SSC CGL 2024 Tier-I · 26 September 2024, Shift 2 · Quantitative Aptitude · Question 5 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 10 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.

The five-digit number 45yz0 is divisible by 40. What is the maximum possible value of (y + z)?

(a)16
(b)18
(c)17
(d)15

Answer

Why

Correct — A. 40 = 8 × 5, and 8 and 5 share no factor, so 45yz0 must pass both tests.

The number ends in 0, so divisibility by 5 is already satisfied.

For 8, look at the last three digits, which are y, z, 0:
100y + 10z = 10(10y + z)
That is a multiple of 8 exactly when 10y + z is a multiple of 4.

So the two-digit block yz must be divisible by 4. Now push y + z as high as it will go:
y + z = 18 gives only 99, and 99 is not a multiple of 4
y + z = 17 gives 89 and 98, neither a multiple of 4
y + z = 16 gives 88, and 88 = 4 × 22 ✓

Check: 45880 ÷ 40 = 1147, so the maximum is 16 → option (a)

Why the others are wrong

(b)
18 — 18 forces y = z = 9, and 45990 ÷ 40 = 1149.75. The block 99 is not a multiple of 4, so the 8-test fails.
(c)
17 — 17 allows only the blocks 89 and 98, and neither is a multiple of 4 — 45890 and 45980 both fail the 8-test.
(d)
15 — 15 does work, through 96: 45960 ÷ 40 = 1149. But the question asks for the maximum, and 88 lifts y + z to 16.

Concept

A composite divisor splits into coprime factors, and a number is divisible by the composite exactly when it clears each factor separately. Here 40 = 8 × 5.

The 5-test costs nothing because the number already ends in 0. The 8-test looks at the last three digits, and because the final digit is 0 those three digits are 10 × (10y + z) — so the real condition is that the two-digit block yz is divisible by 4.

That turns the question into a small search: among the blocks divisible by 4, which has the largest digit sum? Work down from 18, not up from 0.

Testing candidates in descending order of y + z is what makes this quick — the first block that survives is the answer, and 88 survives on the third try.

Key facts

40 = 8 × 5 with gcd(8, 5) = 1, so divisibility by 40 means divisibility by 8 and by 5.
A number is divisible by 8 when the number formed by its last three digits is.
45880 = 40 × 1147, and y = z = 8 gives the largest digit sum, 16.

Study next

Divisibility rules for 8 and 11, and why a composite test factors into coprime partsDigit-fill questions asking for the largest or smallest admissible digitDivisibility by 4 as a test on the last two digits

Common traps

Testing only the last two digits for 8, which is the rule for 4.
Stopping at the first pair that works instead of maximising y + z.
Assuming the largest possible digit sum, 18, must be reachable.

The digit-fill form recurs with different divisors, sometimes with two divisors at once.

See 12 Sep 2024, 09:00, Quant Q.1, which fills 72864*$ for divisibility by both 8 and 5, and 12 Sep 2024, 16:00, Quant Q.10, which fills 91876a2 for 8.

Related PYQs

No directly related past PYQ was found.