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Cryptarithmetic Multiplication 29


1. Which of the following set contains even numbers only ? 
(a) M, S, H (b) Z, S, X (c) Z, M ,U (d) U, E, R
2. Which of the following set contains odd numbers only ? 
(a) Z, M, U (b) S, K, R (c) R, K, H (d) Z, U, X
3. Which of the following set contains prime numbers only ?
(a) Z, S, U (b) U, R, H (c) E, H, X (d) E, U, M
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  1. Salik

    there is no clue…
    split into 3 part 3*1 matrix.
    then hit and trial for S={0-9}
    case 1:
    S=0,then U=even,M=5;
    case 2:
    s=1, U,M={3,7 or 7,3}
    s=2, U,M={3,4},{4,3}………..{8,9},{9,8}
    it will satisfy at S=2, and U=3,M=4……
    7 4 3
    6 7 4
    2 9 7 2
    5 2 0 1
    4 4 5 8
    5 0 0 7 8 2

  2. Ashish Shende

    In this problem
    one clue

    if m=4
    m+1=x=5 put the value then observe and find best option

    put z=7
    E=6 e*z=mm 6*7=42

    7 4 3
    6 7 4
    2 9 7 2
    5 2 0 1
    4 4 5 8
    5 0 0 7 8 2

  3. shehzad

    actually there is a clue, X+M is generating a carry so possible values of M & X can only be (4,5) or (5,6) or (6,7) as X cannot be greater than 7(guess why..?)
    now for set (5,6) M cannot be 5 as it can be seen in the multiplication which leaves only 2 possible values of M and X
    either M=4 & X=5
    or M=6 & X=7
    now use hit and trial

  4. Arvind

    One can answer atleast 2 questions from options only.
    Let’s consider 3rd question where options a) b) d) has U common so Let’s assume ‘U’ is prime number( 2 or odd number-3,7).
    Case 1- if we take U => 2 then, in question 1 option c) and d) must be correct so now consider option C) the probable values for ‘M’ could be 4,6,8 thus (M*M+1) will always give an odd value for Z but Z is even according to our assumption. Therefore option c) is incorrect
    and let’s take option d) and it will be cancelled out.

    Case 2- if ‘U’ is odd i.e either 3 or 7, thus this leads to another conclusion that ‘U’ is odd now we will check question 2 and option a) and d) are correct becoz they contain U as well as Z in common. So Z is again odd that is sure now. Now again move to question 1 and as Z is odd so it can’t be even hence option b) is also wrong, so the only option left is a). and now coming to question 2 back we can conclude that option d) is correct.

    So it’s a kindof calculated risk but then it finally works.

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