Download M1C3-PACKET

Carlisle Math Team
Meet #1 – Category 3
M1C3
Number Theory
Self-study Packet
1. Mystery: Problem solving
2. Geometry: Angle measures in plane figures including supplements and
complements
3. Number Theory: Divisibility rules, factors, primes, composites
4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics
5. Algebra: Simplifying and evaluating expressions, solving equations with 1
unknown including Identities.
For current schedule or information,
see http://www.imlem.org
A Prime Number can be divided evenly only by 1, or itself. A
Composite Number can be divided evenly by numbers other than 1 or
itself.
Number
1
2
3
4
5
6
7
8
9
10
Can be Evenly
Prime, or
Divided By
Composite?
(1 is not considered prime or composite)
1,2
Prime
1,3
Prime
1,2,4
Composite
1,5
Prime
1,2,3,6
Composite
1,7
Prime
1,2,4,8
Composite
1,3,9
Composite
1,2,5,10
Composite
Factors
"Factors" are the numbers you multiply to get another number:
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
“How many factors” problems
How many factors are there of 24? Answer: You can list them, making pairs
whenever you find a factor: 1x24, 2x12, 3x8, 4x6. So, there are 8 factors. In
Meet#2, when we do prime factorization, we learn a short-cut for counting the
factors of a number.
PRIME NUMBER CHART TO 1000
2
31
73
127
181
3
37
79
131
191
5
41
83
137
193
7
43
89
139
197
11
47
97
149
199
13
53
101
151
211
17
59
103
163
223
19
61
107
167
227
23
67
109
173
229
29
71
113
179
233
“Distinct (different) prime factors” problems
What are the distinct prime factors of 100? Answer: 2 and 5. That’s all! Distinct
means we only list each prime factor once, even though there are many of them
in the number. 100 is 2x5 x 2x5.
List all the distinct prime factors of 100,000.
2 and 5.
What are the distinct prime factors of 300,000,000? 2, 3, and 5.
“Finding an unknown digit” problems
The number 8R3 is a multiple of 9. What could R be? Answer: Any multiple of 9
has the sum of digits as 9, so 8+R+3 = 11+R = multiple of 9. What values of R
(0 through 9) make this true? 11+7=18, so R=7. 873 is indeed a multiple of 9.
Greatest / Smallest Primes in a Region
What is the largest prime less than 30?
Answer: 29
What is the smallest prime greater than 100? Answer: 101
(We need to quickly test numbers to see if they’re prime or composite.)
The Divisibility Rules
These rules let you test if one number can be evenly divided by another,
without having to do too much calculation!
A number
is
divisible
by:
2
If:
The last digit is even (0,2,4,6,8)
Example:
128 is
129 is not
381 (3+8+1=12, and 12÷3
= 4) Yes
3
The sum of the digits is divisible by 3
4
The last 2 digits are divisible by 4
5
The last digit is 0 or 5
6
The number is divisible by both 2 and
3
7
217 (2+1+7=10, and 10÷3
= 3 1/3) No
1312 is (12÷4=3)
7019 is
not
175 is
809 is not
114 (it is even, and
1+1+4=6 and 6÷3 = 2)
Yes
308 (it is even, but
3+0+8=11 and 11÷3 =
3 2/3) No
672 (Double 2 is 4, 67If you double the last digit and
subtract it from the rest of the number 4=63, and 63÷7=9) Yes
and the answer is divisible by 7 or 0.
905 (Double 5 is 10, 9010=80, and 80÷7=11 3/7)
(Note: you can apply this rule to that
No
answer again if you want)
109816 (816÷8=102) Yes
8
9
10
11
12
The last three digits are divisible by 8
216302 (302÷8=37 3/4)
No
The sum of the digits are divisible by 9
1629 (1+6+2+9=18, and
again, 1+8=9) Yes
(Note: you can apply this rule to that
answer again if you want)
2013 (2+0+1+3=6) No
The number ends in 0
If you sum every second digit and
then subtract the other digits and the
answer is divisible by 11 or 0
The number is divisible by both 3 and
4
220 is
221 is not
7392 ((7+9) - (3+2) = 11)
Yes
25176 ((5+7) - (2+1+6) =
3) No
648 (6+4+8=18 and
18÷3=6, also 48÷4=12)
Yes
916 (9+1+6=16, 16÷3=
5 1/3) No
“Divisibility” problems:
Is 123 is divisible by 3? Yes, because 1+2+3=6 and 6 is a multiple of 3.
Is 765,000 divisible by 3? Yes, because 7+6+5+0+0+0 = 18 and 18 is divisible
by 3.
Is 212 is divisible by 4? Yes, because 12 is.
Is 77728 divisible by 4? Yes, because 28 is (4*7).
What are the prime factors of 30? Answer: 2, 3, and 5. NOT 6 or 15, as they
aren’t prime.
Is 91 composite? Answer: Yes! 91 = 7x13. This, and 1001, are often tricky.
Category 3
Number Theory
Meet #1, October 2004
1. How many pairs of primes have a sum of 24?
Hint: List the primes up to 24: 2 3 5 7 11 __ __ __ __
Now look for two that add to 24. Keep looking.
2. If the five-digit number 5N82N is divisible by 18, what is the value of N?
Hint: Any multiple of 18 is even. So, what could the low digit be? If it’s a multiple of 9, try the sum of the digits.
3. How many of the positive factors of 660 are odd?
Hint: List all 12 factor pairs, such as 1x660, 2x330, 3x220, 4x____ , … and count the odd factors.
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 3
Number Theory
Meet #1, October 2004
Answers
1. There are three (3) pairs of primes that have a sum of 24. They
are: 19 + 5, 17 + 7, and 13 + 11.
1. 3
2. 6
3. 8
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
Category 3
Number Theory
Meet #1, October 2003
1. 9900  2 a  3b 5 c  7 d 11e
Find the value of a  b  c  d  e .


Hint: Prime factor 9900: Divide by 2 until you can’t. The number of 2’s is a. Now divide what’s left by 3 until you can’t.
The number of 3’s is b, etc.
2. Find the sum of all possible values of the digit N such that the 5-digit number 318N4
is divisible by 12.
Hint: 12 is divisible by 3 and 4, so 318N4 is too. 3 tells you about sum of digits. Is 14 divisible by 4? 24? 34? 44?
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?
Hint: Make a list of primes until you find two differing by 6. 2 3 5 7 11 __ __ __ __ __ __
Consecutive means ‘next to.’
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 3
Number Theory
Meet #1, October 2003
Answers
1. 9900  2 2  32 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

 to be divisble by 3, the sum of its
4. For the 5-digit number 318N4
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 divisible 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
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?
Hint: Any multiple of 15 is also a multiple of 3 and 5. Adding digits should give multiple of 3.
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?
Hint: Start listing primes > 50: 53 59 __ __ __ until you find Twins
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.
Hint: Natural number = 1 2 3 … Sum of digits tells you 19,91 or 28,82, … Now check for product of 2 primes.
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, 2001
1. What is the greatest possible difference of two primes if both of those primes are
between 50 and 100?
Hint: What is the lowest prime > 50, and the greatest prime < 100? Is 51 divisible by 3?
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
Hint: If it’s divisible by 24 and by 21=3*7, then it’s also divisible by 24*7=___.
3. Find the sum of all the different prime factors of 364,000.
Answers
1. _____________
Hint: Do the factors of 364 and 1000 separately. 10 = 2*5, and
2. _____________
3. _____________
different prime factors of 1000 are still only 2 and 5. What are the
prime factors of 364? After you divide by 2 a couple of times, is
what’s left divisible by 7?
Solutions to Category 3
Number Theory
Meet #1, October, 2001
Answers
1. 44
2. 336
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.
3. 27
2. That n is divisible by 24 and 21 are the most helpful clues
3
to start with. Since 24  2  3 and 21  3  7 , we know
3
that n is divisible by 2 , 3, and 7, which are relatively prime.
3
This means that n is a multiple of 2  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  2  3  7 and 17 is a different
3
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
2
2
factors in 364: 364  2  182  2  91  2  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, 2000
1. For how many positive integral values of n will be a whole number?
Hint: 168/n is a whole number if n is a factor of 168. “positive integral values” means 1, 2, 3, …
Try listing all the factors of 168.
2. What is the greatest integer that will always divide the product of four consecutive
integers?
Hint: Try 1*2*3*4 and make a guess, then check with 2*3*4*5 etc. Or, reason using divisibility rules.
3. Find the sum of all the positive integers less than 1000 that are both perfect squares
and perfect cubes.
Hint: Since 103=1000, there are only 9 perfect cubes<1000 so list and see which are also squares: 1 3=1, 23=8, 33=27,
43=64, 53=___, 63=___, 73=___, 83=___, 93=___. Don’t forget 1!
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 a whole
number are the factors of 168. (By “positive integral values”
we mean positive integers or whole numbers.) We can
simply list all the factors of 168 in pairs: , , , , , , , and .
Those are the 16 values of n that will make a whole number.
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: . We
then raise each exponent by one and multiply them like this: .
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, 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: .
3. Numbers that are both perfect squares and perfect cubes
are sixth powers, because. Since the cube root of 1000 is 10,
we need only values under 10 for . Now if is under 10, then
a must be 3 or less. The three sixth powers under 1000 are: ,
, and . Their sum is .