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ElitmusZone » 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

1. ElitmusZone Post author

Hello Naresh,

full solve paper???

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

1. vikash

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.