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.

Which one of the following is the remainder when 74^100 is divided by 9 ?

(a)2
(b)5
(c)3
(d)7

Answer

Why

Correct — D, (d) 7. The booklet prints the exponent as a true superscript, so the number under test is 74 raised to the power 100. Nobody is expected to compute it. Reduce the base first: 74 = 8 x 9 + 2, so 74 leaves 2 on division by 9, and therefore 74 to the power 100 leaves the same remainder as 2 to the power 100. Now find the cycle of the powers of 2 modulo 9. They run 2, 4, 8, 7, 5, 1 and then repeat, because 2 to the power 6 is 64 = 7 x 9 + 1. The cycle length is 6, so only the exponent's remainder on division by 6 matters. 100 = 6 x 16 + 4, which lands on the fourth term of the cycle, and the fourth term is 7. The same result in one line: 2 cubed is 8, which is one less than 9, so 2 to the power 99 leaves 8 (that is, minus 1) and 2 to the power 100 leaves 16 - 9 = 7. Note also the printed options run 2, 5, 3, 7 — not in ascending order. That is the Commission's ordering on this item and it is worth noticing before you start matching your working against a position on the page.

Why the others are wrong

(a)2 — This is the remainder of the BASE, 74, on division by 9 — the first step of the working mistaken for the last. It is also exactly where you land if you take the cycle length to be 3 instead of 6, on the strength of 2 cubed being 8: 8 is congruent to minus 1 modulo 9, not to 1, so the pattern only closes after six steps. Using 3 gives 100 = 3 x 33 + 1 and hands you back the first term of the cycle, 2. A cycle argument is only safe once you have found the exponent at which the residue returns to 1.
(b)5 — 5 is a genuine member of the cycle 2, 4, 8, 7, 5, 1 — it is the fifth term, that is 2 to the power 5 = 32, which leaves 5 — so this option catches an off-by-one in counting the position. 100 leaves 4 on division by 6, not 5. The cheap guard is to check the cycle's own anchor: the sixth term must come out as 1, and if your counting makes 2 to the power 6 anything other than 1, the positions have slipped.
(c)3 — 3 cannot be the remainder of any power of 2 on division by 9, and this can be settled without computing anything. If 2 to the power 100 left 3 on division by 9, then 9 would divide the difference, so 3 would divide it too, and since 3 divides 3 it would have to divide 2 to the power 100 — impossible, because 2 to the power 100 has no factor of 3 in it. The same argument rules out 0 and 6. A remainder modulo 9 that is a multiple of 3 is only available to numbers that are themselves multiples of 3, and 74 is not one.

Concept

This is modular arithmetic, and it rests on one property: remainders survive multiplication. If a leaves remainder r on division by n, then a to the power k leaves the same remainder as r to the power k. So an unmanageable base is replaced by a small one before anything else happens — 74 becomes 2. The second property is periodicity: the successive powers of a number coprime to the modulus must eventually repeat, because there are only finitely many possible remainders, and once a power returns the remainder 1 the whole sequence starts over. For 2 modulo 9 the residues are 2, 4, 8, 7, 5, 1, so the period is 6. Euler's theorem gives the same period as an upper bound without any trial: the count of numbers below 9 that are coprime to 9 is 6, so 2 to the power 6 must leave remainder 1. With the period known, the exponent is reduced modulo the period and only a tiny power is ever evaluated.

Remainder questions are a standing fixture of the quantitative block on EPFO EO/AO papers, and they are chosen precisely because they look impossible and are not. The habit rewarded is a fixed three-step drill — reduce the base, find the period, reduce the exponent — done in that order every time. The one genuinely instructive detail on this particular item is the option ordering: the four numbers are printed 2, 5, 3, 7, and a candidate who assumes numeric options ascend down the page can pick the wrong letter with entirely correct working.

Key facts

74 leaves remainder 2 on division by 9 (74 = 72 + 2), so 74 to the power 100 and 2 to the power 100 leave the same remainder.
The powers of 2 modulo 9 cycle with period 6: 2, 4, 8, 7, 5, 1, because 2 to the power 6 = 64 = 63 + 1.
100 = 6 x 16 + 4, so 2 to the power 100 matches 2 to the power 4 = 16, which leaves 7.
Shortcut: 2 cubed = 8 is congruent to minus 1 modulo 9, so odd multiples of 3 in the exponent give 8 and even multiples give 1.
Euler's theorem: when the base and the modulus share no common factor, the base raised to the count of coprime residues (6, for the modulus 9) leaves remainder 1.
A number not divisible by 3 can never leave a remainder of 0, 3 or 6 on division by 9 — which eliminates one option here before any arithmetic.
The digit-sum test doubles as a quick reduction modulo 9: the digits of 74 sum to 11, whose digits sum to 2.

Reduce the base, find the cycle, then reduce the exponent

Step 1 — reduce the base→74 = 8 x 9 + 2, so 74 leaves 2 on division by 9. Remainders survive multiplication, so 74^100 leaves the same remainder as 2^100. The digit-sum test does the same job in one glance: 7 + 4 = 11, and 1 + 1 = 2.

Step 2 — find the cycle of 2 modulo 9→2, 4, 8, 7, 5, 1 — and then it repeats, because 2^6 = 64 = 7 x 9 + 1. The period is 6. Euler’s theorem gives the same period without any trial: six of the numbers below 9 are coprime to 9.

Step 3 — reduce the exponent→100 = 6 x 16 + 4, so 2^100 lands on the fourth term of the cycle.

Fourth term: 2^4 = 16 = 9 + 7→The remainder is 7.

The same thing in one line→2^3 = 8, which is one less than 9. So 2^99 leaves 8, that is minus 1, and 2^100 leaves 16 − 9 = 7.

One option dies before any arithmetic→3 can never be the remainder of a power of 2 on division by 9: if it were, 3 would divide 2^100, which has no factor of 3 in it. The same argument rules out 0 and 6. Of the rest, 2 is the remainder of the BASE — the first step mistaken for the last — and 5 is the fifth term of the cycle, an off-by-one in counting the position. Note also that the printed options run 2, 5, 3, 7, not in ascending order.

Study next

Cyclicity of unit digits — the same periodic argument applied with modulus 10Euler's totient function and Euler's theorem, and Fermat's little theorem as its prime-modulus special caseRemainders modulo 7, 11 and 13, where the periods are longer and the reduction step matters moreDivisibility tests for 9 and 11 as statements about remainders rather than as tricksThe Chinese Remainder Theorem, for questions that impose two remainder conditions at once

Common traps

Stopping after reducing the base and offering that residue as the answer.
Taking the cycle length from the first power that looks special. 2 cubed = 8 is minus 1 modulo 9, not 1, so the period is 6 and not 3.
Counting the cycle position from the wrong end, which shifts the answer by one place into a neighbouring residue that is also on the list of options.
Assuming the numeric options are in ascending order. On this item they are printed 2, 5, 3, 7.
Forgetting that a remainder must be smaller than the divisor — any option of 9 or more would be impossible here on sight.

EPFO EO/AO quantitative blocks favour one remainder or cyclicity item per paper, always with a base too large to expand and an exponent in the hundreds. The stem is short and the four options are bare numerals, so the entire question is the method. Expect the same shape with a different modulus rather than a different idea.

Related PYQs

No directly related past PYQ was found.

Practice

practice — not a real PYQ
Which one of the following is the remainder when 74 raised to the power 101 is divided by 9 ?
- (a)2
- (b)3
- (c)5
- (d)8
Answer(c) 5
practice — not a real PYQ
Which one of the following is the remainder when 5 raised to the power 100 is divided by 13 ?
- (a)1
- (b)5
- (c)8
- (d)12
Answer(a) 1

Why