If the LCM of two natural numbers A and B is 1260, what is the minimum…

For two natural numbers A and B, A × B = LCM(A, B) × HCF(A, B). When HCF(A, B) = 1 (A and B are coprime), this reduces to A × B = LCM(A, B). For a fixed product, the sum of two positive factors is smallest when the two factors are as close to each other as possible — that is, nearest to the square root of the product.

Applying this to LCM(A, B) = 1260:

1. Factorize 1260 into primes: 1260 = 22 × 32 × 5 × 7.

2. Since A and B must be coprime (HCF(A, B) = 1) for their product to equal exactly 1260, each of the four prime-power blocks — 2², 3², 5, and 7 — must go entirely to one side or the other. If, say, one factor of 2 went to A and another factor of 2 went to B (splitting 2² between them), both A and B would then share that factor of 2, making HCF(A, B) ≥ 2, so HCF(A, B) > 1. Then A × B = LCM(A, B) × HCF(A, B) would exceed 1260, contradicting A × B = 1260. So a whole prime-power block cannot be split between A and B — it must go entirely to one side.

3. List every way of distributing the four blocks between A and B, and note each pair's sum (table below).

4. Target √1260 ≈ 35.5: the distribution whose two resulting numbers are closest to this value gives the smallest sum.

The pair (35, 36) — consecutive integers, so automatically coprime — sits closest to √1260 and gives the smallest total in the table. Check: GCD(35, 36) = 1 and 35 × 36 = 1260, so LCM(35, 36) = 1260, exactly as required. Every other coprime split has a wider gap between its two factors and so a strictly larger sum; a non-coprime pair would need a product greater than 1260 to still reach an LCM of 1260, which only pushes the sum higher.

So the minimum possible value of A + B is 71.