What is the remainder when the product of the 5 smallest prime numbers is divided by 42?

Question

asked Feb 24, 2021 141k views

1 Answer

Gerrit Beuze · Answer 1 · 2021-02-28T20:51:27+0000

Answer:

0

Explanation:

The five smallest prime numbers are 2, 3, 5, 7 and 11.

2 × 3 × 5 × 7 × 11

= 2310

Divide by 42.

2310/42

= 55, Remainder 0