2015 AMC 10B Problem 3

Call the number written twice $a$ and the number written three times $b$. The five numbers add to $100$, so
$$2a + 3b = 100.$$
We are told one of the numbers is $28$, but not which one. So try both.

1. Suppose $a = 28$. Then $2 \cdot 28 = 56$, so $3b = 100 - 56 = 44$ and $b = \dfrac{44}{3}$. That is not a whole number, but the problem says both numbers are integers. So this case is impossible.
2. Suppose $b = 28$. Then $3 \cdot 28 = 84$, so $2a = 100 - 84 = 16$ and $a = 8$. This is a whole number.
3. Check: $8 + 8 + 28 + 28 + 28 = 100$. It works.

The other number is $8$.

The answer is $\boxed{\textbf{(A)}\ 8}$.

The problem hides which role $28$ plays, so the safe move is to test both roles. The word "integer" is what rules one of them out, because it gives a fraction. When a word problem says the numbers are integers, expect that to throw out a case.

Let $a$ be the integer written twice and $b$ the integer written three times, so $2a + 3b = 100$. The number $28$ is either $a$ or $b$.

If $a = 28$: then $3b = 100 - 56 = 44$, so $b = \tfrac{44}{3}$, which is not an integer. Rejected.

If $b = 28$: then $2a = 100 - 84 = 16$, so $a = 8$. This is an integer, and $2 \cdot 8 + 3 \cdot 28 = 16 + 84 = 100$ checks.

The answer is $\boxed{\textbf{(A)}\ 8}$.