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

User Shibankar
by
5.1k points

1 Answer

0 votes

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

User Gerrit Beuze
by
4.6k points
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