What is the remainder when the product of the 5 smallest prime numbers is divided by 42?
1 answer:
Answer:
21
Step-by-step explanation:
The 5 smallest prime numbers are 1, 2, 3, 5, and 7.
So when multiplied it equals 105.
Divide by 42 and you get 2 21/42
So you have a remainder of 21
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