A set of marbles can be divided in equal shares among $2$, $3$, $4$, $5$, or $6$ children with no marbles left over. What is the least number of marbles that the set could have?

  60

Step-by-step explanation:

The least common multiple of 2, 3, 4, 5, and 6 is 3×4×5 = 60.

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2, 3, 5 are primes, there will be at least one factor of each of those in number you're looking for.

4 = 2², so there must be 2 factors of 2 in the number you're looking for.

6 = 2×3, so is already covered by the factors we have listed so far.

  LCM = 2²×3×5 = 60

The least number divisible by numbers 2 through 6 is 60.

1. d.90

2. e. none of these (80)

3. a. 0

4. b. 6

5. e. none of these (10)

6. a. 5