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.

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

Why

Correct — B, (b) 12. Work the count in two stages, because the stem asks for occurrences of the digit and then filters those occurrences by a property of the whole number. Stage one: from 1 to 100 the digit 3 appears 20 times — once in each of 3, 13, 23, 43, 53, 63, 73, 83 and 93, once in each of 30, 31, 32, 34, 35, 36, 37, 38 and 39, and twice in 33. That is 19 numbers carrying 20 appearances. Stage two: strike the numbers that are divisible by 3, because the printed condition (set in bold italic on the booklet) keeps only occurrences sitting inside a number that is not a multiple of 3. Among those 19, the multiples of 3 are 3, 30, 33, 36, 39, 63 and 93 — seven numbers, but eight appearances of the digit, because 33 carries two. So 20 − 8 = 12. A useful check: 33 was the only number with a repeated 3, and it has been struck out, so every surviving number contributes exactly one appearance and the count of surviving numbers equals the count of surviving digits — 13, 23, 31, 32, 34, 35, 37, 38, 43, 53, 73 and 83, twelve of them.

Why the others are wrong

(a)11 — One short. The surviving list is 13, 23, 31, 32, 34, 35, 37, 38, 43, 53, 73, 83 — twelve numbers — and 11 is what a candidate reaches after losing exactly one of them. The thirties are where a number goes missing, because four of that decade's ten members (30, 33, 36, 39) are struck out and the six that survive (31, 32, 34, 35, 37, 38) are easy to strike by association with their neighbours. Sweeping a decade in blocks rather than testing each number is the habit this option punishes.
(c)13 — This is 20 − 7: the right total of digit appearances, less the count of NUMBERS struck out instead of the count of APPEARANCES struck out. Seven numbers are removed (3, 30, 33, 36, 39, 63, 93) but eight appearances go with them, because 33 carries the digit twice. Mixing the two units in one subtraction is the single commonest error on this item, and it costs exactly one.
(d)17 — This sits three short of 20, the number of times the digit 3 appears in 1 to 100 with no condition applied at all. It is the answer of a candidate who counted the digit correctly and then tested only the handful of numbers that look obviously like multiples of 3 — 3 itself, 30, 33 — instead of running the divisibility test across all nineteen numbers that contain the digit. The gap between 17 and the three clustered options tells you the condition has barely been applied.

Concept

This is a digit-counting problem with a filter attached, and the whole difficulty lies in keeping two different objects apart: the DIGIT 3, which is what you are counting, and the NUMBER containing it, which is what the divisibility condition is about. The reliable method is to enumerate rather than to reason in the abstract. Between 1 and 100 the digit 3 occupies the tens place in 30 to 39 (ten numbers) and the units place in 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 (ten numbers); the number 33 is on both lists, so nineteen distinct numbers carry the digit and the digit appears twenty times. Then apply the filter. A number is divisible by 3 when its digit sum is — 30 (3), 33 (6), 36 (9), 39 (12), 63 (9), 93 (12) and 3 itself all pass that test, and no other number on the list does. Removing them removes eight of the twenty appearances.

Part B of the EO/AO General Ability Test opens with a five-item block of quantitative aptitude and reasoning, and this is the reasoning-flavoured end of it: no formula, just a controlled enumeration under time pressure. The habit it rewards is reading the ask exactly. The word 'not' is emphasised in the booklet, but it is a CONDITION on the numbers, not a negative ask — the question is 'How many times', which is as positive an ask as they come. A candidate who reads the emphasis as 'find the exception' has misread a counting problem as an odd-one-out problem.

Key facts

The digit 3 appears 20 times among the whole numbers from 1 to 100 — ten times in the tens place (30 to 39) and ten times in the units place (3, 13, 23, 33, ... , 93).
Nineteen distinct numbers in that range contain the digit 3; only 33 contains it twice, which is why 19 numbers carry 20 appearances.
Divisibility by 3 is tested on the digit sum: a number is a multiple of 3 exactly when its digits add to a multiple of 3.
Of the nineteen numbers containing a 3, seven are multiples of 3 — 3, 30, 33, 36, 39, 63 and 93 — and they carry eight appearances of the digit between them.
20 appearances minus 8 struck appearances leaves 12, and the twelve surviving numbers are 13, 23, 31, 32, 34, 35, 37, 38, 43, 53, 73 and 83.
Because 33 is the only number with a repeated 3 and it is struck out, on this particular question the number of qualifying numbers and the number of qualifying digit appearances happen to coincide.

Count the digit, then strike by the number — two different units

Stage 1 — how often the digit 3 appears from 1 to 100→Tens place: 30 to 39, ten numbers. Units place: 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, ten numbers. 33 stands on both lists, so nineteen distinct numbers carry the digit and it appears twenty times.

Stage 2 — which of those nineteen are multiples of 3→Test the digit sum: a number is a multiple of 3 exactly when its digits add to a multiple of 3. Seven of the nineteen pass — 3, 30, 33, 36, 39, 63 and 93.

The trap: numbers struck, appearances removed→Those seven numbers carry EIGHT appearances of the digit between them, because 33 contains it twice. The filter is about the number; the count is about the digit.

20 − 8 = 12→The surviving appearances, and the answer.

Check by listing→13, 23, 31, 32, 34, 35, 37, 38, 43, 53, 73, 83 — twelve. Since 33 was the only number with a repeated 3 and it has been struck out, every survivor contributes exactly one appearance, so the count of numbers and the count of digits happen to agree here.

Where the wrong options come from→20 − 7 = 13 subtracts the count of NUMBERS instead of the count of APPEARANCES, which is the commonest error and costs exactly one. 17 is 20 less only the obvious multiples 3, 30 and 33. 11 loses one survivor, usually from the thirties, where four of the ten members are struck out and the six that remain are easy to sweep away with them.

Study next

Digit-frequency counting over a range — how many times a given digit appears from 1 to 100, 1 to 1000, and the place-value argument that generalises itDivisibility rules for 2, 3, 4, 5, 6, 8, 9 and 11, and which of them work on the digit sumCounting with a filter: enumerate first, then remove, and keep the units of the two counts straightInclusion-exclusion on overlapping sets, which is what makes 33 appear on both the tens list and the units list

Common traps

Counting numbers when the question asks for appearances of a digit, or the reverse. The two totals differ by exactly the number of repeated-digit cases in the range.
Forgetting that 33 carries the digit twice. Every wrong route through this problem passes through that oversight.
Reading the emphasised 'not' as making this a negative question. It qualifies the numbers; the ask is 'How many times'.
Testing divisibility on the digit rather than on the number containing it.
Sweeping the thirties as a block. Six of those ten numbers qualify and four do not, so the decade has to be tested member by member.

Quantitative items in the opening block of EPFO EO/AO Part B are short-stemmed and terse-optioned, and the wrong options are built to catch a specific slip rather than to be far away from the truth. Here three of the four sit within two of each other. Counting questions in this family almost always attach a condition — divisible by, greater than, ending in — and the marks turn on applying the condition to the right object.

Related PYQs

EPFO_EOAO_2020_Q25Open & attempt →

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(c) Both 1 and 2

The other number-theory item in this paper's opening block: it tests factors of a composite number on either side of its square root, and like this one it is solved by enumerating carefully rather than by recalling a formula.

Practice

practice — not a real PYQ
How many times does the digit 7 appear among the whole numbers from 1 to 100 such that the number where 7 appears is divisible by 7 ?
- (a)3
- (b)4
- (c)5
- (d)6
Answer(b) 4
practice — not a real PYQ
How many whole numbers from 1 to 100 contain the digit 3 at least once ?
- (a)18
- (b)19
- (c)20
- (d)21
Answer(b) 19

Why