Download Cat 3 - Num. Theory

Category 3
Number Theory
Meet #1, October, 2000
1. For how many positive integral values of n will
168
be a whole number?
n
2. What is the greatest integer that will always divide the product of four
consecutive integers?
3. Find the sum of all the positive integers less than 1000 that are both
perfect squares and perfect cubes.
Answers
1. _____________
2. _____________
3. _____________
Solutions to Category 3
Number Theory
Meet #1, October, 2000
Answers
1. 16
2. 24
3. 794
1. The positive integral values of n that will make
168
a whole number are the factors of 168. (By
n
“positive integral values” we mean positive integers
or whole numbers.) We can simply list all the
factors of 168 in pairs: 1 × 168 , 2 × 84 , 3 × 56 , 4 × 42 ,
6 × 28 , 7 × 24 , 8 × 21 , and 1 × 168 . Those are the 16
values of n that will make
168
a whole number.
n
Alternatively, there is a trick for figuring out the
number of factors of a number without actually
listing them. We express 168 in prime factors with
exponents as follows: 168 = 23 ⋅ 31 ⋅ 71 . We then raise
each exponent by one and multiply them like this:
( 3 + 1) × (1 + 1) × (1 + 1) = 4 × 2 × 2 = 16 .
2. The product of four consecutive integers is
guaranteed to contain a multiple of 3, and two
multiples of 2, one of which is also a multiple of 4.
Being careful not to count the same factor of 2 as
both a multiple of 2 and a multiple of 4, we can be
certain of only three factors of 2 along with the one
factor of 3. From these factors we obtain the
product 23 × 3 = 24 , so 24 will always divide the
product of four consecutive integers. Some people
will find this answer simply by multiplying the first
four counting numbers: 1 × 2 × 3 × 4 = 24 .
3. Numbers that are both perfect squares and perfect
3
2
cubes are sixth powers, because a 6 = (a 2 ) = (a 3 ) .
Since the cube root of 1000 is 10, we need only
values under 10 for a 2 . Now if a 2 is under 10, then
a must be 3 or less. The three sixth powers under
1000 are: 16 = 1 , 2 6 = 64 , and 36 = 729 . Their sum is
1 + 64 + 729 = 794 .
Category 3
Number Theory
Meet #1, October, 2001
1. What is the greatest possible difference of two primes if both of those
primes are between 50 and 100?
2. Use the following clues to find the value of n:
• n is a whole number
• n is divisible by 24
• n is greater than 300
• n is not divisible by 17
• n is less than 400
• n is divisible by 21
• the sum of the digits of n is 12
3. Find the sum of all the different prime factors of 364,000.
Answers
1. _____________
2. _____________
3. _____________
Solutions to Category 3
Number Theory
Meet #1, October, 2001
Answers
1. 44
2. 336
3. 27
1. To get the greatest possible difference, we will
want to find the prime closest to 100 and subtract
the prime closest to 50. 99 = 9 × 11 , so it’s not
prime, and 51 = 3 × 17 , so it’s not prime either.
Both 97 and 53 are prime, though, so our greatest
difference is 97 − 53 , which is 44.
2. That n is divisible by 24 and 21 are the most
helpful clues to start with. Since 24 = 23 × 3 and
21 = 3 × 7 , we know that n is divisible by 23 , 3, and
7, which are relatively prime. This means that n is a
multiple of 23 × 3 × 7 = 168 . The only multiple of
168 that is between 300 and 400 is 2 × 168 = 336 , so
that must be it. Checking the other clues, we know
that 336 cannot be divisible by 17, since we created
it by multiplying 2 × 23 × 3 × 7 and 17 is a
(
)
different prime. The final clue says that the sum of
the digits is 12 and this is true of our answer 336.
3. 364,000 = 364 × 1000 and we know that
1000 = 23 × 53 . Now we need to look for other
prime factors in 364:
364 = 2 × 182 = 22 × 91 = 22 × 7 × 13 . So the
different prime factors of 364,000 are 2, 5, 7, and
13. The sum of these numbers is 27.
Category 3
Number Theory
Meet #1, October, 2002
1. If the five-digit number 837A5 is known to be divisible by 15, what is the sum
of all the possible values of the digit A?
2. Twin primes are two primes that have a positive difference of 2, such as 11 and
13. What is the sum of the least twin primes greater than 50?
3. Find the value of n, if
n is a natural number,
n is less than 100,
n is the product of two primes,
the sum of the digits of n is 10, and
the positive difference between the two prime factors of n is a multiple of 13.
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 3
Number Theory
Meet #1, October, 2002
Answers
1. 12
2. 144 or 120
3. 82
1. To be sure a number is divisible by 15, it must be
divisible by both 3 and 5. The five-digit number 837A5
is definitely divisible by 5, since its units digit is a 5. If
it is to be divisible by 3, the sum of its digits must be a
multiple of 3. The known digits have a sum of 8 + 3 + 7
+ 5 = 23. This means that A can have the values 1, 4, or
7. The sum of these possible values is 1 + 4 + 7 = 12.
2. There are no twin primes in the fifties or the sixties.
Thus the least twin primes greater than 50 are 71 and 73.
Their sum is 144.
Editor note: The original meet had 144 as the answer,
but 59 and 61 are also twin primes, so the correct
answer should be 120. Thanks to Zhiping You for
pointing this out.
3. There are too many possible products of two primes
to start with that clue. The natural numbers less than 100
with a sum of digits equal to 10 are 19, 28, 37, 46, 55,
64, 73, 82, and 91. Of these, four of these are the
product of two primes: 46 = 2 • 23, 55 = 5 • 11, 82 = 2 •
41, and 91 = 7 • 13. Since the difference between the
prime factors of 82 is 41 – 2 = 39 and 39 is a multiple of
13, n must be 82.
Category 3
Number Theory
Meet #1, October 2003
1. 9900 = 2 a • 3 b • 5 c • 7 d • 11e
Find the value of a + b + c + d + e.
2. Find the sum of all possible values of the digit N such that the 5-digit number
318N4 is divisible by 12.
3. Melanie’s locker number is the product of the least pair of consecutive primes
that have a difference of 6. What is Melanie’s locker number?
Answers
1. _______________
2. _______________
3. _______________
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Solutions to Category 3
Number Theory
Meet #1, October 2003
Answers
1. 9900 = 2 2 • 3 2 • 5 2 • 7 0 • 111,
so a + b + c + d + e = 2 + 2 + 2 + 0 + 1 = 7.
1. 7
2. 10
3. 667
2. A number that is divisible by 12 must be divisible by both 3
and 4. For the 5-digit number 318N4 to be divisble by 3, the
sum of its digits must be a multiple of 3. 3 + 1 + 8 + 4 = 16 , so
N would have to be 2 or 5 or 8 to make a multiple of 3. This
gives us 31824, 31854, and 31884 to consider. To determine
which among these is divisible by 4, we need only check the
last two digits of each number, since any multiple of 100 is
divisble by 4. 24 and 84 are multiples of 4, but 54 is not. Thus
only 31824 and 31884 are divisible by 12 and the sum of all
possible values of N is 2 + 8 = 10.
3. The least pair of consecutive primes with a difference of 6 is
the pair 23 and 29. Their product is 667, so Melanie’s locker
number must be 667
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Category 3
Number Theory
Meet #1, October 2004
1. How many pairs of primes have a sum of 24?
2. If the five-digit number 5N82N is divisible by 18, what is the value of N?
3. How many of the positive factors of 660 are odd?
Answers
1. _______________
2. _______________
3. _______________
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Solutions to Category 3
Number Theory
Meet #1, October 2004
Answers
1. 3
2. 6
3. 8
1. There are three (3) pairs of primes that have a sum of 24.
They are: 19 + 5, 17 + 7, and 13 + 11.
2. For the five-digit number 5N82N to be divisible by 18, it
must pass the divisibility tests for both 2 and 9. Since there is
an N in the units place (ones place), we know that N has to be
even. To be divisible by 9, the sum of the digits must be a
multiple of 9. Currently the sum of the digits is 5 + N + 8 + 2 +
N = 15 + 2N. For N = 0, we get 15 + 2 × 0 = 15, which is not a
multiple of 9. For N = 2, we get 15 + 2 × 2 = 19. For N = 4, we
get 15 + 2 × 4 = 23. For N = 6, we get 15 + 2 × 6 = 27, which is
a multiple of 9. This tells us that the five-digit number 56826
is divisible by 18. (56826 ÷ 18 = 3157) Note that this is the
only single-digit value of N that will make
15 + 2N equal a multiple of 9. Thus, N = 6.
3. The factor pairs of 660 are listed below. The eight (8) odd
factors are in bold.
1 × 660
2 × 330
3 × 220
4 × 165
5 × 132
6 × 110
10 × 66
11 × 60
12 × 55
15 × 44
20 × 33
22 × 30
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Category 3
Number Theory
Meet #1, October 2005
1. Two prime numbers, each less than 50, have a difference of 35. What is the
sum of these two primes?
2. The letters A and B each represent one digit in the five-digit number AB9BA. If
AB9BA is divisible by 2, 3, 4, 6, 8, and 9, what is the value of 2A + B?
3. How many whole numbers between 80 and 100 are the product of two distinct
primes? In other words, how many whole numbers between 80 and 100 have
exactly four factors?
Answers
1. _______________
2. _______________
3. _______________
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Solutions to Category 3
Number Theory
Meet #1, October 2005
Answers
1. 39
2. 15
1. After 2, all primes are odd. Any two odd primes will have
an even difference. If the difference of two primes is odd, then
2 must be one of the primes. Since our difference is 35, the
other prime must be 35 + 2 = 37. Our two primes are 2 and 37
and their sum is 2 + 37 = 39.
3. 8
2. The sum A + B + 9 + B + A must be a multiple of 9, since
the five-digit number AB9BA is divisible by 9. This means that
A + B = 9. Also, A must be even, since AB9BA is a divisible by
2. The possibilities for the last two digits BA are 18, 36, 54, 72,
and 90. Only 36 and 72 are divisible by 4, so the 5-digit
number is either 63936 or 27972. Since 936 is divisible by 8
and 972 is not, the number must be 63936. So A = 6 and B = 3,
and the value of 2A + B is 2 × 6 + 3 = 15.
3. The prime factorizations of all the numbers between 80 and
100 are listed below. The 8 that are products of two primes are
in bold.
91 = 7 ⋅ 13
81 = 3 4
82 = 2 ⋅ 41
83 = 83
84 = 2 2 ⋅ 3 ⋅ 7
85 = 5 ⋅ 17
86 = 2 ⋅ 43
87 = 3 ⋅ 29
88 = 2 3 ⋅11
89 = 89
90 = 2 ⋅ 3 2 ⋅ 5
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92 = 2 2 ⋅ 23
93 = 3 ⋅ 31
94 = 2 ⋅ 47
95 = 5 ⋅ 19
96 = 2 5 ⋅ 3
97 = 97
98 = 2 ⋅ 7 2
99 = 3 2 ⋅11
Category 3
Number Theory
Meet #1, October 2006
1. What is the least three-digit number that is divisible by exactly three different
prime numbers?
2. The five-digit number 35N2N is divisible by 2, 3, 4, 6, 8, 9, and 12, among
other numbers. What is the value of the digit N?
3. How much greater is the product of the first four primes than the product of the
first three composites?
Answers
1. _______________
2. _______________
3. _______________
www.imlem.org
Solutions to Category 3
Number Theory
Meet #1, October 2006
Answers
1. 102
2. 4
3. 18
Some Incorrect
Answers Seen
1. 105, 110,
120, 1001
Average score: 1.5 answers correct
1. Some people might be fooled by 3 × 5 × 7 = 105, but 2 × 3 ×
17 = 102 is the least three-digit number that is divisible by
exactly three primes.
2. N clearly must be an even digit, so 0, 2, 4, 6, and 8 are
possible. For 35N2N to be divisible by 3, the sum of the digits
must be a multiple of three. So far we have 3 + 5 + N + 2 + N =
10 + 2N. Let’s use the possible values of N to evaluate this
expression. If N = 0, then 10 + 2N = 10 + 0 = 10, which is not
divisible by 3. If N = 2, then 10 + 2N = 10 + 4 = 14, which is
not divisible by 3. If N = 4, then 10 + 2N = 10 + 8 = 18, which
is divisible by 3. Neither N = 6 nor N = 8 gives a multiple of 3,
so 4 is our only candidate. Let’s make sure that the number
35424 is divisible by all the other numbers listed. It is divisible
by 4 since the last two digits 24 are divisible by 4. It is
divisible by 6 since it passes the test for 2 and 3. It is divisible
by 8 since the last three digits 424 are divisible by 8. It is
divisible by 9 since the sum of the digits (18) is divisible by 9.
Finally, it is divisible by 12 since it passes the test for 3 and 4.
The digit N is indeed 4.
3. The product of the first four primes is 2 × 3 × 5 × 7 = 210.
The product of the first three composites is 4 × 6 × 8 = 192.
The difference is 210 – 192 = 18.
2. 9
3. 186, 6
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