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This mock test of Test: Factors And Multiples- 4 for GMAT helps you for every GMAT entrance exam.
This contains 15 Multiple Choice Questions for GMAT Test: Factors And Multiples- 4 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Is the last digit of integer x^{2} – y^{2} a zero?

1) x – y is an integer divisible by 30

2) x + y is an integer divisible by 70

Solution:

QUESTION: 2

How many different prime numbers are factors of the positive integer n?

1) Four different prime numbers are factors of 2n

2) Four different prime numbers are factors of n^{2}

Solution:

QUESTION: 3

If *k* is an integer greater than 1, is *k* equal to 2^{r} for some positive integer *r*?

1) *k *is divisible by 2^{6}

2) *k* is not divisible by any odd integer greater than 1

Solution:

QUESTION: 4

If 500 is the multiple of 100 that is closest to X and 400 is the multiple of 100 closest to Y, then which multiple of 100 is closest to X + Y?

1) X < 500

2) Y < 400

Solution:

QUESTION: 5

If the number x3458623y is divisible by 88, what is the value of x?

Solution:

QUESTION: 6

Is *m* divisible by *n*?

1) *m + n* is divisible by *m – n*

2) *m + n *is divisible by *n*

Solution:

QUESTION: 7

Is *r* a multiple of *s*?

1) *r + 2s* is a multiple of *s*

2) *2r + s* is a multiple of *s*

Solution:

QUESTION: 8

What is the greatest prime factor of 2^{100} – 2^{96}?

Solution:

QUESTION: 9

If *n *and *k *are positive integers, is *n *divisible by 6?

(1) *n *= *k*(*k*+1)(*k*-1)?

(2) *k *– 1 is a multiple of 3.

Solution:

QUESTION: 10

A positive integer *n* is said to be “prime-saturated” if the product of all the different positive prime factors of *n* is less than the square root of *n*. What is the greatest two-digit prime-saturated integer?

Solution:

QUESTION: 11

How many different factors does the integer n have?

1) n = a^{4}b^{3}, where a and b are different positive prime numbers. ?

2) The only positive prime numbers that are factors of n are 5 and 7. ?

Solution:

QUESTION: 12

If the integer *k *is a multiple of 3, which of the following is also a multiple of 3?

Solution:

QUESTION: 13

If x and y are nonzero integers, is 18 a factor of xy^{2}?

1) x is a multiple of 54. ?

2) y is a multiple of 6. ?

Solution:

QUESTION: 14

If both 11^{2} and 3^{3} are factors of the number a * 4^{3} * 6^{2} * 13^{11}, then what is the smallest possible value of 'a'?

Solution:

Step 1 of solving this GMAT Number Properties Question: Prime factorize the given expression

a * 4^{3} * 6^{2} * 13^{11} can be expressed in terms of its prime factors as a * 2^{8} * 3^{2} * 13^{11}

Step 2 of solving this GMAT Number Properties Question: Find factors missing after excluding 'a' to make the number divisible by both 11^{2} and 3^{3}

11^{2} is a factor of the given number.

If we do not include 'a', 11 is not a prime factor of the given number.

If 11^{2} is a factor of the number, 11^{2} should be a part of 'a'

3^{3} is a factor of the given number.

If we do not include 'a', the number has only 3^{2} in it.

Therefore, if 3^{3} has to be a factor of the given number 'a' has to contain 3^{1} in it.

Therefore, 'a' should be at least 11^{2} * 3 = 363 if the given number has 11^{2} and 3^{3} as its factors.

The question is **"what is the smallest possible value of 'a'?"**

The smallest value that 'a' can take is **363**

Choice C is the correct answer.

QUESTION: 15

How many different positive integers exist between 10^{6} and 10^{7}, the sum of whose digits is equal to 2?

Solution:

Method 1 to solve this GMAT Number Properties Question: Find the number of such integers existing for a lower power of 10 and extrapolate the results.

Between 10 and 100, that is 10^{1} and 10^{2}, we have 2 numbers, 11 and 20.

Between 100 and 1000, that is 10^{2} and 10^{3}, we have 3 numbers, 101, 110 and 200.

Therefore, between 10^{6} and 10^{7}, one will have 7 integers whose sum will be equal to 2.

Alternative approach

All numbers between 10^{6} and 10^{7} will be 7 digit numbers.

There are two possibilities if the sum of the digits has to be '2'.

**Possibility 1**: Two of the 7 digits are 1s and the remaining 5 are 0s.

The left most digit has to be one of the 1s. That leaves us with 6 places where the second 1 can appear.

So, a total of __six__ 7-digit numbers comprising two 1s exist, sum of whose digits is '2'.

**Possibility 2**: One digit is 2 and the remaining are 0s.

The only possibility is 2000000.

Total count is the sum of the counts from these two possibilities = 6 + 1 = 7

Choice B is the correct answer.

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