Cryparithmetic Division Problem 6
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A C Q S
_____________
Q B A |B Z Q A R C
X P C
-------
S Y B A
S Z Y P
-------
S A C R
S S X Y
----------
Y C C
Q B A
-------
A Z A
Step 1
The problem can be broken into –
Eq 1
Q B A
x A
-------
X P C
and
Eq 2
Q B A
x C
-------
S Z Y P
Eq 3
Q B A
x Q
-------
S S X Y
Eq 4
Q B A
x S
-------
Q B A
A C Q 1
_____________
Q B A |B Z Q A R C
X P C
-------
1 Y B A
1 Z Y P
-------
1 A C R
1 1 X Y
----------
Y C C
Q B A
-------
A Z A
Values:
S = 1, A = 2
Step 2
Eq 4
Q B A x S ------- Q B A
If we look at equation 4 it is clear that S = 1.
Also,
1 A C R 1 1 X Y -------- Y C
A – 1 = 0
Now, this is only possible if A = 2
- If step C – X is borrowing 1 from A.
- A was 2 and 1 borrowed so A becomes 1
2 C Q 1
_____________
Q B 2 |B Z Q 2 R C
X P C
-------
1 Y B 2
1 Z Y P
-------
1 2 C R
1 1 X Y
----------
Y C C
Q B 2
-------
2 Z 2
Values:
S = 1, A = 2, C = 4
Step 3
Eq 4
Q B A x A ------- X P C
- A x A = C
- Thus C = 2 x 2 = 4
Also, from following subtraction –
1 Y B 2
- 1 Z Y P
-------
1 2 C
After replacing C = 4
1 Y B 2
- 1 Z Y P
-------
1 2 4
The only way 4 can be obtained is –
- If, P = 8
- 2 takes borrow from B and 12 – 8 = 4
2 4 Q 1
_____________
Q B 2 |B Z Q 2 R 4
X 8 4
-------
1 Y B 2
1 Z Y 8
-------
1 2 4 R
1 1 X Y
----------
Y 4 4
Q B 2
-------
2 Z 2
Values:
S = 1, A = 2, C = 4, P = 8
Step 4
Eq 4
Q B 2 x 2 ------- X 8 4
- Now, B x 2 = _8
- This, is only possible when
- B = 2 (Not possible A = 2)
- B = 9
- 9 x 2 = 8 (1 carry to next step)
Also, Q x 2 + 1 (carry from previous step)= X
- Thus, resultant of above is single digit
- Possible values for Q to make this happen
Q = {1, 2, 3, 4}
- Values, 1, 2 and 4 are already taken by S, A and C respectively
- Thus, Q = 3
Also, X = (Q x 2) + 1 carry = 7
X = 7
2 4 3 1
_____________
3 9 2 |9 Z 3 2 R 4
7 8 4
-------
1 Y 9 2
1 Z Y 8
-------
1 2 4 R
1 1 7 Y
----------
Y 4 4
3 9 2
-------
2 Z 2
Values:
S = 1, A = 2, Q = 3, C = 4, X = 7, P = 8, B = 9
Step 5
Eq 4
3 9 2 x 4 ------- 1 Z Y 8
- Clearly, Y = 9 x 4 = 6 (3 carry to next step
- Y = 6
- Also, Z = 3 x 4 + 3(carry) = 12 + 3 = 15
- Z = 5
Finally, step
1 2 4 R
1 1 7 6
----------
6 4
Here,
- R – 6 = 4
- Only possible when R takes 1 borrow from 4
- (1R) – 6 = 4
- Thus, R = 0
2 4 3 1
_____________
3 9 2 |9 5 3 2 0 4
7 8 4
-------
1 6 9 2
1 5 6 8
-------
1 2 4 0
1 1 7 6
----------
6 4 4
3 9 2
------
2 5 2
Final Step
These are the final values for equation.
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