SSC CGL 2024 Tier-I · 9 September 2024, Shift 1 · Quantitative Aptitude · Question 11 of 100

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.

Which of the following can be the value of ‘k’ so that the number 217924k is divisible by 6?

(a)2
(b)6
(c)0
(d)4

Answer

Why

Correct — A. Divisibility by 6 needs both the 2-test and the 3-test, because 6 = 2 × 3 and 2 and 3 share no factor.

2-test: k is the last digit, so k must be even.
3-test: 2+1+7+9+2+4 = 25, so 25 + k must be a multiple of 3.
The multiples of 3 from 27 upward give k = 2, 5 or 8.

Even and in {2, 5, 8}: only 2 and 8, and 8 is not offered.
k = 2 → option (a), since 2179242 = 6 × 363207.

Why the others are wrong

(b)
6 — 6 is even, so it clears the 2-test, but 25 + 6 = 31, which is not a multiple of 3 — so 2179246 fails the 3-test and is not divisible by 6.
(c)
0 — 0 is even, yet 25 + 0 = 25 leaves remainder 1 on division by 3, so 2179240 is divisible by 2 but not by 3.
(d)
4 — 4 is even, but 25 + 4 = 29 is not a multiple of 3, so 2179244 clears the 2-test and fails the 3-test — exactly the half-check that makes this option tempting.

Concept

A composite divisor may be split into factors only when those factors are coprime. 6 splits safely into 2 and 3, so passing both tests is enough.

The 2-test reads the last digit. The 3-test reads the digit sum, because 10 ≡ 1 (mod 3), so every digit contributes its own value.

Here the fixed digits already sum to 25, and the unknown k has to carry that sum to the next multiple of 3 while staying even.

The stem says 'can be', not 'is'. Both k = 2 and k = 8 make the number divisible by 6, and the option list offers just one of them.

Key facts

Divisible by 6 means divisible by 2 and by 3, since 2 and 3 are coprime.
The 3-test uses the digit sum: 2+1+7+9+2+4+k = 25 + k.
25 + k is a multiple of 3 for k = 2, 5, 8, and only k = 2 and k = 8 are also even.

Study next

Divisibility rules for 7, 11 and 13Divisibility by 4 and 8 from the last two and last three digitsProblems with two unknown digits, where both rules must hold at once

Common traps

Checking only the digit-sum rule and choosing any k that makes it a multiple of 3
Splitting 12 into 2 and 6 — they are not coprime, so 12 needs the 4-test and the 3-test
Adding the fixed digits as 24 rather than 25, which shifts every candidate for k

SSC hides one digit in a long number and asks which option 'can be' its value. Read that phrase literally — several digits usually work, and your job is to find the one on the list.

Related PYQs

No directly related past PYQ was found.