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Cryptarithmetic Multiplication 24


1. Value of F ? 
(a) 5 (b) 6 (c) 7 (d) 8
2. Value of E + W + T ? 
(a) 15 (b) 16 (c) 17 (d) 18
3. Value of H ? 
(a) 4 (b) 5 (c) 6 (d) 7
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  1. Lokesh Kumar

    First divide into 3 parts by unit digit method
    Take 2 part
    F Y H
    * W
    E K Z Z

    By unit digit method check for
    Z=1 all cases doesn’t satisfy (check yourself)
    Z=2 make all the cases and start finding that cases will satisfy or not
    You will find that H=6 and W=7 is looking appropriate.
    Start solving take the value of T first start and u will see it will take t=3,9 start solving and rest value you can find yourself. I can provide the whole solution but it would be better, if you try yourself.

  2. Rohan Saxena

    From option we can obtain that
    h is not 5 from basic rule thus h=4,6,7
    but f is not 5 because h*t=f
    so f=6,7,8
    now FYH(4,6,7)
    * T
    when you will verify this you get f=8, h=6
    all the value can obtain by this

  3. rohan418

    Since E+Z shoud not give carry,we have possible value of F as 9,8,7,6,5,4,3.
    (1)Lets first consider F=9,and solve.
    ZETF …………..(1)
    the product of two no can only be 9 when we take 3*3=9 and 7*7=49.We cannot take these no in consideration because two letters cannot have same no.
    So Fcannot be 9….
    (2)Lets take F=8,and solve.
    (a)considering first(2*4=8)
    T=2 and H=8
    putting this in equation (1).
    we get,Y=6,Z=1,E=7.
    EKZZ ……………..(2)
    Z=1 while having H=8(we cannot get such combination) is not possible from equation(2).
    same is the case with H=2,T=8.
    (b)Now,considering second(2*9=18)
    putting this in equation (1) we cannot get any soution from equation(you can check).
    (c)Now,lets consider third(3*6=18)
    putting this in equation(1) we get,
    using these results in equation(2) we get….
    using these all solution in the question we get..
    F=8,Y=4,T=3,H=6,Z=2,E=5,W=7,K=9,X=0 ANS.

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