Prop of No. - Error Log

Question 1

Which of the following has 15^80 as a factor?

A. 15^60
B. 30^40
C. 40^80
D. 60^60
E. 90^80

Answer 1

E. 90^80

Prime factors of 15^80 is 3^80 and 5^80. Only the prime factorization of 90^80 also contains 5^80 and 3^80.

Question 2

What is the units digit of 17^83 * 13^82 * 11^87?

Answer 2

D. 7

The exponent pattern for the number set gives us 3 * 9 * 1 = 27. So, there must be a 7 in the units digit.

Question 3

What is the number units digit of 357^110?

Answer 3

9

108 is the closest multiple of 4 before 110. Because 7 has an exponent pattern of 7,9,3,1 we know that 357^108 has a units digit of 1 and 357^110 would have to have a units digit of 9.

Question 4

If N = 1+4^31, what is the remainder when N is divided by 3?

Answer 4

2.

Put 4 to the power of 1 and solve and we get 5/3. Put 4 to the power of 2 and we get 17/3 which is 5 + 2/3. So, this shows a pattern where the remainder is always 2.

Question 5

If a,b,c, and d are integers; w,x,y,and z are prime numbers; w<x<y<z; and (w^a)(x^b)(y^c)(z^d) = 660, what is the value of (a+b) - (c+d)?

Answer 5

1

Prime factorize 660 and we get (2^2)(3)(5)(11). When the bases are ordered smallest to largest we plug the exponents into the equation and get (2+1) - (1+1) = 1

Question 6

To create a rectangular box, one dimension of a cube is increased by 1 cm, another dimension is decreased by 1 cm, and the third dimension remains unchanged. If the dimensions of the cube are integer lengths in cm, which of the following could be the volume in cm^3?

A. 238,266
B. 238,267
C. 238,268
D. 238,269
E. 238,270

Answer 6

A. 238,266

This results in 3 consecutive numbers. Any 3 consecutive numbers would be divisible by 3!. 3! = 6 and A is the only answer choice that is divisible by 6.

Question 7

What is the greatest value of positive integer x such that 2^x is a factor of 100^80?

Answer 7

160

Prime factorize 100^80 and we get (10^2)^80 => 10^160 => 2^160 * 5^160

Question 8

When X is divided by 63, the remainder is 27. What is the remainder when x is divided by 7?

Answer 8

6

• Put the first sentence into formulaic form: x/63 = Q + 27/63.
• Solve for x: x= 63Q + 27.
• Put X over 7: x/7 = (63Q + 27)/7
• Simplify: x/7 = 9Q +27/7
• Remainder: 27/7 = 3 + 6/7

Question 9

What is the remainder when 3^123 is divided by 5?

Answer 9

2

Units digit exponent pattern is 3,9,7,1

The corresponding pattern when we divide those by 5 is: 0 R 3, 1 R 4, 1 R 2, and 0 R 1.

The 3,4,2,1 remainder pattern repeats every fourth iteration. This means 3^120 divided by 5 has a remainder of 1 and therefore 3^123 divided by 5 would have a remainder of 2.

Question 10

If (30! * 30!)/30^n is an integer, what is the largest possible value of integer n?

Answer 10

14

Look at the prime factors of the denominator 2^n, 3^n, 5^n. Because 5 is the largest prime factor there will be less of them in 30!, so the answer is only dependent on 5.

Use factorial division shortcut:
30/5 = 6
30/(5^2) = 1 (ignore the remainder)
30/(5^3) = 0

This means there are seven 5s in 30! Thus there are another seven 5s in the next 30! The answer is 14

Question 11

How many digits are in the integer 8^7 * 25^12?

Answer 11

24

Use the number of 5 and 2 pairs to determine the number of 0s at the end of the integer. We have 21 pairs of 5 and 2 with another three 5s remaining. We multiple 555 to get 125 which has 3 digits. This means there are 21 + 3 digits.

Question 12

A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?

Answer 12

6

Translate the first sentence to:
n = 9Q + 6

Translate the 2nd sentence to:
m = 12Z + 4

Multiplied together we have:
108QZ + 36Q + 72Z + 24

Divide by 18. It divides into evenly except into 24. 24/18 gives us a remainder of 6

Question 13

When n is divided by 19, the remainder is 6. When n is divided by 11, the quotient is equal to z, and the remainder is 6. Z must be divisible by which number?

Answer 13

19

First sentence: n = 19Q + 6
Second sentence: n = 11Z + 6

Set them equal to each other and solve for Z:
19Q=11Z or Q=11Z/19 so Z must be divisible by 19

Question 14

When X is divided by 143, the remainder is 45. Which of the following numbers, if added to X, would result in a number divisible by 13?

Answer 14

7

First sentence:
x/143 = Q + 45/143

Solve for x and divide by 13:
x = (143Q + 45)/13 –> 11Q + 45/13

to make 45/13 an integer we add 7 to 45

Question 15

If x = 989 and y = 991, what is the remainder of xy/9?

Answer 15

8

990 is divisible by 9, so 989 must have a remainder of 8 and 991 must have a remainder of 1.

8*1 = 8

Question 16

If W = (63^33) * (36^195), what is the remainder of w/10?

Answer 16

8

Using our exponent rules, we know that 63^33 has a units digit of 3 and 36^195 has a units digit of 6.

3*6 = 18 so the units digit is 8.
Finally, 8/10 will give us a remainder of 8

Question 17

What is the units digit of 26! + 50! + 4! + 4! ?

Answer 17

8

The 5 and 2 pairs in the first 2 numbers let us know their units’ digit is 0.

4x3x2 = 24 –> the last 2 numbers have a 4 in the unit’s digit

4+4 = 8

Question 18

If x = 25!/20!, how many trailing 0s are in x?

Answer 18

2

25!/20! = 25x24x23x22x21

When finding the numbers of 5 and 2 pairs, we are limited by the number of 5s. Because there are only two 5s, we know there are two trailing 0s.

Question 19

What is the units digit of:
(5! 4! + 6! 5!)/31 ?

Answer 19

0

Simplify the numerator:
5! (4! + 6!)/31

5! 4! (1 + 6 * 5)/31

5!4!

Units digit will be 0 because any factorial greater or equal to 5! has a 5*2 pair.

Question 20

If the original mass of an element, in grams, is a whole number and the element has decayed by exactly 94 percent in some time period, which of the following could be the final mass of that element?

A. 27 grams
B. 28 grams
C. 29 grams
D. 31 grams
E. 32 grams

Answer 20

27 grams

N = new weight; W = old weight

Only 6 percent (6/100) of old element remains. Therefore:
N = (6/100)W
N = (3/50)W
50N = 3W
W = 50N/3

Answer must be divisible by 3.

Question 21

How many positive integers divide evenly into 520?

Answer 21

16

Prime factorization

Add 1 to each exponent

4 * 2 * 2 = 16

Question 22

If x, y, and z are positive integers such that x^2 + y^2 + z^2 = 64,470 which of the following could be the values of x, y, and z?

1. x = 115, y = 146, z = 173
2. x = 114, y = 142, z = 171
3. x = 110, y = 108, z = 179

Answer 22

1 only

Square the units digit of each possible answer and add them together. Only the first answer choice would have a units digit of 0.

Question 23

If a, b, c, d, and x are nonzero numbers, which of the following conditions must be true for
(ax)^61(bx)^21 (cx)^14 (dx)^6 >0?

A. x>0
B. a * b > 0
C. c * d > 0
D. a * d > 0
E. a * b < 0

Answer 23

B. a * b > 0

The cx and dx terms are irrelevant because any term raised to an even exponent will be positive.

When we look at (ax)^61 and (bx)^21 we can expand the terms to:
a^61 x^61 x^21 b^21

When combining like terms we see that x is x^82 meaning it is irrelevant

That leaves us with the a and b terms, so a * b must be positive (greater than 0).

Question 24

What (50!50!50!)/50^n is an integer, what is the largest possible value of integer n?

Answer 24

18

We take prime factors of the denominator which is 2 and 5^2.

We take the larger prime factor (5) and do 50/5, 50/25, 50/125.

After adding those together we get 12. Because 50! occurs 3 times in the numerator we multiply 12*3 and then divide by 2 (because 5 is squared when we prime factorize 50).

Question 25

If 2/a + 3/a + 4/a is an integer, what must also be an integer?

A. 12/a
B. 3/a
C. 6/a
D. 9/a
E. 15/a

Answer 25

D. 9/a

When we add all of the fractions together we have 9/a. Should be self-explanatory…

Question 26

X cannot equal 0 and
(-y^2 - xy)/x = y + x, the value represented by which of the following expressions is equal to its own opposite?

Answer 26

(y + x)^2

The only number equal to its opposite is 0.

When we simplify the equation we get y^2 + 2xy + x^2 = 0

According to quadratic properties that is the same as (y + x)^2 = 0

Question 27

How many leading zeros are in the decimal form of 1/7! ?

Answer 27

3

When we multiply out 7! we do (5 * 2) * (7 * 6 * 5 * 4 * 3)

This is the same as 10 * 504 = 5,040

If X is an integer with k digits, then 1/x will have k-1 leading zeros unless x is a perfect power of 10. 4 digits minus 1 equals 3 leading zeros.

Question 28

If a and b are integers and 250a = b^2, all of the following values must divide evenly into b^2, except

A. 20
B. 40
C. 250
D. 625
E. 1,250

Answer 28

B. 40

If b squared is equal to 250a, 250a must be a perfect square. We prime factorize 250 and any value that if prime factorized would create even exponents when combined with the prime factors of 250 will create a perfect square.

Only 40 would not create a perfect square.

Question 29

What is the 57th digit to the right of the decimal point in the decimal equivalent of 13/55 ?

Answer 29

6

Do long division and notice that the value is 0.2363636…Every odd digit after the decimal (except for the initial 2) is 6.

Question 30

If q = 40! + 1, which of the following cannot be a prime factor of q?

1. 11
2. 19
3. 37

Answer 30

None of these numbers can be a prime factor of q.

The prime numbers between 2 and 37 inclusive are within 40! However, 2 consecutive numbers cannot share the same prime factors. So 40! and 40! + 1 cannot share any prime factors.

Question 31

If n is an integer and (9! + 8! + 7!)/3^n is an integer, what is the largest possible value of n?

Answer 31

6

Factor 7! in the numerator and we get 7! (9*8+8+1) –> 7!(81)

Now expand 7!(81) to 7654321*3^4

There are 6 threes in the numerator (don’t forget that 6 is 3*2), so n can’t be greater than 6.

Question 32

To create a rectangular box, one dimension of a cube is increased by 1 cm, another is decreased by 1 cm, and the third dimension remains unchanged. If the dimensions of the cube are odd integer lengths in cm, which of the following could be the volume of the rectangular solid created?

A. 99
B. 132
C. 198
D. 231
E. 336

Answer 32

E. 336

Notice we are dealing with 3 consecutive numbers, with the lowest and highest of these three numbers being even.

Because (n-1) and (n+1) are even, one of the numbers will be divisible by 4 and another will be divisible by 2. That means that the answer must be divisible by 8.

Only 336 is divisible by 8.

Question 33

If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?

Answer 33

67

Remember that n^3 - n can also be written as n (n^2 - 1), which is the same as n(n-1)(n+1). This means we are looking at 3 consecutive numbers.

If n is odd, n-1 and n+1 are even. That means every time n is odd, n^3 - n will be divisible by 12. There are 90 two digit numbers, half of those (45 are possible answers).

If n is even, then only digit multiples of 4 are possible. We can use the formula: ((Highest # divisible by the given number - Lowest # divisible by the given number)/Given Number)+1

So 96-12=84 –> 84/4 + 1 = 22

45 + 22 = 67