SSC CGL 2024 Tier-I · 17 September 2024, Shift 1 · Quantitative Aptitude · Question 20 of 98

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Find the least value of k for which a 7-digit number 249k876 is divisible by 6.

(a)5
(b)0
(c)7
(d)1

Answer

Why

Correct — B. A number is divisible by 6 only when it clears both the 2-test and the 3-test.

2-test: the last digit of 249k876 is 6, which is even whatever k turns out to be. So k has no say in this half.

3-test: 2 + 4 + 9 + 8 + 7 + 6 = 36, so the digit sum is 36 + k.
36 is already a multiple of 3, so 3 divides 36 + k exactly when 3 divides k.
Values of k that work: 0, 3, 6, 9.

The least of those is 0, and 0 is on the list of choices → option (b). Check: 2490876 is even, and its digits add to 36.

Why the others are wrong

(a)
5 — 5 makes the digit sum 41. 41 is not a multiple of 3, so 2495876 is even but still fails the 3-test.
(c)
7 — 7 makes the digit sum 43, again not a multiple of 3. And 7 could not be the answer regardless, because the question asks for the least value.
(d)
1 — 1 makes the digit sum 37, one past the multiple 36. It is the smallest non-zero choice, which is what catches a reader who assumes 'least' rules out 0.

Concept

You may split a divisibility test into factors only when those factors are coprime. 6 = 2 × 3 and 2 and 3 share no factor, so passing both tests is enough.

The same move fails for 8: a number divisible by 2 and by 4 need not be divisible by 8, because 2 and 4 share a factor. 12 clears both and is not a multiple of 8.

The 3-test itself works because 10, 100, 1000 and so on all leave remainder 1 on division by 3, so a number and its digit sum leave the same remainder.

'Least value' includes 0 here. The blank sits in the middle of 249k876, not at the front, so a leading-zero objection does not arise and 2490876 is an ordinary 7-digit number.

Key facts

6 = 2 × 3, and because 2 and 3 are coprime a number divisible by both is divisible by 6.
A number is divisible by 3 exactly when its digit sum is divisible by 3, whatever the order of the digits.
The digits of 249k876 apart from k add to 36, which is itself a multiple of 3.
0 is a multiple of every non-zero integer, so k = 0 passes the 3-test.

Study next

Divisibility by 8 and 11, where the test uses place value rather than a plain digit sumDigit-fill items with two blanks, such as a number divisible by both 8 and 5Reading the remainder on division by 9 straight off the digit sum

Common traps

Assuming 'least value' rules out 0 and settling for 1.
Stopping after the even-last-digit check, which every choice passes.
Splitting a composite into factors that share a divisor, as in testing 8 by testing 2 and 4.

SSC prints a 6- or 7-digit number with one digit blanked and asks which value makes it divisible by a small composite, so the work is a digit sum plus a glance at the last digit.

The same 6-test is set at 9 Sep 2024, 09:00, Quant Q.11 on 217924k, and at 11 Sep 2024, 16:00, Quant Q.12 on 723*56*.

Related PYQs

No directly related past PYQ was found.