 3 × 112, that is 726.
2004
Q20
Answer E
Let N be the two-digit number ‘ab’, that is N = 10a + b. So the sum of N and its ‘reverse’ is
10a + b + 10b + a = 11a + 11b = 11 (a + b). As 11 is prime and a and b are both single digits, 11
(a + b) is a square if, and only if, a + b = 11. So the possible values of N are 29, 38, 47, 56, 65, 74,
83, 92.
2007
Q12
Damien wishes to find out if 457 is a prime
number. In order to do this he needs to
check whether it is exactly divisible by
some prime numbers. What is the smallest
number of possible prime number divisors
that Damien needs to check before he can
be sure that 457 is a prime number?
A
8
B
9
C
10
D
11
E
12
The smallest number of possible prime divisors of 457 that Damien needs to check is the
number of prime numbers less than or equal to the square root of 457. Since 212 < 457 < 222,
he needs to check only primes less than 22. These primes are 2, 3, 5, 7, 11, 13, 17 and 19.
JAF note: Such efficiency tricks like this come into play when programming!
Level 5
1
2
3
Question
Three people each think of a number which
is the product of two different primes.
Which of the following could be the product
of the three numbers which are thought of?
A
120
B
144
C
240
D
3000
E
12100
Just one of the following is a prime number.
Which one is it?
A
10002 + 1112
B
5552 + 6662
C
20002 − 9992
D
10012 + 10022
E
10012 + 10032
The statement “There are exactly four
integer values of n for which (2n + y)/ (n − 2)
is itself an integer” is true for certain values
of y only.
For how many values of y in the range 1 ≤ y ≤
20 is the statement true?
A
0
B
7
C
8
D
10
E
20
Solution
Answer E
120 = 23 × 3 × 5; 144 = 24 × 32; 240 = 24 × 3 × 5;
3000 = 23 × 3 × 53 12100 = 22 × 52 × 112.
The product of the three numbers must have 6 prime factors, not necessarily all different,
but with no prime factor repeated more than 3 times. Of these, only 12100 satisfies this
condition. The three numbers are 10 (2 × 5), 22 (2 × 11) and 55 (5 × 11).
Teaching Points (JAF): An interesting question is finding the form of all possible solutions. If
we pick 6 primes, the choice of numbers is valid if no prime is repeated more than 3 times. If
we have only two distinct prime factors, each prime must occur three times (i.e. a3 x b3), and
therefore each person can pick an a and a b to end up with different primes.
Answer A
B cannot be prime since 111 is clearly a factor of it; C cannot be prime since it equals 2999 ×
1001 (difference of two squares); the units digit of D will be 5 and therefore 5 must be a
factor of it whilst the units digit of E is 0 and therefore it cannot be prime either since both 2
and 5 will be factors of it.
JAF note: This question is essentially combining all the various mini number-theory tricks
from other questions: i.e. difference of two squares and observing the units digit.
Answer B
2n  y 2n – 4 y  4
y4


 2
(n  2)
n–2
n–2 n–2
n–2
Thus (2n + y)/ (n − 2) is an integer if and only if (y + 4)/ (n − 2) is an integer. As n varies, the
integer values taken by (y + 4) / (n − 2) are all the integers which divide exactly into y + 4.
There are exactly 4 of these if and only if y + 4 is prime. They are {(y + 4), −(y + 4), 1, −1}.
Thus the integer values of the expression will be 2 + y + 4 = y + 6; 2− (y + 4) = −(y + 2); 3 and
1. For 1 ≤ y ≤ 20, the values of y for which y + 4 is prime are 1, 3, 7, 9, 13, 15 and 19.
Ref
1999
Q15
1999
Q21
1999
Q23
4
5
How many pairs of positive integers (x, y)
satisfy the equation x – 17  y ?
A
0
B
1
C
2
D
17
E
infinitely many
Answer E
For how many values of n are both n and
n –1
1 3
n –1 k
 where k = 0, 1, 2, 3, …
is 0, ,1, , ... i.e.
2 2
n 1 2
n 1
k n –1 n 1 – 2
2
4

1–
Let 
. Then k = 2 –
. For k to be an integer, the only
2 n 1
n 1
n 1
n 1
possible values of n + 1 are −4, −2, −1, 1, 2, 4; i.e. the possible values of n are −5, −3, −2, 0, 1,
3. The corresponding values of k are 3, 4, 6, −2, 0, 1. However, k cannot be negative and
therefore n = 0 is not valid. Therefore the five possible values of n are −5, −3, −2, 1, 3.
2001
Q22
4
A
C
E
 n 1 


 n 1 
integers?
1
B
3
D
5
2
4
Squaring both sides of the equation gives y = x − 2 17x  17 . The righthand side will be a
positive integer if x is of the form 17n2 for an integer n ≥ 2, and the equation then becomes
y = 17n2 − 34n + 17 = 17 (n − 1)2.
As there are infinitely many n ≥ 2, there are infinitely many pairs of positive integers (x, y)
which satisfy the equation e.g. (68, 17), (153, 68), (272, 153) etc.
 n –1 


4  n 1  is an integer if the value of
2000
Q24
6
Let N be a positive integer less than 102002.
When the digit 1 is placed after the last digit
of N, the number formed is three times the
number formed when the digit 1 is placed in
front of the first digit of N.
How many different values of N are there?
A
1
B
42
C
333
D
667
E
2002
Answer C
Let N have x digits, so that x ≤ 2002. When the digit 1 is placed at its end, N becomes 10N +
1. When 1 is placed in front of it, N becomes 10x + N. Therefore: 10N + 1 = 3 (10x + N) i.e.
7N = 3 × 10x − 1. So we need to find which of the numbers 2, 29, 299, 2999, 29999, …. are
divisible by 7. The first such number is 299999 (corresponding to x = 5), giving N = 42857 and
we check that 428571 = 3 × 142857.
The next such numbers correspond to x = 11, x = 17, x = 23, x = 29 and the largest number in
the given range corresponds to x = 1997.
The number of different values of N, therefore, is 1 + (1997 − 5) ÷ 6 = 333.
2002
Q25
7
How many pairs of positive integers (x, y) are
1 2
3
solutions of the equation + =
?
x y 19
A
0
B
1
C
2
D
3
E
more than 3
Answer D
1 2 3
  .
x y 19
2003
Q25
i.e. 38x + 19y = 3xy.
i.e. 9xy − 114x − 57y + 38 × 19 = 38 × 19.
i.e. (3x − 19) (3y − 38) = 2 × 192.
The factors of 2 × 192 are 1, 2, 19, 38, 361 and 722 and 3x − 19 has to be one of these.
If 3x − 19 = 1, 19 or 361, then x is not an integer. If 3x − 19 = 2, then x = 7 and 3y − 38 = 361
giving y = 133. If 3x − 19 = 38, then x = 19 and 3y − 38 = 19 giving y = 19 as well. If 3x − 19 =
722, then x = 247 and 3y − 38 = 1 giving y = 13.
1 2 3
(The equation   always has exactly 3 solutions when x and y are positive integers
x y p
and p is a prime greater than or equal to 5. Can you prove this result?)
8
The number N is exactly divisible by 7. It has
4008 digits. Reading from left to right, the
first 2003 digits are all 2s, the next digit is n
and the last 2004 digits are all 8s.
What is the value of n?
A
4
B
5
C
0 or 3 D
2 or 9
E
1 or 8
First note that 1, 11, 111, 1111, 11111 are not divisible by 7 but that 111111 = 15873 × 7. So
any number in which all the digits are the same is divisible by 7 if the number of digits is a
multiple of 6. Therefore the 2004-digit number 888…888 is a multiple of 7. Similarly, the
2004-digit number 222…222 is a multiple of 7, as is the 2004-digit number 222…229 since it
differs from 222…222 by 7. Furthermore, the 2004-digit number 222…22n is not a multiple
of 7 if n is any digit other than 2 or 9. Now N = 222…22n × 102004 + 888…888, so if N is
divisible by 7 then 222…22n is divisible by 7 and we deduce that n = 2 or 9.
9
Positive integers x and y satisfy the equation: Answer C
1
1
x + 1 y – x – 1 y = 1.
x
y – x–
y  1.
2
2
2
2
Which of the following is a possible value of
1
1
1
y?
y – 2 x2 – y  x –
y 1,
Therefore x +
2
4
2
A
5
B
6
C
7
D
8
1
2
that is 2x − 1 = 2 x – y .
E
9
4
1
y ),that is y = 4x − 1.
4
So y must be 1 less than a multiple of 4.
Of the values offered, the only possibility is 7. This gives x = 2, but it is necessary to check
Therefore 4x2 − 4x + 1 = 4 (x2 −
2004
Q23
2004
Q25
that these values are solutions of the original equation since they were derived from an
argument which involved squaring equations. It is not difficult to show that x = 2, y = 7,
satisfy the second equation in the above solution. We now need to confirm that
1
1
2
7 – 2–
7 does indeed equal 1 rather than –1, which we are able to do since
2
2
it is clear that
2
1
7 is greater than
2
2–
1
7 .
2
10
What is the sum of the values of n for which
n2 – 9
both n and
are integers?
n –1
A
−8
B
−4
C
0
D
4
E
8
Answer E
Note that n2 − 1 is divisible by n − 1. Thus:
n2 – 9 n2 – 1
8
8

–
 n 1–
(n  1).
n –1
n –1 n –1
n –1
n2 – 9
So, if n is an integer, then
is an integer if and only if n − 1 divides exactly into 8. The
n –1
possible values of n − 1 are −8, −4, −2, −1, 1, 2, 4, 8, so n is −7, −3, −1, 0, 2, 3, 5, 9. The sum of
these values is 8.
(Note that the sum of the 8 values of n – 1 is clearly 0, so the sum of the 8 values of n is 8.)
2005
Q21
11
Given that S = (x + 20) + (x + 21) + (x + 22)
+…+ (x + 100), where x is a positive integer,
what is the smallest value of x such that S is
a perfect square?
A
1
B
2
C
4
D
8
E
64
Answer C
2005
Q22
1
There are 81 terms in the series, so, using the formula S  n(a  l ) for an arithmetic
2
series:
81
S=
(x + 20 + x + 100) = 81 (x + 60) .
2
Now 81 is a perfect square, so S is a perfect square if and only if x + 60 is a perfect square. As
x is a positive integer, the smallest possible value of x is 4.
12
The factorial of n, written n!, is defined by n!
= 1 × 2 × 3 ×…× (n − 2) × (n − 1) × n.
For how many positive integer values of k
less than 50 is it impossible to find a value of
n such that n! ends in exactly k zeros?
A
0
B
5
C
8
D
9
E
10
Answer D
When n! is written in full, the number of zeros at the end of the number is equal to the
power of 5 when n! is written as the product of prime factors, because there is at least that
high a power of 2 available. For example, 12! = 1 × 2 × 3 × … × 12 = 210 × 35 × 52 × 7 × 11.
This may be written as 28 × 35 × 7 × 11 × 102, so 12! ends in 2 zeros, as 28 × 35 × 7 × 11 is not
a multiple of 10.
We see that 24! ends in 4 zeros as 5, 10, 15 and 20 all contribute one 5 when 24! is written
as the product of prime factors, but 25! ends in 6 zeros because 25 = 5 × 5 and hence
contributes two 5s. So there is no value of n for which n! ends in 5 zeros. Similarly, there is
no value of n for which n! ends in 11 zeros since 49! ends in 10 zeros and 50! ends in 12
zeros. The full set of values of k less than 50 for which it is impossible to find a value of n
such that n! ends in k zeros is 5, 11, 17, 23, 29, 30 (since 124! ends in 28 zeros and 125! ends
in 31 zeros), 36, 42, 48.
2005
Q24
British Maths Olympiad
Legal note: While model solutions can be purchased online, I have not accessed these, and all solutions (at least what I’ve got round to so far!)
are my own, unless their source is otherwise stated.
1
2
3
4
5
6
Question
One number is removed from the set of
integers from 1 to n. The average of the
remaining numbers is 40.75. Which integer
was removed?
Solution
Yet to attempt.
Ref
2010
Determine the sequences a0, a1, a2, … which
satisfy all of the following conditions:
11.
for every integer
,
12.
is a rational number and
13.
for some I, j with
Find all positive integers n such that both
n+2008 divides n2 + 2008 and n+2009 divides
n2 + 2009.
Find four prime numbers less than 100
which are factors of 332 – 232.
Yet to attempt.
2008
Yet to attempt.
2008
JAF’s solution: (332 – 232) = (316 + 216)(316 – 216)
And continuing to factorise the difference of two squares we get:
(316 + 216) (38 + 28) (34 + 24) (32 + 22)(3 + 2)(3 – 2)
That gives us our first prime factors of 5 and 13 immediately. 34 + 24 = 81+16 = 97 which is
also prime. 38 + 28 is 812 + 162 = 6561 + 256 = 6817 = 17 x 401. So 17 is our fourth prime
factor, and we’re done.
Yet to attempt.
2006
JAF’s solution: The equations are: p + 3 = nj and j + n = 3p. Rearranging we get:
2004
Let n be an integer. Show that, if
is an integer, then it is a perfect
square.
Each of Paul and Jenny has a whole number
of pounds.
He says to her: “If you give me £3, I will have
n times as much as you.”
And we’re interested in integers n when j is an integer. Note that 3j-1 grows more rapidly
2006
7
8
9
10
11
She says to him: “If you give me £n, I will
have 3 times as much as you”.
Given that all these statements are true and
that n is a positive integer, what are the
possible values for n?
Determine the least possible value of the
largest term in an arithmetic progression of
seven distinct primes.
Let s be an integer greater than 6. A solid
cube of side s has a square hole of side x<6
drilled directly through from one face to the
opposite face (so the drill removes a cuboid).
The volume of the remaining solid is
numerically equal to the total surface area of
the remaining solid. Determine all possible
integer values of x.
Determine the least natural number n for
which the following result holds:
No matter how the elements of the set {1, 2,
… , n} are coloured red or blue, there are
integers, x, y, z, w in the set (not necessarily
distinct) of the same colour such that x + y +
z = w.
Find, showing your method, a six-digit
integer n with the following properties: (i) n
is a perfect square, (ii) the number formed
by the last three digits of n is exactly one
greater than the number formed by the first
three digits of n. (Thus n might look like
123124, although this is not a square.)
than j+9, so we’ll find eventually as j increases that 3j – 1 > j + 9 and we can no longer have
integer solutions (since the denominator is bigger than the numerator). Integer solutions
can only exist when 3j – 1 ≤ j + 9, which gives j ≤ 5. This is quite handy as there’s now only 5
values of j we need to try! It works when j=1, giving us n=5 and p=2. It also works when j=5,
giving us n=1 and p=2. And we’re done.
Yet to attempt.
2004
Yet to attempt.
2010
Yet to attempt.
2004
JAF’s solution: Suppose the first 3 digits are a, b and c. Then the number is 100,000a +
10,000b + 1,000c + 100a + 10b + c + 1 = 1001(100a + b + c) + 1, which needs to be a square
number.
Thus for some integer n, 1001(100a + 10b + c) = n2 – 1 = (n+1)(n-1).
To finish writeup, but I found the numbers to be 183184, 328329, 528529 and 715716. The
number squared in each case is 428, 573, 727 and 846.
1993
Let m = (4p − 1)/3, where p is a prime
number exceeding 3. Prove that 2m−1 has
Yet to attempt.
1993
Round
12
13
remainder 1 when divided by m.
Starting with any three digit number n (such
as n = 625) we obtain a new number f(n)
which is equal to the sum of the three digits
of n, their three products in pairs, and the
product of all three digits. (i) Find the value
of n/f(n) when n = 625. (The answer is
an integer!) (ii)Find all three digit numbers
such that the ratio n/f(n)=1.
How many perfect squares are there (mod
2n)?
Yet to attempt.
Solution from http://schoolexercisebooks.blogspot.co.uk/2010/09/30th-britishmathematical-olympiad-1994.html
Answer: n odd (2n-1 + 5)/3, n even (2n-1 + 4)/3.
We find the first few:
2: 0, 1
4: 0, 1
8: 0, 1, 4
16: 0, 1, 4, 9
32: 0, 1, 4, 9, 16, 17, 25
64: 0, 1, 4, 9, 16, 17, 25, 33, 36, 41, 49, 57
a2 = b2 mod 2n iff (2a)2 = (2b)2 mod 2n+2 (JAF note: see Key Theory), so the number of even
squares mod 2n+2 equals the number of squares mod 2n. This suggests looking separately at
the odd and even squares. Let un be the number of odd squares mod 2n and vn be the
number of even squares mod 2n.
(2a + 1)2 = (2b + 1)2 mod 2n is equivalent to 4a2 + 4a = 4b2 + 4b mod 2n, which is equivalent to
a2 + a = b2 + b mod 2n-2 or (a - b)(a + b + 1) = 0 mod 2n-2. But a - b and a + b + 1 have sum 2a +
1 which is odd, so they have opposite parity, so either a - b = 0 mod 2n-2 or (a + b + 1) = 0
mod 2n-2. Thus just 3 other odd numbers have the same square as (2a + 1), namely (2a + 2n-1
+ 1), -(2a + 1) and -(2a + 2n-1 + 1). So there are 2n-3 different odd squares mod 2n. In other
words, vn = 2n-3.
Now un = un-2 + vn-2 = un-4 + vn-4 + vn-2 etc. So the total number of squares u2m + v2m = v2m + v2m2m-3
+ 22m-5 + ... + 2 + 2 = 2(4m-2 + 4m-3 + ... + 1) + 2 = 2(4m-1 - 1)/3 + 2 =
2 + v2m-4 + ... + v4 + u4 = 2
n-1
(2 + 4)/3.
2
1994
1994
Round
2
14
15
Find the first integer n > 1 such that the
average of 12, 22, 32, . . . , n2
is itself a perfect square.
Find all solutions in positive integers a, b, c
to the equation
a! b! = a! + b! + c!
Similarly, the total number of squares for n odd is (2n-1 + 5)/4.
Yet to attempt.
Solution here: http://sciencecentral1234.blogspot.co.uk/2012/02/two-solutions-toproblem-5-from-bmo-1.html (the unique solution is a=3, b=3, c=4).
1994
Round
2
2002/
3