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Cryptarithmetic Problem 7

Cryptarithmetic_Problem_7

1. Which of the following forms the right angled triangle ? 
(a) S, G, Q (b) S, G, P (c) Q, P, A (d) R, S, C
2. Value of C ? 
(a) 5 (b) 6 (c) 7 (d) 8
3. Value of 2Q + D ? 
(a) 15 (b) 16 (c) 17 (d) 18

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14 comments

    1. ElitmusZone Post author

      Hello Naresh,
      We didn’t understood your requirements.

      full solve paper???

      Please elaborate your comments.
      Thanks for co-operations

  1. saumya

    Here in this ques while predicting the value of D how can we say that carry over Q will be 1 only … It can be 2 or 3…

    1. aishwary

      take A as maximum ,let say 9 ….it will pass 2 to the next line…now you take any value of P and G between (1 to 8) it will never pass more than 1 carry to the next step….

    1. Aman

      @Varun
      If u take D=5 then multiplying it by S would give C…C has to be either be 0 or 5 (anythng multiplied by 5 gives 0 or 5)..C cnnot be 5 as D is already 5…and it cannot be 0 also as R=0 already..
      So we cnnot take D=5.

    2. gaurav kumar

      look in multiplication in the last Q+1=D
      look in 3rd question 2Q+D we have to solve….
      2Q+D= 2Q+Q+1=3Q+1
      no check options 3Q+1 will be 16 only so Q will be 5 and D will be 6

  2. bhawinee

    i dont understand a question asked in cryptairthmetic, even after finding out all the values, i am not able to answer this question.

    The que is : Which of the following forms the right angled triangle ?
    (a) S, G, Q (b) S, G, P (c) Q, P, A (d) R, S, C
    how to answer this problem please elaborate.

    1. vikash

      what will be the answer of this question please explain.

      Which of the following forms the right angled triangle ?
      (a) S, G, Q (b) S, G, P (c) Q, P, A (d) R, S, C

        1. Ravi

          In this problem why Q is fixed by value 5 why it is not considered as either 2,4,6,8

          We can also fix the value of D

  3. Anurag Pal

    You get R = 0 and can deduce that Q = 5. From this, it can be safely concluded that Q + 1 = D. So, D = 6.

    The explanation for this is adding two numbers will never generate a carry more than 1. Take the worst case: if A = 7 the carry over to the next place ( G + P = S) will be maximum 2. Again, assuming the worst we can take ( G + P ) = 9 + 8 = 17. Even a carry of 2 won’t result in a sum greater than 19(meaning maximum carry to the leftmost digit is 1).

    Now we take
    Q + S + A = _R
    or
    5 + S + A = _0
    or
    S + A = _5

    This means S + A = 5 or S + A = 15.
    1. For sum 15, the two options are ( 7 + 8 ) and ( 6 + 9 ) but only 7 + 8 is valid as per our conclusion Q = 5.

    2. For sum 5, the two options are ( 1 + 4 ) and ( 2 + 3 ).

    Solving this by trial and error will give you the values of all letters.

  4. Anurag Pal

    Edited:

    You get R = 0 and can deduce that Q = 5. From this, it can be safely concluded that Q + 1 = D. So, D = 6.

    The explanation for this is adding two numbers will never generate a carry more than 1. Take the worst case: if A = 7 the carry over to the next place ( G + P = S) will be maximum 2. Again, assuming the worst we can take ( G + P ) = 9 + 8 = 17. Even a carry of 2 won’t result in a sum greater than 19(meaning maximum carry to the leftmost place is 1).

    Now we take
    Q + S + A = _R
    or
    5 + S + A = _0
    or
    S + A = _5

    This means S + A = 5 or S + A = 15.
    1. For sum 15, the two options are ( 7 + 8 ) and ( 6 + 9 ) but only 7 + 8 is valid as per our conclusion D = 6.

    2. For sum 5, the two options are ( 1 + 4 ) and ( 2 + 3 ).

    Solving this by trial and error will give you the values of all letters.

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