\(n\) candies with \(20 < n < 50\)

If Marge divides the candies equally among \(5\) children, she will have \(2\) candies remaining.

so \(n = 5x + 2\) so candies can be \(22, 27, 32, 37, 42, 47\)

If she divides the candies among 6 children, she will have 1 candy remaining.

so \(n = 6y + 1\) so candies can be \(25, 31, 37, 43, 49\)

The common value among both is \(37\)

\(n = 37\)

If she divides among \(7\) children, each will get \(5\) candies with \(2\) candies remaining.

Answer C