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Math Help - 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.
    0\times{0}=0. This works, but evidently isn't a possibility as multiplying anything by 0 will give 0.
    1\times{1}=1 This works, but again isn't a posibility for Q due to it being too small.
    2\times{2}=4 This doesn't work.
    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:

    (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:

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

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

    Plug in values of R until 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 P,Q,R are different digits such that:

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

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

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

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

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

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


    If Q = 5, we have:

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

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


    Hence: Q = 6 and we have:

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

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

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


    Therefore, P = 7.

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

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

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