CAPF AC 2021 Q78: General Mental Ability – Quantitative Aptitude
The smallest number which when added to 10000 becomes divisible by 20, 24 and 30 is
- (a) 120
- (b) 800
- (c) 80
- (d) 83
Answer: (c) 80
Explanation
Smallest number to add for divisibility
Given: find the smallest number which when added to 10000 makes it divisible by 20, 24 and 30.
How to think: a number divisible by all three must be divisible by their LCM.
Find the LCM, then see how far 10000 is from the next multiple.
| Step | Working |
|---|---|
| Prime factors | 20 = 2² × 5; 24 = 2³ × 3; 30 = 2 × 3 × 5 |
| LCM | 2³ × 3 × 5 = 120 |
| Divide | 10000 = 120 × 83 + 40 |
| Number to add | 120 − 40 = 80 |
Check: 10000 + 80 = 10080 = 120 × 84.
Answer: (c) 80.
Why each option is right or wrong
- (a) Incorrect. 120 is the LCM, not the number to be added.
- (b) Incorrect. 800 also works (10800 = 120 × 90) but it is not the smallest.
- (c) CORRECT. The remainder is 40, so 120 − 40 = 80 must be added.
- (d) Incorrect. 83 is the quotient when 10000 is divided by 120.
Key point
Smallest number to add = divisor − remainder. Smallest number to subtract = remainder. For several divisors, use their LCM first.
Remember
Number to add = LCM − remainder (120 − 40 = 80).
Common mistakes
- Marking the remainder (40) or the LCM (120) instead of 120 − 40.
- Choosing 800, which works but is not the smallest such number.
- Using the HCF of 20, 24 and 30 where the LCM is needed.
Exam tip
Educated guess: 10000 + answer must end in a zero and be divisible by 3. 10080 has digit sum 9, so it passes; 10120 has digit sum 4, so it fails; 10083 is odd.
Sources
- NCERT Class VI Mathematics — Playing with Numbers (LCM and divisibility)
- NCERT Class X Mathematics — Real Numbers