Problem 658

Let $A$, $B$, $C$, and $D$ be the number of coins that Amr, Bai,
Cindy, and Derek received, respectively. Then $A+B+C+D=N$. The three conditions given on
how the coins were distributed correspond to the equations $$\begin{align*} A &= \dfrac{1}{3}(B+C+D) \\ B &= \dfrac{1}{5}(A+C+D) \\ C &= \dfrac{1}{7}(A+B+D)\end{align*}$$ Scaling these three
equations by $3$, $5$, and $7$, respectively, we get $$\begin{align*} 3A &= B+C+D \tag{1} \\ 5B &= A+C+D \tag{2} \\ 7C &= A+B+D \tag{3}\end{align*}$$ Subtracting Equation $(1)$ from Equation $(2)$ gives $5B-3A = (A+C+D) - (B+C+D)$ which can be
simplified to $6B=4A$ or $B=\dfrac{2}{3}A$.

Subtracting Equation $(1)$ from
Equation $(3)$ gives $7C-3A = (A+B+D) - (B+C+D)$ which can be
simplified to $8C=4A$ or $C=\dfrac{1}{2}A$.

Substituting $B=\dfrac{2}{3}A$ and
$C=\dfrac{1}{2}A$ into Equation $(1)$ gives $3A
= $\dfrac{2}{3}A+\dfrac{1}{2}A+D$$, or$$D=\dfrac{11}{6}A\$.$

We have now expressed each of $B$,
$C$, and $D$ in terms of $A$, so we will substitute these
expressions into the equation $A+B+C+D=N$ to get $A+\dfrac{2}{3}A+\dfrac{1}{2}A+\dfrac{11}{6}A=N$.

Simplifying the left side, we have $4A=N$. Since $A$ and $N$ are integers, this tells us that $N$ must be a multiple of $4$.

The largest multiple of $4$ that is
less than $100$ is $N=96$, so we guess that this is the
answer.

If $N=96$, then $4A=N$ implies $A=24$, and using the equations for the
other three variables in terms of $A$, we get $B=16$, $C=12$, and $D=44$. One can check that these four
integers satisfy the conditions given in the problem.