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# Test: Divisibility - 1

## 10 Questions MCQ Test Quantitative Aptitude for GMAT | Test: Divisibility - 1

Description
This mock test of Test: Divisibility - 1 for UPSC helps you for every UPSC entrance exam. This contains 10 Multiple Choice Questions for UPSC Test: Divisibility - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Divisibility - 1 quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Test: Divisibility - 1 exercise for a better result in the exam. You can find other Test: Divisibility - 1 extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1

Solution:
QUESTION: 2

Solution:
QUESTION: 3

### How many positive integer values of N are possible if 21 is divisible by N?

Solution:
QUESTION: 4

If a number N is divisible by both 2 and 8, then which of the following statements must be true?

I.           N is divisible by 4

II.         N is divisible by 6

III.      N is divisible by 16

Solution:
QUESTION: 5

N = abc where a, b and c are the hundreds, tens and units digit respectively. If a, b and c are non-zero consecutive numbers such that a < b < c, then which of the following must be true?

I.            N is always divisible by 2

II.            N is always divisible by 3

III.            N is divisible by 6 only if b is odd.

Solution:
QUESTION: 6

If t is a positive integer and 8t is divisible by 96, what will be the remainder when t3 is divided by 108?

Solution:

Given: 108 = 22 × 33

Since 8t is divisible by 96, we may write

8t = 96k, where k is a positive integer

• 23t = 25 × 3× k
• t = 22 × 3× k
• t3 = 26 × 33 × k3
• t3 = ( 22 × 33 )( 24  × k3)
• t3 = 108( 24  × k3)

This shows that t3 is completely divisible by 108, implying that the remainder is 0.

QUESTION: 7

If 32455 × 3145208 × K2 is divisible by 3, which of the following could be the value of K?

Solution:

Step 1: Question statement and Inferences

32455 × 3145208 × K2 is divisible by 3

• Either 32455 or 3145208 or K2 must be divisible by 3

Step 2: Finding required values

Given:

A number is divisible by 3 when sum of its digits are divisible by 3

• Sum of the digits of 32455 = 3+2+4+5+5 = 19 not divisible by 3
• Sum of the digits of 3145208 = 3+1+4+5+2+0+8 = 23 not divisible by 3
• Therefore 32455 and 3145208 are not divisible by 3.

Hence, K2 should be divisible by 3

• K should be divisible by 3 (3 is prime, Kn and K will have same prime factors)

Step 3: Calculating the final answer

Checking for all the options:

• Sum of the digits of 6000209 = 6+0+0+0+2+0+9 = 17, not divisible by 3
• Sum of the digits of 6111209 = 6+1+1+1+2+0+9 = 20, not divisible by 3
• Sum of the digits of 6111309 = 6+1+1+1+3+0+9 = 21, divisible by 3
• Sum of the digits of 6111109 = 6+1+1+1+1+0+9 = 19, not divisible by 3
• Sum of the digits of 6111809 = 6+1+1+1+8+0+9 = 26, not divisible by 3

Only 6111309 is divisible by 3

QUESTION: 8

n = 234yzn

is a positive integer whose tens and units digits are y and z respectively. It is given that n is divisible by 4, 5 and 9. Find n.

Solution:
QUESTION: 9

What is the remainder when the positive three-digit number 1yz is divided by 7?

(1)  y + z = 7

(2)  y -2 is a non-zero positive number divisible by 3

Solution:
QUESTION: 10

If t is a positive integer, can t2 + 1 be evenly divided by 10?

(1)  916 × t leaves a remainder of 1 when divided by 2

(2)  916 × t leaves a remainder of 2 when divided by 5

Solution:

Steps 1 & 2: Understand Question and Draw Inferences

t2 + 1 can be evenly divided by 10 if t2 + 1 = 10 *k, where k is a positive integer

• We need to check if the last digit of t2 + 1 is 0

Step 3: Analyze Statement 1

916 × t leaves a remainder of 1 when divided by 2

• t is odd
• last digit of t can be 1, 3, 5, 7 or 9

Not Sufficient.

Step 4: Analyze Statement 2

916 × t leaves a remainder of 2 when divided by 5

• last digit of 916 × t can be 2 or 7
• last digit of t can be 2 or 7

Not Sufficient

Step 5: Analyze Both Statements Together (if needed)

Inference from statement 1: last digit of t can be 1, 3, 5, 7 or 9

Inference from statement 2: last digit of t can be 2 or 7

Inference from statement 1 and statement 2: last digit of t can be 7

--> Last digit of t2 + 1 = Last digit of (9 + 1) = 0

Hence, t2 + 1 can be evenly divided by 10

Statement 1 and Statement 2 together are sufficient to answer the question.