UPSC Prelims 2019 · CSAT
Question 75: Basic Numeracy
An 8-digit number 4252746B leaves remainder 0 when divided by 3. How many values of B are possible ?
AnswerUPSC’s answer is (c).
A number is divisible by 3 when its digit sum is. The known digits 4, 2, 5, 2, 7, 4 and 6 add to 30, which is already a multiple of 3, so B must itself be a multiple of 3: 0, 3, 6 or 9. That gives 4 possible values. Remember that 0 and 9 both count.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) Wrong: undercounts; 0, 3, 6 and 9 all work.
- (b) Wrong: undercounts; it misses 0 or 9.
- (c) Right: B can be 0, 3, 6 or 9.
- (d) Wrong: overcounts; only multiples of 3 work.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.