In the multiplication shown, $P, Q$ and $R$ are all different digits so that

$$
\begin{array}{r}
P P Q \\
\times \quad Q \\
\hline R Q 5 Q
\end{array}
$$

What is the value of $P+Q+R$ ?
(A) 20
(B) 13
(C) 15
(D) 16
(E) 17
First, we try to figure out what $\operatorname{digit} Q$ is.

Since the product is not equal to $0, Q$ cannot be 0 . Since the product has four digits and the top number has three digits, then $Q$ (which is multiplying the top number) must be bigger than 1 .

Looking at the units digits in the product, we see that $Q \times Q$ has a units digit of $Q$.

Since $Q>1$, then $Q$ must equal 5 or 6 (no other digit gives itself as a units digit when multiplied by itself).

But $Q$ cannot be equal to 5 , since if it was, the product $R Q 5 Q$ would end " 55 " and each of the two parts $(P P Q$ and $Q)$ of the product would end with a 5 . This would mean that each of the parts of the product was divisible by 5 , so the product should be divisible by $5 \times 5=25$. But a number ending in 55 is not divisible by 25 .

Therefore, $Q=6$.

So the product now looks like

$$
\begin{array}{r}
P P 6 \\
\times \quad 6 \\
\hline R 656
\end{array}
$$

Now when we start the long multiplication, $6 \times 6$ gives 36 , so we write down 6 and carry a 3 . When we multiply $P \times 6$ and add the carry of 3 , we get a units digit of 5 , so the units digit of
$P \times 6$ should be 2.

For this to be the case, $P=2$ or $P=7$.

We can now try these possibilities: $226 \times 6=1356$ and $776 \times 6=4656$. Only the second ends " 656 " like the product should.

So $P=7$ and $R=4$, and so $P+Q+R=7+6+4=17$.

ANSWER: (E)