UPSC Prelims 2021 · CSAT

Question 36: Basic Numeracy

Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum. Which of the following is/are correct ?

1.S is always divisible by 74.
2.S is always divisible by 9.

Select the correct answer using the code given below :

a1 only

b2 only

cBoth 1 and 2

dNeither 1 nor 2

AnswerUPSC’s answer is (c).

The non-zero single digits that are multiples of 3 are 3, 6 and 9, so the numbers use these three digits once each. There are 6 such numbers. Each digit appears twice in each place, so S = 2 × (3 + 6 + 9) × 111 = 2 × 18 × 111 = 3996. Now 3996 = 74 × 54, so it is divisible by 74, and 3996 = 9 × 444, so it is divisible by 9. Both statements are correct. Remember: when digits can be arranged in every order, each digit appears equally often in every place.

1. 3996 = 74 × 54.
2. 3996 = 9 × 444; its digit sum 27 is also a multiple of 9.

How you could have got this

If you were sure statement 1 is correct, (b) and (d) go: each of them leaves 1 out.
What you needed to know: 3996 = 74 × 54.
Then, if you were sure statement 2 is correct, (a) goes: it leaves 2 out.
What you needed to know: 3996 = 9 × 444; its digit sum 27 is also a multiple of 9.

Only (c) is left. This one needs every statement.

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) Statement 1 is correct, but 2 is also correct: 3996 = 9 × 444.
(b) Statement 2 is correct, but 1 is also correct: 3996 = 74 × 54.
(c) Right: S = 3996 is divisible by both 74 and 9.
(d) S = 3996 is divisible by both, so this fails.

Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.

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Source: UPSC, Civil Services (Preliminary) Examination 2021, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.