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Thread: Multiplication Values

  1. #1
    Junior Member MoniMini's Avatar
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    Post 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!
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  2. #2
    Super Member Quacky's Avatar
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    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.
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  3. #3
    Super Member
    earboth's Avatar
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    Re: Multiplication Values

    Quote Originally Posted by MoniMini View Post
    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 $\displaystyle P \in \{1,2,...,9\}$
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  4. #4
    Super Member

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

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

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