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

Which of the following is a multiple of 6?

Solution:

QUESTION: 2

How many multiples of 7 are there between 14 and 140, inclusive?

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 = **2**^{2} × **3**^{3}

Since 8t is divisible by 96, we may write

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

**2**^{3}t =**2**^{5}×**3**× k- t =
**2**^{2}×**3**× k - t
^{3}=**2**^{6}×**3**^{3 }× k^{3} - t
^{3}= (**2**^{2}×**3**^{3 })(**2**^{4}^{ }× k^{3}) - t
^{3}= 108(**2**^{4}^{ }× k^{3})

This shows that t^{3 }is completely divisible by 108, implying that the remainder is 0.

**Answer: Option (A) **

QUESTION: 7

If 32455 × 3145208 × K^{2} is divisible by 3, which of the following could be the value of K?

Solution:

**Step 1: Question statement and Inferences**

32455 × 3145208 × K^{2} is divisible by 3

- Either 32455 or 3145208 or K
^{2}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, K^{2} should be divisible by 3

- K should be divisible by 3 (3 is prime, K
^{n}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

**Answer: Option (C) **

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 t^{2} + 1 be evenly divided by 10?

(1) 91^{6} × t leaves a remainder of 1 when divided by 2

(2) 91^{6} × t leaves a remainder of 2 when divided by 5

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

t^{2 }+ 1 can be evenly divided by 10 if t^{2} + 1 = 10 *k, where k is a positive integer

- We need to check if the last digit of t
^{2 }+ 1 is 0

__Step 3: Analyze Statement 1__

91^{6} × 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__

91^{6} × t leaves a remainder of 2 when divided by 5

- last digit of 91
^{6}× 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 t^{2} + 1 = Last digit of (9 + 1) = 0

Hence, t^{2 }+ 1 can be evenly divided by 10

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

** Answer: Option (C)**

### Divisibility Test for 5

Video | 02:50 min

### Divisibility Test for 10

Video | 03:34 min

### Divisibility Test for 3, 6, 9

Video | 03:44 min

- Test: Divisibility - 1
Test | 10 questions | 20 min

- Test: Divisibility And Remainders- 1
Test | 20 questions | 40 min

- Divisibility Test - MCQ Test
Test | 5 questions | 10 min

- Test: Divisibility - 2
Test | 10 questions | 20 min

- Divisibility Test - Practice Test
Test | 5 questions | 10 min