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# Test: Factors And Multiples- 4

## 15 Questions MCQ Test Quantitative Aptitude for GMAT | Test: Factors And Multiples- 4

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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. The solved questions answers in this Test: Factors And Multiples- 4 quiz give you a good mix of easy questions and tough questions. GMAT students definitely take this Test: Factors And Multiples- 4 exercise for a better result in the exam. You can find other Test: Factors And Multiples- 4 extra questions, long questions & short questions for GMAT on EduRev as well by searching above.
QUESTION: 1

Solution:
QUESTION: 2

Solution:
QUESTION: 3

### If k is an integer greater than 1, is k equal to 2r for some positive integer r? 1) k is divisible by 26 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 2100 – 296?

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 = a4b3, 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 xy2?

1)  x is a multiple of 54. ?

2)  y is a multiple of 6. ?

Solution:
QUESTION: 14

If both 112 and 33 are factors of the number a * 43 * 62 * 1311, 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 * 43 * 62 * 1311 can be expressed in terms of its prime factors as a * 28 * 32 * 1311

Step 2 of solving this GMAT Number Properties Question: Find factors missing after excluding 'a' to make the number divisible by both 112 and 33

112 is a factor of the given number.
If we do not include 'a', 11 is not a prime factor of the given number.
If 112 is a factor of the number, 112 should be a part of 'a'

33 is a factor of the given number.
If we do not include 'a', the number has only 32 in it.
Therefore, if 33 has to be a factor of the given number 'a' has to contain 31 in it.

Therefore, 'a' should be at least 112 * 3 = 363 if the given number has 112 and 33 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 106 and 107, 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 101 and 102, we have 2 numbers, 11 and 20.
Between 100 and 1000, that is 102 and 103, we have 3 numbers, 101, 110 and 200.

Therefore, between 106 and 107, one will have 7 integers whose sum will be equal to 2.

Alternative approach

All numbers between 106 and 107 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.