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Cryptarithmetic Multiplication 11

Cryptarithmetic-Multiplication_11

1. Value of H + S + P ? 
(a) 10 (b) 11 (c) 12 (d) 13
2. Which of the following forms a triangle ? 
(a) H, S, P (b) H, T, O (c) D, U, O (d) U, T, P
3. Value of 2D + P ? 
(a) 14 (b) 15 (c) 16 (d) 17

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7 comments

    1. satheesh

      if we take case2 then p=5 and p+p=10 ie O=0 WHICH IS NOT TRUE BCOZ IF O=0 THEN EVERYTHING MULTIPLIED BY ZERO WILL BE ZERO

  1. Shujayat Ali1

    @Sandeep…. actually we are not using here pythagoras theorem to check formation of triangle.

    We are applying here…. (“”in a triangle the sum of two sides must be greater than the third side””)

    a) H=3,S=8,P=2 (not forming triangle because H+P is not greater than S)
    b) H=3,T=5,O=4 (here it is following the rule..in a triangle the sum of two sides must be greater than the
    third side)
    c) D=6,U=1,O=4 (it is not following the rule…in a triangle the sum of two sides must be greater than the third
    side)
    d) U=1,T=5,P=2 ((it is not following the rule…in a triangle the sum of two sides must be greater than the third
    side)

  2. vikas solanki

    we can proceed like this also AND BEST WAY I THINK TO LEARN THESE MULTIPLICATION IS DONOT LOOK FOR OPTION and just follow unit digit method

    as we can easily see U is =1
    and d*p=p
    possible of values for P and D is
    P{5} D{3,7,9,}
    P{2,4,8} D{6}
    WE CAN EASILY REJECT FIRST SET OF VALUES IF YOU WILL SEE
    P+P=0
    AND IF P=5 THEN O SHOULD BEE ZERO. BUT O CANT BE ZERO IF YOU OBSERVE THE MULTIPLIACTION
    SO PROCEED WITH SECOND SET OF VALUES AND HIT ANS TRIAL WITH
    P{2,4,6}
    D{6}

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