Enforcement Officer / Accounts Officer (EO/AO) 2020 recruitment test, advertisement 421. Transcribed from the Commission's own booklet (series A, NMJX-U-GTA); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

Let n ( > 1) be a composite natural number whose square root is not an integer. Consider the following statements : 1. n has a factor which is greater than 1 but less than the square root of n. 2. n has a factor which is greater than the square root of n but less than n. Which of the statements given above is/are correct ?

(a)1 only
(b)2 only
(c)Both 1 and 2
(d)Neither 1 nor 2

Answer

Why

Correct — C, (c) Both 1 and 2. Because n is composite, it can be written as n = a × b with both a and b strictly between 1 and n. Take a to be the smaller of the two, so a ≤ b. Now a cannot be larger than the square root of n: if it were, then b, being at least as large as a, would also exceed the square root, and their product a × b would exceed n. And a cannot EQUAL the square root of n, because the stem has ruled that out — the square root of n is not an integer, so no whole number squares to n. So a is strictly greater than 1 and strictly less than the square root of n, which is statement 1. Statement 2 then follows for free from the same factorisation: b = n ÷ a, and dividing n by something smaller than its square root leaves something larger than its square root, so b exceeds the square root of n; and b is smaller than n because a is greater than 1. So both statements hold, and they hold together rather than separately — every factor a below the square root is paired with the factor n ÷ a above it. Test it on 15: the factors are 1, 3, 5, 15, the square root is about 3·87, and 3 sits below it while 5 sits above.

Why the others are wrong

(a)1 only — This says only statement 1 holds. It cannot be right, and not because of anything special about statement 2 — the two statements are the same fact seen from opposite ends. Divisors of n come in pairs, a and n ÷ a, whose product is n; if one member of a pair is below the square root the other must be above it. So the moment statement 1 supplies a factor a with 1 < a < √n, the number n ÷ a is a factor with √n < n ÷ a < n, which is statement 2. There is no composite number for which the first is true and the second false.
(b)2 only — This says only statement 2 holds, and it fails for the mirror-image reason. If n has a factor b with √n < b < n, then n ÷ b is also a factor, it is greater than 1 (because b is less than n) and it is less than √n (because b is greater than √n) — which is statement 1. The pairing runs in both directions, so 1 and 2 stand or fall together and no option that separates them can be the answer.
(d)Neither 1 nor 2 — This says neither statement holds, which would make n prime — and the stem has already told you n is composite, so it has a factor other than 1 and itself by definition. The one situation in which both statements really do fail is a perfect square of a prime, such as 9, whose factors are 1, 3 and 9: the only factor between 1 and 9 is 3, which is exactly the square root, so nothing lies strictly on either side of it. That is precisely the case the stem excludes by saying the square root of n is not an integer, and noticing why that clause is there is the whole point of the question.

Concept

The idea being tested is the pairing of divisors about the square root. Every factor a of n has a partner, n ÷ a, and the two multiply back to n. If a is smaller than √n then its partner must be larger, and if a is larger then its partner must be smaller — because if both were below √n their product would fall short of n, and if both were above it their product would overshoot. The only way a factor can be its own partner is when a × a = n, that is when n is a perfect square and a is its square root. So for a non-square n the factors split cleanly into two equal-sized groups, one on each side of √n, with nothing sitting on the line; and once n is composite, neither group can be empty. That is exactly what the two statements assert. This is also the reason a non-square number always has an even count of divisors while a perfect square has an odd one — the square root is the unpaired divisor. And it is the reason that testing a number for primality only requires trial division up to its square root: if there were any factor above √n, there would already have been one below it, and the search would have found it first.

This is one of only a handful of items on this paper that print a numbered statement list and ask which statements are correct, and it rewards a different reflex from the rest of the quantitative block. There is nothing to compute — the four options are 1 only, 2 only, both, neither — so the work is in deciding whether each statement is true for EVERY number the stem allows, not merely for the first example that comes to mind. The productive habit is to look for the clause that has been inserted deliberately. Here it is 'whose square root is not an integer'. A stem does not carry a condition like that for decoration: it is there because the statements fail without it, and finding the case it excludes (a prime squared, such as 9, 25 or 49) is faster than any amount of general reasoning.

Key facts

A composite number greater than 1 has at least one factor other than 1 and itself, by definition.
Factors pair off as a and n ÷ a; if one is below the square root of n, the other is above it.
A factor can equal the square root only when n is a perfect square, so a non-square n has no factor sitting exactly on √n.
Consequence: a perfect square has an odd number of divisors and every other number has an even number.
Consequence: to test whether n is prime, trial division by numbers up to √n is enough.
The stem's exclusion of perfect squares matters — for n = 9 both statements fail, because its only middle factor, 3, IS the square root.
Worked instance: n = 15, √15 ≈ 3·87; the factor 3 lies below it and the factor 5 above it, and 3 × 5 = 15.
Worked instance: n = 221 = 13 × 17, √221 ≈ 14·87; 13 satisfies statement 1 and 17 satisfies statement 2.

The two statements are one fact seen from opposite ends

Statement 1 is true→n is composite, so n = a × b with both factors strictly between 1 and n. Take a as the smaller. It cannot exceed √n, because then b would exceed it too and the product would overshoot n; and it cannot EQUAL √n, because the stem has ruled out an integer square root. So 1 < a < √n.

Statement 2 follows for free→b = n ÷ a. Dividing n by something smaller than its square root leaves something larger than its square root, so b > √n; and b < n because a > 1. Both statements therefore hold, which is option (c).

Why they cannot be separated→Divisors come in pairs, a and n ÷ a, whose product is n. If one member of a pair lies below the square root, the other must lie above it. That is why ‘1 only’ and ‘2 only’ are both impossible: the pairing runs in both directions, so the two statements stand or fall together.

What the stem’s exclusion is doing→For a perfect square of a prime, such as 9, both statements fail: the only factor between 1 and 9 is 3, which IS the square root, so nothing lies strictly on either side of it. Ruling out an integer square root is exactly what makes the answer ‘both’.

Worked instances→n = 15, √15 ≈ 3·87 — the factor 3 lies below it and 5 above it, and 3 × 5 = 15. n = 221 = 13 × 17, √221 ≈ 14·87 — 13 satisfies statement 1 and 17 satisfies statement 2.

Two consequences worth banking→A perfect square has an odd number of divisors and every other number an even number, because the square root is the one unpaired divisor. And testing a number for primality needs trial division only up to its square root: any factor above it would already have been found below it.

Study next

Primality testing by trial division up to the square root, and why that bound is enoughThe number-of-divisors formula from the prime factorisation, and why perfect squares are the odd ones outThe Sieve of Eratosthenes, which starts crossing out at the square of each prime for the same reasonFundamental Theorem of Arithmetic — unique factorisation into primesHCF and LCM through prime factorisation, and the identity HCF × LCM = product of the two numbers

Common traps

Testing a statement on one convenient example and generalising. The statements have to hold for every n the stem admits.
Ignoring the clause 'whose square root is not an integer'. Drop it and both statements fail at n = 9, 25 or 49.
Reading 'factor' as 'prime factor'. The statements are about factors of any kind — for n = 24 the factor 4 is not prime and still counts.
Treating statements 1 and 2 as independent when they are two views of the same pairing, which makes the 'only' options structurally impossible.
Overlooking that the strict inequalities matter: statement 1 says greater than 1 and less than √n, so 1 itself never qualifies.

Numbered-statement items on EPFO EO/AO papers usually pair two claims that look independent and are not, and the four options are always the same four: one only, the other only, both, neither. On the quantitative side the claims tend to be about divisibility, factors, remainders or ordering of numbers. The reliable method is to test each statement against the extreme or degenerate case the stem seems to be guarding against, because the guard clause tells you where the examiner expected candidates to fall.

Related PYQs

EPFO_EOAO_2020_Q21Open & attempt →

How many times does the digit 3 appear between 1 and 100 such that the number where 3 appears is not divisible by 3 ?

(a) 11
(b) 12
(c) 13
(d) 17

Answer(b) 12

The other number-theory item in this paper's opening block: it counts occurrences of the digit 3 under a divisibility condition, and like this one it is settled by careful enumeration and case analysis rather than by a formula.

Practice

practice — not a real PYQ
Which one of the following is the smallest prime factor of 221 ?
- (a)3
- (b)7
- (c)13
- (d)17
Answer(c) 13
practice — not a real PYQ
A composite natural number n has a square root that is not an integer. What is the least possible number of distinct positive divisors of n ?
- (a)2
- (b)3
- (c)4
- (d)6
Answer(c) 4

Why