How many three-digit numbers are possible such that the difference between the original number and the number obtained by reversing the digits is 396? (no digit is repeated)
- (a)4
- (b)5
- (c)50
- (d)40
Correct — D, 40. Write the number as 100a + 10b + c. Reversing it gives 100c + 10b + a, and the difference is 99(a − c). Setting 99(a − c) = 396 gives a − c = 4, so the middle digit b drops out entirely and never affects the difference. Now count. Since the reversal is itself described as a number obtained by reversing the digits of a three-digit number, c cannot be 0, so the possible pairs are (a, c) = (5,1), (6,2), (7,3), (8,4) and (9,5) — five of them. For each, b may be any digit from 0 to 9 except the two already used, which leaves 8 choices. The total is 5 × 8 = 40.
- (a)4 — This is the value of a − c, not a count of numbers. It is the answer to the first line of the working mistaken for the answer to the question.
- (b)5 — The number of valid (a, c) pairs, with the middle digit forgotten. Each of those five pairs still has eight admissible middle digits.
- (c)50 — Five pairs times ten middle digits — the count you get by ignoring the bracketed instruction that no digit is repeated. Dropping the two digits already used takes 10 down to 8 and 50 down to 40.
Every three-digit number and its reversal differ by a multiple of 99, because (100a + 10b + c) − (100c + 10b + a) = 99(a − c). The tens digit cancels, so the difference depends only on the gap between the first and last digits. Similarly, the sum of a three-digit number and its reversal is always 101(a + c) + 20b, and the difference of a two-digit number and its reversal is 9 times the digit gap.
The option set is a map of the working, and it is worth reading it that way. Option (a) is a − c, option (b) is the pair count, option (c) is the pair count times ten, and only option (d) applies the no-repeat restriction. There is one more thing the options tell you. Had the units digit been allowed to be 0, there would be six pairs — (4,0) as well — and the answer would have been 48, which is not printed. The absence of 48 confirms that the reversal is meant to remain a three-digit number, which is how the examiner signals the intended reading without spelling it out. Note also that a is automatically at least 5 once c is at least 1 and a = c + 4, so the leading-digit condition needs no separate check.
- For a three-digit number, original − reversed = 99(a − c), where a and c are the hundreds and units digits.
- Difference 396 ⟹ 99(a − c) = 396 ⟹ a − c = 4; the middle digit is irrelevant to the difference.
- With the reversal required to be a three-digit number, c ≥ 1 and the pairs are (5,1), (6,2), (7,3), (8,4), (9,5) — five in all.
- No digit repeated leaves 8 admissible middle digits per pair, so the count is 5 × 8 = 40.
- Allowing c = 0 would add the pair (4,0) and give 48, which is not an option — confirming the intended reading.
Each wrong option is one of these lines stopped too early.
- Answering with the digit gap or the pair count instead of the number of three-digit numbers.
- Ignoring 'no digit is repeated' and answering 50.
- Allowing the units digit to be zero, which would make the reversal a two-digit number.
Asked as a counting item disguised as an arithmetic one — the algebra takes one line and the marks are in the enumeration.
How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9 such that the digits are in ascending order?
- (a) 80
- (b) 81
- (c) 83
- (d) 84
Answer(d) 84
The same family — count the three-digit numbers meeting a digit condition. There the condition fixes the order, so each choice of three digits yields exactly one number and the count is simply 9C3 = 84; here the condition fixes a difference and the free middle digit has to be counted separately.
The value at the tenth place of a three-digit number is twice the value at the unit's place. The value at the hundredth place is 75% of the value at the unit's place. How many such numbers are possible?
- (a) 0
- (b) 1
- (c) 2
- (d) 3
Answer(c) 2
CAPF returned to the same shape two papers later: relations between the three digits of a number, then a count of how many numbers satisfy them. Setting the digits out as separate variables and testing each admissible value is the method both items reward.
- practice — not a real PYQ
The difference between a two-digit number and the number obtained by reversing its digits is 45. The difference between the two digits is
- (a)3
- (b)4
- (c)5
- (d)9
Answer(c) 5 — the difference is always 9 times the digit gap, and 45 ÷ 9 = 5.
- practice — not a real PYQ
How many three-digit numbers with no repeated digit are such that the number exceeds its reversal by 297?
- (a)5
- (b)6
- (c)48
- (d)56
Answer(d) 56 — 99(a − c) = 297 gives a − c = 3, so c runs from 1 to 6 giving 7 pairs, each with 8 middle digits: 7 × 8 = 56.