SSC CGL 2024 Tier-I · 26 September 2024, Shift 3 · Quantitative Aptitude · Question 17 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.

7¹⁵ + 7¹⁶ + 7¹⁷ is divisible by which of the following given numbers?

(a)5
(b)2
(c)4
(d)3

Answer

Why

Correct — D. Take the lowest power out as a common factor.

7¹⁵ + 7¹⁶ + 7¹⁷ = 7¹⁵(1 + 7 + 7²)
= 7¹⁵ × (1 + 7 + 49) = 7¹⁵ × 57

57 = 3 × 19, so the expression is 7¹⁵ × 3 × 19.

Its prime factors are 7, 3 and 19, so 3 divides it → option (d).

Why the others are wrong

(a)
5 — Divisibility by 5 needs a factor of 5. 7¹⁵ supplies only 7s and 57 = 3 × 19, so no 5 appears anywhere in the product.
(b)
2 — Every power of 7 is odd, and three odd numbers add to an odd number. An odd number has no factor of 2.
(c)
4 — 4 needs two factors of 2, and the total is odd for the same reason as option (b). No factor of 2 exists to supply even one.

Concept

A sum of consecutive powers factorises: pull out the lowest power, and what remains is a small number you can factor by hand.

7¹⁵ + 7¹⁶ + 7¹⁷ = 7¹⁵(1 + 7 + 49), and 1 + 7 + 49 = 57 = 3 × 19.

Every divisor of the whole expression is therefore built from 7, 3 and 19. Testing an option means asking whether its prime factors appear in that list — 3 does, 2 and 5 do not.

The powers never need evaluating. 7¹⁵ is a thirteen-digit number, and the question is answerable without knowing a single one of its digits.

That is the point of the large exponent: it makes computation useless and factorisation the only route.

Key facts

aⁿ + aⁿ⁺¹ + aⁿ⁺² = aⁿ(1 + a + a²).
1 + 7 + 49 = 57, and 57 = 3 × 19.
A number is divisible by d only if every prime factor of d appears in its own factorisation.

Study next

Factorising sums of consecutive powers of one baseReading divisibility straight off a prime factorisationUnit-digit cycles of powers, which settle divisibility by 2 and 5 at a glance

Common traps

Factoring out 7¹⁷, the largest power, which leaves fractions instead of whole numbers.
Adding the exponents 15, 16 and 17 instead of factorising the terms.
Assuming 7¹⁶ is even because 16 is even — the exponent's parity says nothing about the value's.

SSC's number-system items here ask which number divides an expression rather than what the expression equals. The exponent is chosen large enough that evaluation is pointless and the bracket small enough to factor in your head.

Related PYQs

No directly related past PYQ was found.