CAPF AC 2025 Q58: General Mental Ability – Quantitative Aptitude
- (a) 2
- (b) 3
- (c) 4
- (d) 5
Answer: (d) 5
Explanation
Unit digit of a long sum
Given: the sum 3×1 + 4×2×1 + 5×3×2×1 + … + 20×18×17×…×2×1.
The k-th term is (k + 2) × k! for k = 1 to 18.
How to think: rewrite each term using factorials.
(k + 2) × k! = (k + 1) × k! + k! = (k + 1)! + k!.
| Part | Terms | Unit digit |
|---|---|---|
| Sum of (k + 1)! | 2! + 3! + 4! + … + 19! | 2 + 6 + 24 = 32 from the first three; every factorial from 5! upward ends in 0 |
| Sum of k! | 1! + 2! + 3! + … + 18! | 1 + 2 + 6 + 24 = 33 from the first four; the rest end in 0 |
Total of the surviving unit digits: 32 + 33 = 65, so the unit digit is 5.
Bank key corrected: the stored key was (c) 4; the right answer is (d) 5.
Answer: (d) 5.
Why each option is right or wrong
- (a) Incorrect. The surviving units add to 65, which ends in 5, not 2.
- (b) Incorrect. The unit digit is 5, not 3.
- (c) Incorrect. 4 would come from leaving out a term or adding wrongly.
- (d) CORRECT. The unit digit of the sum is 5.
Key point
(k+2)·k! = (k+1)! + k!. For n ≥ 5, n! ends in 0. Surviving units: (2!+3!+4!) + (1!+2!+3!+4!) = 32 + 33 = 65, unit digit 5.
Remember
Factorials from 5! end in 0, so add only the first few terms: unit digit 5.
Common mistakes
- Computing every term fully instead of using factorial patterns.
- Forgetting that factorials from 5! onwards contribute nothing to the unit digit.
- Leaving out 1! or counting 19! as the last term of the second sum.
Exam tip
Educated guess: only the first few terms matter because later ones end in 0. Adding the first four terms by hand: 3 + 8 + 30 + 144 = 185, which ends in 5, and the later terms add only zeros.
Sources
- NCERT Class XI Mathematics, Chapter 7 (Permutations and Combinations: factorial notation)
- NCERT Class VI Mathematics, Chapter 3 (Playing with Numbers)