Download (1) The integer 6 has four positive divisors: 1, 2, 3 and 6

(1)
The integer 6 has four positive divisors: 1, 2, 3 and 6. What is the smallest
positive integer with exactly five positive divisors?
(2)
How many odd perfect square factors does 24 × 36 × 510 × 79 have?
(3)
How many positive perfect cube factors does 24 × 36 × 510 × 79 have?
(4)
What is the units digit of 1! + 3! + 5! + 7! + 9! + 11! ?
(5)
What is the number of positive factors of 648?
(6)
What is the largest integer whose cube is less than 10,000?
(7)
What is the sum of all of the multiples of 6 from -100 through 200?
(8)
What is the largest perfect square factor of 1512?
(9)
What is the sum of the two least natural numbers that each have exactly six
distinct, positive factors?
(10)
A relatively prime day is one where the month number (e.g. January = 1,
February = 2, etc.) and date of the month (1, 2, . . . , 31) have no common factor other
than 1. Which month has the fewest relatively prime days?
(11)
How many digits are in the value of the following expression:
22001 × 51950 ÷ 427 ?
(12)
What is the least natural number that has exactly three factors?
(13)
For a certain natural number n, n2 gives a remainder of 4 when divided by 5,
and n3 gives a remainder of 2 when divided by 5. What remainder does n give when divided
by 5?
(14)
What is the least whole number that is divisible by 7, but leaves a remainder
of 1 when divided by any integer 2 through 6?
(15)
What is the smallest positive three-digit number that is a multiple of 4 and
whose digits’ sum is 23?
(16)
A group of N students, where N < 50, is on a field trip. If their teacher puts
them in groups of 8, the last group has 5 students. If their teacher instead puts them in
groups of 6, the last group has 3 students. What is the sum of all possible values of N?
(17)
What is the remainder when 10! is divided by 27 ? (Reminder: If n is a
positive integer, then n! stands for the product 1 · 2 · 3 · · · · · (n − 1) · n.)
(18)
A circular target has scoring regions of 5 and 7 points. What is the largest
score that cannot be obtained by throwing any number of darts that land on the target?
7
5
7
5
(19)
(20)
How many prime positive integers are divisors of 555?
The greatest common divisor of positive integers m and n is 6. The least
common multiple of m and n is 126. What is the least possible value of m + n?
(21)
A whole number, N, is chosen so that
is the value of N ?
N
3
is strictly between 7.5 and 8. What
How many positive perfect squares less than 300 are multiples of 9?
(22)
(23)
The product of the base seven numbers 247 and 307 is expressed in base
seven. What is the base seven sum of the digits of this product?
(24)
Four consecutive positive integers have a product of 840. What is the
largest of the four integers?
(25)
A composite number is a number that has two or more prime factors. The
number 87 can be expressed as the sum of two composite numbers in many ways. What is
the minimum positive difference between two such numbers?
(26)
What is the least possible positive integer with exactly five distinct positive
factors?
(27)
What is the remainder when the sum of the first 100 positive integers is
divided by 9?
(28)
For what value of n is the four-digit number 712n, with units digit n,
divisible by 18?
(29)
What is the median of all values defined by the expression 2x − 1, where x is
a prime number between 0 and 20?
(30)
How many positive integer factors of 22 × 32 × 5 are multiples of 12?
Copyright MATHCOUNTS Inc. All rights reserved
Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Answer
16
120
96
7
20
21
2550
36
30
June
1950
4
3
301
788
66
0
23
3
60
23
5 perfect squares
6
7
3
16
1
8
1087
4
Problem ID
C343
2B03
CC13
D3C2
35001
22A2
C3B5
4A03
D3B5
A513
ABC3
1113
AC42
52AC
DDC2
0BD3
DCB3
C4B5
DB55
0A13
50D2
53A2
DBD3
ABA2
0DC1
ABAC
53001
24BC
A2AC
1103
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1) 16
ID: [C343]
No solution is available at this time.
(2) 120
ID: [2B03]
No solution is available at this time.
(3) 96
ID: [CC13]
No solution is available at this time.
(4) 7
ID: [D3C2]
We observe that for all n > 5, n! has a units digit of 0, because 5! has a factor of 5 and 2,
which become a factor of 10. So, the terms in the sum, 5!, 7!, 9!, and 11! all have a 0 for
the units digit. And, 1! + 3! = 1 + 6 = 7 is the units digit of the sum.
(5) 20
ID: [35001]
No solution is available at this time.
(6) 21
ID: [22A2]
No solution is available at this time.
(7) 2550
ID: [C3B5]
No solution is available at this time.
(8) 36
ID: [4A03]
Let’s find the prime factorization of 1512: 1512 = 23 · 189 = 23 · 33 · 7. The only two
squares of primes that divide 1512 are 22 = 4 and 32 = 9. Therefore, the largest perfect
square factor of 1512 is 22 · 32 = (2 · 3)2 = 36 .
(9) 30
ID: [D3B5]
No solution is available at this time.
(10) June
ID: [A513]
Since exactly 1 in every n consecutive dates is divisible by n, the month with the fewest
relatively prime days is the month with the greatest number of distinct small prime divisors.
This reasoning gives us June (6 = 2 · 3) and December (12 = 22 · 3). December, however,
has one more relatively prime day, namely December 31, than does June, which has only 30
days. Therefore, June has the fewest relatively prime days.
(11) 1950
ID: [ABC3]
We have
22001 × 51950 ÷ 427 = 22001 ÷ 254 × 51950
= 21947 × 51950
= (2 × 5)1947 × 53
= 125 × 101947
Since 125 × 101947 has three non-zero digits followed by 1947 zeros, it has a total of 1950
digits.
(12) 4
ID: [1113]
No solution is available at this time.
(13) 3
ID: [AC42]
If two numbers give the same remainder when divided by 5, they are said to be equivalent,
modulo 5. From n2 to n3 , we have multiplied by n. Since n2 is equivalent to 4 (modulo 5),
and n3 is equivalent to 2 (modulo 5), we are looking for an integer n for which 4 · n is
equivalent to 2, modulo 5. Notice that if n is greater than 4, then we can replace it with
its remainder when divided by 5 without changing whether it satisfies the condition.
Therefore, we may assume that 0 ≤ n < 5. Trying 0, 1, 2, 3, and 4, we find that only 3
times 4 leaves a remainder of 2 when divided by 5.
(14) 301
ID: [52AC]
If n leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, then n − 1 is divisible by all
of those integers. In other words, n − 1 is a multiple of the least common multiple of 2, 3,
4, 5, and 6. Prime factorizing 2, 3, 4, 5, and 6, we find that their least common multiple is
22 · 3 · 5 = 60. Thus the possible values for an integer n which is one more than a multiple
of 2, 3, 4, 5, and 6 are 61, 121, 181, 241, 301 and so on. Checking them one at a time,
we find that the least of these integers which is divisible by 7 is 301 .
(15) 788
ID: [DDC2]
23 = 5 + 9 + 9 = 6 + 8 + 9 = 7 + 7 + 9 = 7 + 8 + 8 = .... The six smallest three-digit
numbers whose digits’ sum is 23 are 599, 689, 698, 779, 788, and 797 and since 88 is
divisible by 4, 788 is divisible by 4 as well. Therefore, the smallest three-digit number that
is a multiple of 4 and whose digits’ sum is 23 is 788 .
(16) 66
ID: [0BD3]
We are given that N ≡ 5 (mod 8) and N ≡ 3 (mod 6). We begin checking numbers which
are 5 more than a multiple of 8, and we find that 5 and 13 are not 3 more than a multiple
of 6, but 21 is 3 more than a multiple of 6. Thus 21 is one possible value of N. By the
Chinese Remainder Theorem, the integers x satisfying x ≡ 5 (mod 8) and x ≡ 3 (mod 6)
are those of the form x = 21 + lcm(6, 8)k = 21 + 24k, where k is an integer. Thus the 2
solutions less than 50 are 21 and 21 + 24(1) = 45, and their sum is 21 + 45 = 66 .
(17) 0
ID: [DCB3]
10! is divisible by 2, 4 = 22 , 6 = 2 · 3, 8 = 23 , and 10=2 · 5, so 10! is divisible by 28 .
Therefore, the remainder when 10! is divided by 27 is 0 .
(18) 23
ID: [C4B5]
No solution is available at this time.
(19) 3
ID: [DB55]
When we find the prime factorization of 555, we end up with 3 · 5 · 37, which means we
have 3 prime positive divisors.
(20) 60
ID: [0A13]
Since the GCD of m and n is 6, m = 6x and n = 6y for some integers x and y . Note that
minimizing m + n = 6x + 6y = 6(x + y ) is equivalent to minimizing x + y .
The LCM of m and n is 126 = 2 · 32 · 7 = 6 · 3 · 7, so one of x and y is divisible by 3 and
one is divisible by 7. Then we can minimize x + y by setting x and y to be 3 and 7 in some
order. Therefore, the least possible value of m + n is 6(3 + 7) = 60 .
(21) 23
ID: [50D2]
N
7.5 < 3 < 8 ⇒ 22.5 < N < 24. Since N is a whole number, N = 23 .
(22) 5 perfect squares
ID: [53A2]
2
Since 9 = 3 , any perfect square that is a multiple of 9 must be in the form 9n2 for some
positive integer n. For 1 ≤ n ≤ 5, 9n2 is less than 300 and for n ≥ 6, 9n2 is greater than
300. Therefore there are 5 positive perfect squares less than 300 that are multiples of 9.
(23) 6
ID: [DBD3]
We can ignore the 0 digit for now, and find the product of 247 × 37 . First, we need to
multiply the units digit: 47 × 37 = 1210 = 157 . Hence, we write down a 5 and carry-over
the 1. Evaluating the next digit, we need to multiply 27 × 37 + 17 = 710 = 107 . Thus, the
next digit is a 0 and 1 is carried over. Writing this out:
1
247
× 37
1057
We can ignore the 0 in 307 , since it does not contribute to the sum. Thus, the answer is
1+0+5= 6.
Notice that the base seven sum of the digits of a number leaves the same remainder
upon division by 6 as the number itself.
(24) 7
ID: [ABA2]
We have 840 = 23 · 3 · 5 · 7. From this prime factorization, it is clear that the product of
four consecutive positive integers is 840 = 22 · 5 · (2 · 3) · 7 = 4 · 5 · 6 · 7. The largest of the
four integers is 7 .
(25) 3
ID: [0DC1]
The minimum difference between two numbers whose sum is 87 is achieved when the
numbers are as close as possible to 87 ÷ 2 = 43.5. These numbers are 43 and 44, but 43 is
prime, so we consider the next pair, 42 and 45, both of which are composite. Thus, the
minimum positive difference is 45 − 42 = 3 .
(26) 16
ID: [ABAC]
No solution is available at this time.
(27) 1
ID: [53001]
No solution is available at this time.
(28) 8
ID: [24BC]
We know that 18 = 9 · 2, so in order for the four digit number to be divisible by 18 it must
also be divisible by 9 and 2. In order for a number to be divisible by 9, the sum of its digits
must be divisible by 9. Thus, 7 + 1 + 2 + n, or 10 + n, must be divisible by 9. Since 18 is
the smallest multiple of 9 that is greater than 10, n = 18 − 10 = 8 .
(29) 1087
ID: [A2AC]
No solution is available at this time.
(30) 4
ID: [1103]
No solution is available at this time.
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