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Cryptarithmetic Problem 3

Cryptarithmetic-Multiplication_3

1. Value of A + S + T + A + R + T ? 
(a) 21 (b) 26 (c) 24 (d) 25
2. Find the value of 4R + T ? 
(a) 19 (b) 20 (c) 21 (d) 22
3. Find the value of 2A + R ? 
(a) 5 (b) 6 (c) 7 (d) 8

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15 comments

  1. nitish

    In hint why A=1 is the only possible value of A.
    A can be 6 as well, then {T,G,O}= {2 ,4 ,8} can take any values in these three,
    Please explain.

  2. Utsav Jain

    Don’t GIve Up!!!!!

    571
    * 326
    ————————–
    3426
    1142
    1713
    ————————–
    186146

  3. charan kumar

    in this probelm your solution directly assumed as a = 1but why you can also take as a = 6 and then so you can assume as t, o,g, can belong to 2,4,8 as a*t = t, a*o = o, a*g= g implies that a can be 6 and so if t may be one among {2,4,8} so that 6 *2= _2 (last digit), 6*4 = _4 last digit, 6*8=_8 last digit then why cant we assume as a = 6 and solve the problem, you mentioned directly that a= 1 is the only alternative.

  4. Harsh Mishra

    Guys, I also got the same thought. I guessed that this could help:
    Note that in the second column from right the addition goes O+O=R i.e., 2O=R. Also if you’ve got A=6 then G, O, T must be in {2,4,8}.
    Now,
    2O = R
    Then possible values of R can be
    2*2=4, or
    2*4=8, or
    2*8=16 (ending with 6)
    which contradicts all the values of G, O, T and A as well.

  5. silambarasan

    you are all correct A={6} T,O,G={2,4,8}

    But if take sum at G+A+A=T

    CASE 1:
    A=6 , G=8
    8+6+6=T (G+A+A=T)
    T=20 (CASE IS WRONG BECAUSE T IS NOT BECOME 0)

    CASE 2:
    A=6, G=4
    4+6+6=T (G+A+A=T)
    T=16 (T IS NOT BECOME 6 BECAUSE WE ALSO TAKE A =6)

    CASE 3:
    A=6 G=2
    2+6+6=T (G+A+A=T)
    T=14 (TRUE T IS BECOME 4)

    WE DECLARE T=4 & G=2 & O=8(BECAUSE 8 ONLY REMAINING IN THAT POSSIBILITIES SHOWN ABOVE)
    THEN,
    WE GO ANOTHER SUM O+O=R
    8+8=16 (R IS NOT BECOMES 6 BECAUSE A=6 WE ALREADY TAKEN)

    SO THEY CANNOT TAKE A=6 IN THIS PROBLEM

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