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Cryptarithmetic Problem 9

Cryptarithmetic_Problem_9

1. Value of  2B + M ?
(a) 5 (b) 6 (c) 7 (d) 8
2. Value of  M + A + R + L + E + Y ?
(a) 19  (b) 20 (c) 21 (d) 22
3. Value of Y ?
(a) 6 (b) 7 (c) 8 (d) 9

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2 comments

  1. SAURABH HIRWANI

    heres the short answer:
    After getting .
    Case-I- When O={5} and B={3, 7, 9} Rule 3
    Case-II When B={6} and O={2, 4, 8}.
    Now observe that since O is the unit digit of O*B and B(at the unit place of first number)*O(at tens of second number)= O , i.e. the previous multiplication B*B (both at unit place) gives a digit and not a number , i.e Y is a digit ,i.e B^2 = Y , we get
    Case-III {B,Y}={2,4},{3,9}.
    B=3 satisfy case III as well as one of Case I/II . thus B=3 ,O=5 .rest you can get .

  2. Rayhan

    When you already know the combinations, the why do it in a complex way…!! Idiotic…
    B*O = _ O
    Possibilities :-
    3*5= _,5
    7*5=_,5
    9*5=_,5
    6*2=
    6*4=
    6*8=
    So rather than trying solving you can get answer by checking :-
    353
    353
    ——–
    .
    .
    .
    Now try multiplying the patterns….
    Funny thing is…u’ll be getting the sum done on the first go….
    The guy above said the same thing….this is practical….
    Notice the units tens digit n units digit in the 1st n 2nd line respectively…

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