Test: Divisibility - 1
10 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Test: Divisibility - 1
Which of the following is a multiple of 6?
How many multiples of 7 are there between 14 and 140, inclusive?
How many positive integer values of N are possible if 21 is divisible by N?
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
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.
If t is a positive integer and 8t is divisible by 96, what will be the remainder when t3 is divided by 108?
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.
Answer: Option (A)
If 32455 × 3145208 × K2 is divisible by 3, which of the following could be the value of K?
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
Answer: Option (C)
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.
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
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
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.
Answer: Option (C)
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