A divisibility that always holds — Aptitude Question
A divisibility that always holds: For every natural number n, which is the largest of these that always divides 6n²… Answer and worked solution on CodeKairo.
- Topic: Number System
- Section: Quantitative Aptitude
- Difficulty: Medium
- Time target: 75 seconds
Question
For every natural number n, which is the largest of these that always divides 6n² + 6n?
Options
- 6
- 12
- 18
- 24
Show the answer and the worked solution
Correct answer: B. 12
Worked solution
- 6n² + 6n = 6n(n + 1).
- n and n + 1 are consecutive, so one of them is even and n(n + 1) is always even.
- Therefore 6n(n + 1) is always a multiple of 6 × 2 = 12 — 6 divides it too, but 12 is the larger guarantee.
- Check n = 1: 12; n = 2: 36; n = 3: 72 — all multiples of 12, and 36 is a multiple of neither 18 nor 24, ruling both out. ✓
Answer: 12.
The idea
Factorise, then use the fact that a product of consecutive integers carries guaranteed factors: two consecutive always give a 2, three consecutive always give a 6. Testing one or two small values rules out the larger options.
Shortcut. n = 1 gives 12, so any number that always divides the expression must divide 12 — 18 and 24 are out. 12 is the fuller guarantee, since n(n + 1) is always even.
Common trap. 24 — n = 3 and n = 4 give 72 and 120, both multiples of 24, but n = 1 gives 12 and n = 2 gives 36, neither of which 24 divides.