SSC CGL 2024 Tier-I · 12 September 2024, Shift 3 · Quantitative Aptitude · Question 10 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.

Find the smallest value that must be assigned to number ‘a’ in order for 91876a2 to be divisible by 8.

(a)3
(b)0
(c)1
(d)2

Answer

Why

Correct — A. A number is divisible by 8 when its last three digits are. Everything to the left of them is a multiple of 1000, and 1000 = 8 × 125, so only the block 6a2 decides this.

a = 0 → 602 = 8 × 75 + 2
a = 1 → 612 = 8 × 76 + 4
a = 2 → 622 = 8 × 77 + 6
a = 3 → 632 = 8 × 79, remainder 0

The smallest digit that clears the test is 3 → option (a).

Why the others are wrong

(b)
0 — a = 0 leaves 602, and 602 = 8 × 75 + 2. Zero is the smallest digit on offer, which is why it tempts, but smallest only counts among the values that actually divide.
(c)
1 — a = 1 gives 612, which passes the divisibility-by-4 test (last two digits, 12) and still fails by 8: 612 = 8 × 76 + 4. Stopping at the 4 rule is the usual slip.
(d)
2 — a = 2 gives 622 = 8 × 77 + 6. The number is even and its last two digits are 22, but 8 judges the whole three-digit block, not the last digit or the last pair.

Concept

Divisibility by 8 is a last-three-digits test, and the reason is positional: every place value from the thousands upward is a multiple of 1000, and 1000 = 8 × 125.

So 91876a2 splits into 91876000 plus 600 + 10a + 2. The first part is divisible by 8 whatever the leading digits are, so the whole number is divisible by 8 exactly when 602 + 10a is.

Since 10 leaves remainder 2 on division by 8, each step of a lifts the remainder by 2 and the pattern repeats every four digits — which is why a = 3 and a = 7 both work, and the question asks for the smaller.

The paper names the unknown digit a, sitting inside the numeral 91876a2, right next to answer labels that also run (a) to (d). The digit and the option letter have nothing to do with each other.

Key facts

A number is divisible by 8 when its last three digits form a multiple of 8, because 1000 = 8 × 125.
The divisibility-by-4 test uses only the last two digits, which is why 612 passes it and still fails by 8.
632 = 8 × 79, and the next digit that also works is a = 7, giving 672 = 8 × 84.

Study next

Divisibility rules for 7, 11 and 13 on five- and six-digit numbersFinding the missing digit that makes a number divisible by 9 or by 11Reducing a large number modulo 8 by looking only at its last three digits

Common traps

Applying the last-two-digits rule for 4 and stopping there.
Reading smallest value as smallest option printed and marking 0 without testing it.
Dividing the whole seven-digit number when only the 6a2 block matters.

SSC asks divisibility either as a fill-the-missing-digit item like this one, or as pick-the-number-that-satisfies-several-rules. The second form is at Quant Q.3 of this same paper, where the number has to clear 2, 5, 10 and 11 together.

Related PYQs

No directly related past PYQ was found.