ElitmusZone » Cryptarithmetic Multiplication 15

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
You might be interested in this article
Cryptarithmetic Problem 16
Cryptarithmetic Problem 14
Cryptarithmetic Tutorial
Cryptarithmetic Multiplication Problem

 

Share this article Share on FacebookShare on Google+Email this to someoneTweet about this on TwitterPin on Pinterest

7 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

  3. Soumya Chowdhury

    W P D * I = K F P P
    Now possibility of getting P=4, we can have two cases
    Case 1: P=4, D=8 and I=3
    Case 2: P=4, D=3 and I=8
    So W 4 8 * 3 = K F 4 4 & W 4 3 * 8 = K F 4 4
    Case 1:
    Now, W 4 8 * 3 , the possible values of K F are 10, 13, 16, 19, 22, 25 & 28.
    Now K can’t be 1 because anything multiplied by 1 will result same. Multiplication of W P D by K don’t give same result. So 10 to 19 are rejected.
    22 cant be as K not equal to F
    Now for 25 value of W ill be 8 which is not possible.
    28 cant be because F and D can’t be same.
    So case 1 is rejected.
    Case 2:
    Now W 4 3 * 8 , the possible values of K F are 11, 19, 27, 35, 43, 51, 59, 67 & 75
    K can’t be 1. So 11 & 19 are rejected.
    For 27 & 35, value of W will be 3 & 4 respectively both of which are assigned to D and P respectively. So 27 & 35 are rejected
    43 cant be as 4 and 3 are already assigned to P and D.
    ANything multiplied by 5 would result in 0 or 5. D * K = M. So if K=5 then M can’t be 5. Now P + M = F and there is no carry from previous side. So M is not 0. Since M is not equal to 0 or 5. So K can’t be 5. So 51 and 59 are rejected.
    For 51, value of W=8. But I=8. So 51 is rejected.
    Now for K F = 7 5, W =9. So it is accepted

Leave a Reply

Your email address will not be published. Required fields are marked *