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Cryptarithmetic Multiplication 15

Multiplication_Problem_15

1. Value of W ? 
(a) 6 (b) 7 (c) 8 (d) 9
2. Value of 2F + P ? 
(a) 12 (b) 13 (c) 14 (d) 15
3. Value of 3G + F 
(a) 20 (b) 21 (c) 22 (d) 23
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5 comments

  1. shehzad

    take ques 3. and we know that according to ques F+1=G.
    i.e. F and G are consecutive numbers.
    now check optns in ques 3 by taking any two consecutive no. only option (d) satisfy the condition which means G=6 and F=5
    now the question is just a piece of cake

  2. vikas solanki

    FOR PRACT. PURPOSE AND IMPROVE YOUR CALCULATION TRY NOT TO LOOK FOR OPTION JUST USE HIT AND TRIAL
    HERE THERE IS NO CLUE BUT WE CAN GUESS VALUE OF G >=6
    WHY??
    ANS::

    G +G =D AND LEAVING A REMAINDER FOR F… SO G+G SHOLUD BE 8 WITH CARRY FROM PREVIOUS AS 2..
    G+G SHOULD BE 10…MEANS G =5 BUT THIS IS NOT POSSIBLE AS IN MULTIPLICATION G*D=I THEN I WILL BE ZERO 0R 5 ..BUT I CANT BE ZERO(CRYPT RULE )AND CANT BE 5 SO G SHOULD BE GRATER THAN OR EQUAL TO 6
    POSSIBLE VALUES ARE {6,7,8,9}

    CHECK W P D
    *K
    _________
    G G Z M 4TH PLACE WILL BE >=6 I.E. G AND IT CAN ONLY POSSIBLE IF K>=7
    ————- SO POSSIBLE VALUES OF K{7,8,9}

    NOW YOU CAN DO HIT AND TRIAL

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