Assistant Provident Fund Commissioner and Enforcement Officer / Accounts Officer (APFC and EO/AO) 2025 recruitment test. Transcribed from the Commission's own booklet (series A, PENA-A-PTS); answer worked out by our faculty — UPSC has published no answer key for this sitting. This question is in Part B. Answer last reviewed 25 August 2026.

How many three-digit numbers are there that have the middle digit as the sum of the first digit and the third digit ?

(a)40
(b)45
(c)50
(d)55

Answer

Why

Correct — B, (b) 45.

Name the digits. Let the three-digit number be written as a b c, where a is the hundreds digit, b the tens digit and c the units digit. The condition is b = a + c.

The restrictions. Since the number has three digits, a runs from 1 to 9 — it cannot be 0. The units digit c runs from 0 to 9. The middle digit b is not free: it is fixed by a and c. But b must itself be a single digit, so a + c cannot exceed 9.

So the whole problem reduces to counting the pairs (a, c) with a from 1 to 9, c from 0 to 9, and a + c ≤ 9. Every such pair produces exactly one number, and different pairs produce different numbers.

Count them by fixing a. a = 1: c may be 0 to 8 — 9 values a = 2: c may be 0 to 7 — 8 values a = 3: 7 values a = 4: 6 a = 5: 5 a = 6: 4 a = 7: 3 a = 8: 2 a = 9: c = 0 only — 1 value

Total = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45.

That last sum is the sum of the first nine natural numbers, which is 9 × 10 ÷ 2 = 45 — worth recognising rather than adding term by term.

Examples to see that the count is right: with a = 1 the numbers are 110, 121, 132, 143, 154, 165, 176, 187 and 198 — nine of them, exactly as the count says. At the other end, a = 9 gives only 990.

One point of care: nothing forbids c from being 0, and nothing forbids two digits from being equal — 110 and 220 are perfectly good answers to the question as asked. Only the middle digit is constrained, and only by having to be a single digit.

The same count from the other direction, as a check. Fix the MIDDLE digit at b: the pairs that produce it are those with a + c = b and a at least 1, so a can be 1, 2, ..., b — that is b pairs. Summing over b from 1 to 9 gives 1 + 2 + ... + 9 = 45 again. Two different ways of slicing the same count agreeing is the best assurance available on a counting question, and both slices are quick.

Why the others are wrong

(a)40 — Forty comes from dropping some of the edge cases — most often the numbers in which the units digit is 0, such as 110, 220 and 330, on the unstated assumption that a three-digit number cannot end in zero. It can; only the leading digit is barred from being zero. Excluding the nine numbers with c = 0 would in fact leave 36, and other partial exclusions produce figures in the low forties.
(c)50 — Fifty is what comes of letting the hundreds digit be 0 as well, or of allowing a + c to reach 10 or more. Both are ruled out: a leading zero would make the number two-digit, and a middle digit of 10 is impossible. It is a plausible round number and is placed next to the correct answer for that reason.
(d)55 — Fifty-five is the sum of the first TEN natural numbers, 10 × 11 ÷ 2 — the answer to this count if the hundreds digit were allowed to take ten values, from 0 to 9, instead of nine. It is a precise and revealing error: the arithmetic is right and the constraint that a three-digit number cannot begin with zero has been forgotten.

Concept

Counting problems about digits are handled by naming the digit positions, listing the restrictions on each, and then counting systematically rather than by trying to enumerate the numbers themselves. The restrictions that recur are that the leading digit of an n-digit number cannot be zero, that every digit lies between 0 and 9, and that any digit determined by others must still satisfy that range. Once the problem is reduced to counting pairs or triples subject to an inequality, fixing one variable and counting the rest — as done here — is the reliable method. The sum of the first n natural numbers, n(n+1)/2, appears constantly in such counts and should be recognised on sight: 45 for nine terms, 55 for ten.

Digit-counting questions appear in the reasoning block of these papers, and their options are always built from the small family of numbers that different mistakes produce — here 40, 45, 50 and 55, four values within fifteen of one another. That closeness means the item cannot be answered by estimation; it has to be counted. The compensation is that it can be counted in under a minute once the pairs, rather than the numbers, are the object of the count.

The four options lie within fifteen of one another, which rules out estimation and forces an exact count. The efficient way to count is to shift the object of the counting: do not enumerate three-digit numbers, enumerate the PAIRS of outer digits that generate them, since the middle digit is determined once the other two are chosen. That reduction turns an unmanageable list into a nine-line table. Each wrong option here corresponds to mishandling one restriction — a leading zero admitted, a trailing zero excluded, or the single-digit cap on the middle digit forgotten.

Key facts

A three-digit number has a hundreds digit from 1 to 9 and other digits from 0 to 9.
The condition 'middle digit is the sum of the first and third' fixes b once a and c are chosen.
b must remain a single digit, so a + c ≤ 9.
Counting by fixing a gives 9, 8, 7, ..., 1 possibilities.
The total is 9 + 8 + ... + 1 = 9 × 10 ÷ 2 = 45.
Numbers such as 110 and 990 are valid; a zero units digit is not excluded.
The sum of the first n natural numbers is n(n+1)/2 — 45 for nine terms and 55 for ten.
Allowing a leading zero would inflate the count to 55, which is one of the offered options.

Study next

Counting numbers with digit restrictions, including palindromes and numbers with repeated digitsPermutations and combinations applied to digits, with and without repetitionDivisibility rules and counting multiples in a rangeThe sum of the first n natural numbers, squares and cubesSystematic enumeration as a technique when a formula is not obvious

Common traps

Allowing the hundreds digit to be zero, which gives 55.
Excluding a zero units digit, which loses nine valid numbers.
Forgetting that the middle digit must remain a single digit, so a + c cannot exceed 9.
Trying to list the numbers themselves instead of counting the pairs that generate them.

Digit-counting items appear once in most papers and are always answered by the same routine: name the positions, write the constraint on each, reduce to a count of pairs, and use the standard summation. Estimation is useless here because the wrong options are deliberately placed within a few units of the answer.

Related PYQs

EPFO_APFC_2023_Q74

Suppose that x and y are distinct variables that take values from {1, 2, 3, 4, 5, 6}. What is the probability that the value of the expression xy + x + y is even?

(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/5

Answer(d) 1/5

A counting item from the earlier paper worked the same way — a probability question over pairs of distinct values drawn from a small set, where the answer comes from enumerating admissible pairs.

Practice

practice — not a real PYQ
How many three-digit numbers are there in which the units digit is equal to the sum of the hundreds digit and the tens digit ?
- (a)36
- (b)45
- (c)50
- (d)55
Answer(b) 45
practice — not a real PYQ
How many three-digit numbers have all three digits the same ?
- (a)9
- (b)10
- (c)81
- (d)90
Answer(a) 9

Why