# Thread: Multiplication Values

1. ## Multiplication Values

I dunno what to do about this.....
Please Help!

In a multiplication shown here, P, Q and R are all different digits such that
P P Q
x Q
--------------
R Q 5Q
What is the value of P + Q + R?
Options are
a) 20
b) 13
c) 15
d) 17

Any help will be acknowledged!

Thanks a lot!

2. ## Re: Multiplication Values

Look at the units column. Which units exist that, when multiplied by themselves, give a number ending in the same digit?
E.g.
$\displaystyle 0\times{0}=0$. This works, but evidently isn't a possibility as multiplying anything by $\displaystyle 0$ will give $\displaystyle 0$.
$\displaystyle 1\times{1}=1$ This works, but again isn't a posibility for Q due to it being too small.
$\displaystyle 2\times{2}=4$ This doesn't work.
$\displaystyle 3\times{3}=9$ Again, this doesn't fit.

Keep going along this route. The possibilities for Q vanish quickly, and then consider the possibilities for P given the options for Q.

3. ## Re: Multiplication Values

Originally Posted by MoniMini
I dunno what to do about this.....
Please Help!

Options are
a) 20
b) 13
c) 15
d) 17

Any help will be acknowledged!

Thanks a lot!
This calculation can be written as:

$\displaystyle (100P+10P+Q) \cdot Q = 1000R+100Q+50+Q$

You'll get the same last digit as the multiplicator if it is

• 0 - but then the result must be 0 too
• 1 - but then the result must be the multiplicand
• 6

So the multiplication becomes:

$\displaystyle (100P+10P+6) \cdot 6 = 1000R+600+50+6$

$\displaystyle P = \frac{50}{33}R+\frac{31}{33}, ~~ R \in \{0,1,...,9\}$

Plug in values of R until $\displaystyle P \in \{1,2,...,9\}$

4. ## Re: Multiplication Values

Hello, MoniMini!

Most "alphametics" can be solved without Algebra.

In this multiplication $\displaystyle P,Q,R$ are different digits such that:

. . $\displaystyle \begin{array}{cccc}_1 & _2 & _3 & _4 \\ & P & P & Q \\ & \times && Q \\ \hline R & Q & 5 & Q \end{array}$

What is the value of $\displaystyle P+Q+R\,?$

. . $\displaystyle (a)\;20 \qquad (b)\;13 \qquad (c)\;15 \qquad (d)\;17$

In column-4,$\displaystyle Q\!\cdot\!Q$ ends in $\displaystyle Q$ . . . which is written:.$\displaystyle Q\!\cdot\!Q \to Q$

The choices for $\displaystyle Q$ are: $\displaystyle 0, 1, 5, 6.$

We can see that $\displaystyle Q \ne 0$ and $\displaystyle Q \ne1.$

If $\displaystyle Q = 5$, we have:

. . $\displaystyle \begin{array}{cccc}_1 & _2 & _3 & _4 \\ & P & P & 5 \\ & \times && 5 \\ \hline R & 5 & 5 & 5 \end{array}$

In column-3, we have: $\displaystyle 5\!\cdot\!P + 2 \to 5$ . . . which is impossible.

Hence: $\displaystyle Q = 6$ and we have:

. . $\displaystyle \begin{array}{cccc}_1 & _2 & _3 & _4 \\ & P & P & 6 \\ & \times && 6 \\ \hline R & 6 & 5 & 6 \end{array}$

In column-3, we have: $\displaystyle 6\!\cdot\!P + 3 \to 5$
. . Hence: $\displaystyle P\,=\,2\text{ or }7.$

If $\displaystyle P = 2$, we have:.$\displaystyle 226 \times 6 \,=\,1356$
. . and the product does not satisfy the pattern $\displaystyle R\,6\,5\,6.$

Therefore, $\displaystyle P = 7.$

Solution: .$\displaystyle \begin{array}{cccc}& 7 & 7 & 6 \\ & \times && 6 \\ \hline 4 & 6 & 5 & 6 \end{array}$

. . $\displaystyle P + Q + R \;=\;7+6+4 \;=\;17\;\;\text{ answer (d)}$