UPSC Prelims 2019 CSAT Paper II · Q75 of 79 Basic Numeracy easy

An 8-digit number 4252746B leaves remainder 0 when divided by 3. How many values of B are possible ?

  1. (a) 2
  2. (b) 3
  3. (c) 4 ✓ UPSC's answer
  4. (d) 6

Why the answer is (c)

• The divisibility rule for 3 states that a number is divisible by 3 if the sum of its digits is divisible by 3.

• Calculate the sum of the known digits: 4 + 2 + 5 + 2 + 7 + 4 + 6 = 30.

• The total sum of the digits is 30 + B.

• For the number to be divisible by 3, (30 + B) must be a multiple of 3.

• Since 30 is already divisible by 3, B itself must be divisible by 3.

• The possible single-digit values for B (0-9) that are divisible by 3 are 0, 3, 6, and 9, which gives 4 possible values.

Why the other options are wrong

(a) 2
Option (a) is incorrect because there are four valid digits (0, 3, 6, 9), not two.
(b) 3
Option (b) is incorrect because it misses one of the valid digits (0, 3, 6, 9) that satisfy the condition.
(d) 6
Option (d) is incorrect because only digits divisible by 3 are valid, and there are only four such digits in the range 0-9.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2019, held on 2 June 2019. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.

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