Question

1. The smallest 3-digit number
divisible by 3 as well as 5. *
*
Your answer​

Answer 1

Answer:

Assuming that you meant integers evenly divisible by 3 and 5 (without a remainder),

3 and 5 are prime numbers so any number evenly divisible by both is evenly divisible by their product, 15.

For three digit integers, that would be all multiples of 15 from 7 x 15 to 66 x 15, or 60 multiples of 15 from 105 to 990.

On the other hand, if you are considering any real number with three digits and allowing remainders as a literal interpretation of your question suggests, then an infinite collection of real numbers from -999 to -100, -9.99 to -9.91, -9.89 to -8.81, etc, -.001 to - .999, +001 to + .999, and +100 to 999 fulfills the requirement.

Answer 2

120

is your answer

120 ÷ 3 = 40

120 ÷ 5 = 24

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