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Cryptarithmetic Multiplication 12

Cryptarithmetic-Multiplication_12

1. Value of B ?
(a) 2 (b) 4 (c) 6 (d) 8
2. Value of W/2 ?
(a) 1 (b) 2 (c) 3 (d) 4
3. Sum of E + S + D + D ?
(a) 16 (b) 17 (c) 18 (d) 19

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8 comments

  1. vikas solanki

    proceed without options to learn better
    just hint

    AS D+A=D ,IT MEASN A=0
    E * W=A MEANS
    A=MUST BE A PART OF 10,20,30,0R 40
    10=2*5
    20=4*5
    30=6*5
    40=8*5
    MEANS ONE OF THEM MUST BE 5 AND OTHER WILL BE AN EVEN NUMBER
    IF W =5 THEN
    FROM MULTIPLICATION B P W
    * N
    ______
    D ….D MUST BE 0 OR 5…D CANT BE ZERO AS A IS ZERO AND D CANT BE 5 ALSO
    SO W IS NOT 5 ….E WILL BE 5…AND POSSIBLE VALUES OF W{2,4,6,8}

    1. Soumya Chowdhury

      D+E+M=D where E=5.
      Now for this equation we can have two things D+5+3+carry2 or D+5+4+carry1.
      In these two ways D+E+M. Assuming the maximum value of D=9,
      then D+E+M=9+5+3+caary2=19 or D+E+M=9+5+4+caary1=19. SO 19 is the maximum possible value and carry is 1 hence

  2. Soumya Chowdhury

    D+A=D which means A=0
    ANything multiplied by 5 gives 0 or 5. Since A=0 then W*E=0. Either of E and W is 5 and (2,4,6,8)
    Now W*N=D and W*M=B. Therefore W cant be 5. So hence E=5 and W=2,4,6,8

    Now 5+D+M=D. HEre there can be two cases 5+D+M+carry1 or 5+D+M+carry2. Assuiming the max value of D=9, two cases are possible:
    5+9(D)+4(M)+carry1 or 5+9(D)+3(M)+carry2. The result in both case is 19 and hence carry is 1. So P+5+carry1=0

    B 4 W
    x 5
    4 D M O
    In this case B= 8 or 9 and W= 2,6,8 (4 is already assigned to P)
    Now if B=9 and W=2,6 or 8, the values of D=7 in all three values of W. But D=W*N where W=2,6,8. Multiplication of any number by 2,6,8 can give 7. HEnce B cant be 9. So b isdefintely 8.

    Now B=8 and W=2,6. Now if W=2 then D also becomes 2. Hence W and B cant be 2 at the same time. So value of W is definitely 6

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