Download (2) Multiply 10.5 by 1.05 and write the product as a

a
b
(1)
Find a pair of integers a and b such that
(2)
Multiply 10.5 by 1.05 and write the product as a common fraction.
(3)
Express 3.7̄ as a common fraction.
(4)
(5)
(6)
(7)
(8)
(9)
= .09.
If the factors of twenty-four and the multiples of four are removed from the
set of integers from 1 through 50, how many integers remain in the set?
What is the next prime number after 89?
The number 13 is prime. If you reverse the digits you also obtain a prime
number, 31. What is the larger of the pair of primes that satisfies this condition and has a
sum of 110?
Find the sum of the prime factors of 1122.
Name the greatest whole number less than 100 that has an odd number of
positive factors.
Suppose A and B are non-zero digits and A5 · B1 = 1995. Find A + B
(10)
Find the smallest positive integer divisible by 10, 11, and 12.
(11)
What is the remainder when 5 to the 999,999th power is divided by 7?
(12)
If x y = y x , and x and y are unequal positive integers, what is the smallest
possible value of x + y ?
(13)
What is the difference between the largest and smallest prime factors of
15,015?
(14)
Express as a fraction in lowest terms: 0.1 + 0.01
(15)
Express .28 as a common fraction.
(16)
What is the least positive integer by which you could multiply 180 to get a
product that is a perfect cube?
(17)
In how many zeros does the number represented by 10! end?
(18)
Given 2a = 32 and ab = 125 find ba .
(19)
A test for divisibility by 7 tells us that the number 10n + d is divisible by 7
exactly when n − 2d is divisible by 7, where d is a digit. Use this rule to determine whether
777784 is a multiple of 7.
(20)
(21)
Find the value of x such that the four-digit number x 15x is divisible by 18.
What is the sum of the median and mean of all natural number divisors of
16? Express your answer as a mixed number.
(22)
What numeral is in the 100th decimal place in the decimal representation of
6
7?
(23)
(24)
Simplify: (1 12 )−2 + (1 21 )−1 + (1 21 )0 + (1 12 )1 + (1 21 )2
The product of 2 consecutive two-digit numbers is 8010. What is the sum
of the two numbers?
(25)
(26)
Find the sum of the smallest and largest prime factors of 10101.
The product of two consecutive negative integers is 272. Find the larger of
the two integers.
36
x+3
is an integer?
(27)
How many positive integer values of x are there such that
(28)
How many positive divisors does the number 2 × 32 × 53 have?
(29)
What common fraction (that is, a fraction reduced to its lowest terms) is
equivalent to .325?
(30)
Find distinct digits A and B such that A47B is as large as possible and
divisible by 36. Name the number.
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
a = 1, b = 10
441/40
34/9
34 integers
97
73
33
81
11
660
6
6
10
4/33
28/99
150
2
243
yes
6
14 1/5
1
211
36
179
40
−16
6
24 divisors
161/495
5472
Problem ID
D32B
502B
1C1B
50DB
D23A1
1DAA1
15AB
5C412
BAB41
1C412
345A1
2C3A1
3C4D1
21BB
BDC41
D42B
31DB
4B4D1
424A1
C23B
35C51
133B
335D1
DB5B
5C0C
0BCB
4BAB
CBCB
DBCA
A1DB
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1) a = 1, b = 10
ID: [D32B]
No solution is available at this time.
(2) 441/40
ID: [502B]
No solution is available at this time.
(3) 34/9
ID: [1C1B]
To express the number 3.7 as a fraction, we let x = 3.7, so 10x = 37.7 and:
−
This shows that 3.7 =
10x = 37.77777 . . .
x = 3.77777 . . .
9x = 34
34
.
9
(4) 34 integers
ID: [50DB]
No solution is available at this time.
(5) 97
ID: [D23A1]
No solution is available at this time.
(6) 73
ID: [1DAA1]
No solution is available at this time.
(7) 33
ID: [15AB]
No solution is available at this time.
(8) 81
ID: [5C412]
We claim that a number has an odd number of positive factors if and only if it is a perfect
square. Indeed, for all non-square numbers x , we can pair each factor f with another factor
x
must be an even number of factors. For perfect squares, this argument fails
f , so there
√
only for x, so there are an odd number of factors for perfect squares. Thus we seek the
lowest perfect square below 100, which is 81 .
(9) 11
ID: [BAB41]
No solution is available at this time.
(10) 660
ID: [1C412]
Factoring all three numbers, we find that 10 = 2 · 5, 11 = 11, and 12 = 22 · 3. Taking the
highest power of each, we see that the least common multiple of the three numbers is
22 · 3 · 5 · 11 = 60 · 11 = 660 .
(11) 6
ID: [345A1]
To find the remainder when 5999999 is divided by 7, we look at the first few powers of 5
modulo 7:
50 ≡ 1,
51 ≡ 5,
52 ≡ 5 · 5 ≡ 25 ≡ 4,
53 ≡ 5 · 4 ≡ 20 ≡ 6,
54 ≡ 5 · 6 ≡ 30 ≡ 2,
55 ≡ 5 · 2 ≡ 10 ≡ 3,
56 ≡ 5 · 3 ≡ 15 ≡ 1
(mod 7).
Since 56 ≡ 1 (mod 7), the remainders become periodic, with period 6. Since 999999 ≡ 3
(mod 6), 5999999 ≡ 53 ≡ 6 (mod 7).
(12) 6
ID: [2C3A1]
No solution is available at this time.
(13) 10
ID: [3C4D1]
Since 15,015 ends in 5, it’s divisible by 5. Using the divisibility by 3 rule, we know that since
the sum of the digits of 15,015 is divisible by 3, 15,015 is divisible by 3. Finally, since the
alternating sum of the digits of 15,015 is divisible by 11, the divisibility by 11 rule implies
that 15,015 is divisible by 11. We use these 3 observations to write 15, 015 = 3 · 5 · 11 · 91.
At this point, we try to factor 91 into primes and get that 91 = 7 · 13, so we can write
15, 015 = 3 · 5 · 7 · 11 · 13. We have now factored 15,015 into primes, so we observe that the
difference between the largest, which is 13, and the smallest, which is 3, is 13 − 3 = 10 .
(14) 4/33
ID: [21BB]
We begin by realizing that 0.1 = 0.11, so 0.1 + 0.01 = 0.11 + 0.01 = 0.12. (Note that this
can be done because there is no carrying involved.)
To express the number 0.12 as a fraction, we call it x and subtract it from 100x :
100x = 12.121212 . . .
−
x = 0.121212 . . .
99x = 12
This shows that 0.12 = 12
99 .
But that isn’t in lowest terms, since 12 and 99 share a common factor of 3. We can
4
12
, which is in lowest terms.
reduce 99
to
33
(15) 28/99
ID: [BDC41]
If x = .28, then 100x = 28.28. Notice that we can eliminate the repeating decimal by
28
subtracting .28 from 28.28. We have 100x − x = 99x = 28, so x = 99
. The repeating
28
.
decimal can be expressed as the fraction
99
(16) 150
ID: [D42B]
No solution is available at this time.
(17) 2
ID: [31DB]
No solution is available at this time.
(18) 243
ID: [4B4D1]
We note that 32 = 2 · 2 · 2 · 2 · 2 = 25 , so a = 5. This leaves us with
5b = 125 = 5 · 5 · 5 = 53 , which means that b = 3. Therefore our answer is ba = 35 = 243 .
(19) yes
ID: [424A1]
No solution is available at this time.
(20) 6
ID: [C23B]
The prime factorization of 18 is 32 · 2, so in order for a number to be divisible by 18 it must
be divisible by both 3 and 2. First, in order for a number to be divisible by 3, the sum of its
digits must be divisible by 3. In the case of x 15x , this means that x + 1 + 5 + x = 2x + 6
must be divisible by 3. Since the constant term (6) is already a multiple of 3, 2x must be
divisible by 3, which means that x itself must be a multiple of 3 as well. Second, in order
for a number to be divisible by 2, the units digit must be an even number. In this case, the
divisibility rule for 2 implies that x must be an even number. Thus, since we know that it
must be an even single-digit multiple of 3, the only possible value for x is 6 .
(21) 14 1/5
ID: [35C51]
No solution is available at this time.
(22) 1
ID: [133B]
The decimal representation of 67 is 0.857142, which repeats every 6 digits. Since 100
divided by 6 has a remainder of 4, the 100th digit is the same as the fourth digit following
the decimal point, which is 1 .
(23)
211
36
ID: [335D1]
We rewrite 1 12 as 23 and use the fact that ( ba )−c = ( ba )c . Our sum is then
( 32 )2 + 32 + 1 + 23 + ( 32 )2 = 49 + 32 + 1 + 23 + 49 . Since we have fractions with denominators of
4 and 9, we multiply each fraction as necessary to put everything over a common
24
36
54
81
denominator of 4 · 9 = 36. This leaves our sum as 16
36 + 36 + 36 + 36 + 36 , which simplifies
211
.
our final answer of
36
(24) 179
ID: [DB5B]
No solution is available at this time.
(25) 40
ID: [5C0C]
10101 is clearly not divisible by 2 or 5. The sum of 10101’s digits is 3, so it is divisible by
3, but not by 9. 10101 = 3 · 3367. 3367/7 = 481 and 481/7 = 68 57 so 10101 = 3 · 7 · 481
and 481 is not divisible by any prime number less than 11. Applying the divisibility test for
11, we have 4 − 8 + 1 = −3, which is not divisible by 11, so 481 is not divisible by 11 either.
481/13 = 37 and 37 is prime, so the prime factorization of 10101 is 10101 = 3 · 7 · 13 · 37.
So, the sum of its smallest and largest prime factors is 3 + 37 = 40 .
(26) −16
ID: [0BCB]
No solution is available at this time.
(27) 6
ID: [4BAB]
No solution is available at this time.
(28) 24 divisors
ID: [CBCB]
No solution is available at this time.
(29) 161/495
ID: [DBCA]
To express the number 0.325 as a fraction, we call it x and subtract it from 100x :
100x = 32.5252525 . . .
−
x = 0.3252525 . . .
99x = 32.2
161
.
495
(Note: This last fraction is in lowest terms, because 161 = 7 · 23 and 495 = 32 · 5 · 11.)
This shows that 0.325 =
32.2
99
=
322
990
=
(30) 5472
ID: [A1DB]
No solution is available at this time.
Copyright MATHCOUNTS Inc. All rights reserved