A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?
Answer & explanation
Answer: (c) 294
If A gets k pens, the others get k/2, k/3, k/4, k/5 and k/6, and each must be even. So k must be a multiple of 4, 6, 8, 10 and 12; the least such k is 120, giving 120 + 60 + 40 + 30 + 24 + 20 = 294 pens.
- Let A get k pens. Then B, C, D, E and F get k/2, k/3, k/4, k/5 and k/6.
- Each share must be an even whole number, so k must be divisible by 4, 6, 8, 10 and 12.
- LCM(4, 6, 8, 10, 12) = 120, so the least k is 120.
- Shares: 120, 60, 40, 30, 24, 20 — all even. Total = 294.
- Check: k = 60 would give 147 pens, but then D gets 15, an odd number, so 147 fails.
Remember · ‘k/n must be even’ means k is a multiple of 2n; take the LCM of all such 2n.
Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·