Question

All possible two-factor products are formed from the numbers 1, 2, 3, …,100. How many numbers out of the total obtained are multiples of 3?

Answer 1

The total number of two-factor products = ¹⁰⁰C₂

Out of the numbers 1, 2, 3, …, 100; the multiples of 3 are 3, 6, 9, …, 99; that is, there are 33 multiples of 3, and therefore there are 67 non-multiples of 3.

Therefore, the number of two-factor products which are not multiples of 3 = ⁶⁷C₂

The required number = ¹⁰⁰C₂ - ⁶⁷C₂

= 4950 – 2211 = 2739 Alternatively, the number of two-factor products formed when both factors are multiples of 3 = ³³C₂ and the number of two-factor products formed when one is a multiple of 3 and the other a non-multiple of 3 = ³³C₁ x ⁶⁷C₁

It either case the product is a multiple of 3. Therefore, the required number = 528 + 2211 = 2739