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 |

```
C G D
x B Q S
A Q S C
G A S R
Q P A A
D S B R S C
As, S + R = S
Therefore, R=0 Detailed Explanation- Rule 1
Put R=0 and write the problem again,
C G D
x B Q S
A Q S C
G A S 0
Q P A A
D S B 0 S C
Here we have one more clue,
C G D
x Q
G A S 0
D x Q = _0
i.e. value of Q=5
Possible ways of getting 0 at Unit Digit
Put Q=5
C G D
x B 5 S
A Q S C
G A S 0
5 P A A
D S B 0 S C
[To get unit digit as 0 you have to multiply 5 with any even number]
[If you multiply 5 to any number, you will only get [0,5] as their last digit.(5*even=_0 and 5*odd=_5)]
Now proceed further
```

1. A 2. D 3. B C=8, G=4, D=6, B=7, Q=5, S=3, A=2, P=9, R=0 8 4 6 x 7 5 3 2 5 3 8 4 2 3 0 5 9 2 2 6 3 7 0 3 8

(The solution has been given considering you as a beginner in Cryptarithmetic)C G D x B Q S A Q S C G A S R Q P A A D S B R S C As, S + R = S Therefore, R=0 Detailed Explanation- Rule 1 Put R=0 and write the problem again, C G D x B Q S A Q S C G A S 0 Q P A A D S B 0 S C Here we have one more clue, C G D x Q G A S 0 D x Q = _0 i.e. value of Q=5 and D={2, 4, 6, 8} Possible ways of getting 0 at Unit Digit Put Q=5 C G D x B 5 S A 5 S C G A S 0 5 P A A D S B 0 S C [To get unit digit as 0 you have to multiply 5 with any even number] [If you multiply 5 to any number, you will only get [0,5] as their last digit.(5*even=_0 and 5*odd=_5)] Further, C G D x B 5 S A 5 S C G A S 0 5 P A A D S B 0 S C 5 + 1 (carry) = D Therefore, value of D=6, now rewrite the problem after replacing the value of D=6 C G 6 x B 5 S A 5 S C G A S 0 5 P A A 6 S B 0 S C At this stage, split the problem in 3 parts for collecting the more clues... 1. C G 6 x S A 5 S C 2. C G 6 x 5 G A S 0 3. C G 6 x B 5 P A A Taking (2) [you can see three variables have been replaced by digits. i.e. you have less number of variable in Case (2)] (2) C G 6 x 5 G A S 0 You have to start hit and trial with the possible values of G Here you can see the possible values of S={3, 8} Explanation : If you multiply any number by 5 then you will only get 0 and 5 at last digit. 0 + 3 (carry) =_3 5 + 3 (carry) =_8 [It will take some time to understand the concept. Please read..... again!] You have start hit and trial with both the possible values of S={3, 8} Firstly take S=3 put in the main problem and rewrite it. i.e. Q=5, S=3, R=0, D=6 C G 6 x B 5 3 A 5 3 C G A 3 0 5 P A A 6 3 B 0 3 C Here you can see value of C=8 put C=8 and rewrite the problem 8 G 6 x B 5 3 A 5 3 8 G A 3 0 5 P A A 6 3 B 0 3 8 Now you can easily solve the problem further G=4, A=2, P=9 8 4 6 x 7 5 3 2 5 3 8 4 2 3 0 5 9 2 2 6 3 7 0 3 8 [Relax - it's going to take some time to understand the whole concept. If you are facing any difficulty. Please go through the Cryptarithmetic Tutorial.]

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pleased send me the full solve paper,,

Hello Naresh,

We didn’t understood your requirements.

full solve paper???

Please elaborate your comments.

Thanks for co-operations

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…

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….

Why did u take only as Q=5 and D={2,4,6,8} ?

we can also take D=5 and Q={2,4,6,8}….

plz reply

@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.

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

Do we have ny test series for e litmus

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.

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

Hi Vikash,

https://en.wikipedia.org/wiki/Pythagorean_theorem

You have to check which one is following the Pythagoras theorem.

In Q1.

G=4, Q=5, S=3

a2 + b2 = c2 (Please read as a square + b square= c square)

Please let us know if more clarifications is required.

Best Regards,

Team ElitmusZone

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

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.

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.