Problem 126

Solution 1:

The sum of the original five digits is $5+2+8+7+9=31$.

The largest multiple of $4$ that is
less than $31$ is $28$, and $28$ is $3$ less than $31$.

However, $3$ is not among the list of
five digits, and thus Ruhab cannot erase a $3$.

The next largest multiple of $4$ that
is less than $31$ is $24$.

Since $31-24=7$, and $7$ does appear in the original list of
five digits, then if Ruhab erases the $7$, the sum of the remaining four digits
is a multiple of $4$.

(We may confirm that $5+2+8+9=24$.)

Solution 2:

We may erase each of the five digits one at a time and in each case,
determine the sum of the remaining four digits.

Doing so, we get $$5+2+8+7=22; \ 5+2+8+9=24; \ 5+2+7+9=23;$$ $$5+8+7+9=29; \ 2+8+7+9=26$$ Of these sums, only $24$ is a multiple of $4$ and so Ruhab erased the digit $7$.