Question 622939
I actually competed at the National MATHCOUNTS competition, back in 2007 and 2008 when they were held in Ft. Worth and Denver (didn't come close to the countdown round though...).


Note that the number of pennies determines uniquely the value, modulo 5. Assume we have no pennies. Using nickels and dimes, we can make the values 25, 30, ..., 50 cents (six values).


If we have one penny, we can make 21, 26, ..., 41 cents (five values).


If we have two pennies, we can make 17, 22, 27, 32 cents (four values).


It turns out that as we add one more penny, the number of possible values decreases by 1 (because we have one fewer dime/nickel). Therefore the answer is 6+5+4+3+2+1 = 21.