eLitmus Cryptarithmetic Problem – 3
Cryptarithmetic Questions asked in eLitmus 3
Ques. For the following Cryptarithmetic find the answers to the below questions?
P O P
M U T
---------
S O I U
G M T U -
U I R O - -
-----------------
U U S M U U
| 1. | Value of G + U + T ? | |||
| (a) 3 | (b) 10 | (c) 6 | (d) 9 | |
| 2. | Value of S + G ? | |||
| (a) 3 | (b) 9 | (c) 5 |
(d) 8 | |
| 3. | Value of 2U ? | |||
| (a) 4 | (b) 6 | (c) 8 | (d) 2 | |
It is highly suggested to go through these before solving
Row 1 P O P Row 2 M U T Row 3 --------- Row 4 S O I U Row 5 G M T U - Row 6 U I R O - - Row 7 ----------------- Row 8 U U S M U U
Step 1
- As I + U = U
- This is only possible when, I = 0
- Also, there can not be a carry from previous step as
- Max values of U can be 9. Thus, no carry over.
Thus I = 0
Row 1 P O P Row 2 M U T Row 3 --------- Row 4 S O 0 U Row 5 G M T U - Row 6 U 0 R O - - Row 7 ----------------- Row 8 U U S M U U
Step 2
Read Rule 3 here before next step
- P x U = _U
- Can also be written as
- U x P = _U (is of form X x Y = _X)
According to Rule 3, this is only possible when –
- P = {6} & U = {2, 4, 8}
- U = {5} & P = {3, 7, 9}
Let us try for values (P,U) = (6, 2) and start hit and trial method –
Replace all the values and start hit and trial again –
Row 1 6 O 6 Row 2 M 2 T Row 3 --------- Row 4 S O 0 2 Row 5 G M T 2 - Row 6 2 0 R O - - Row 7 ----------------- Row 8 2 2 S M 2 2
Step 3
After replacing with assumed hit and trial values –
- 6 x T = _2
- This is only possible when T = 7
- As 6 x 7 = 42
- As of now, it doesn’t cause any conflict this, moving further, we create the sub problem as this –
Row 1 6 O 6 Row 2 2 Row 3 --------- Row 4 G M 7 2 Note - 0(digit) and O(alphabet) are different.
- 6 x 2 = GM
- We can easily predict the following
- G is 1 as 6 x 2 results in 12
- M is 2 + carry (from previous step)
Also,
- O x 2 + 1(carry from previous step 6 x 2) = _7
- Possible values for O can be 3 or 8
If, O = 3
- 6 X 2 = GM
- O x 2 from previous step is 3 x 2 = 6
- This, doesn’t generate any carry from previous step, thus, M = 2
- This is not possible as U value is already assumed to be 2
Let us try with O = 8
- O x 2 will generate 1 carry to next step as result will be 17 (1 carry from previous step)
- Thus, 6 X 2 + 1 carry = GM
- M = 3 and G = 1
Rewriting the question again
Thus,I = 0 M = 3, G = 1, O = 8, P = 6, U = 2, T = 7 till now
Pending values are S R
Row 1 6 8 6 Row 2 3 2 7 Row 3 --------- Row 4 S 8 0 2 Row 5 1 3 7 2 - Row 6 2 0 R 8 - - Row 7 ----------------- Row 8 2 2 S 3 2 2
Step 4
R value can be found out as follows with this sub problem
Row 1 6 8 6 Row 2 3 Row 3 --------- Row 4 2 0 R 8
- 6 x 3 from units places gives 18 and 1 carry to next step
- 8 x 3 from tens places gives us 24 + 1 (Carry from previous step) this 25
- Value of R = 5
Thus, I = 0, G = 1, U = 2, M = 3, R = 5, P = 6, , T = 7, O = 8, till now
Finally, finding the value of S is also very easy
S = either 9 or 4.
We do hit and trial and find out its 4.
Row 1 6 8 6 Row 2 3 2 7 Row 3 --------- Row 4 S 8 0 2 Row 5 1 3 7 2 - Row 6 2 0 R 8 - - Row 7 ----------------- Row 8 2 2 S 3 2 2
Answers:
Answer 1: Valu of G + U + T
G = 1
U = 2
T = 7
G + U + T = 1 + 2 + 7 = 10
Answer 2: Value of S + G
S = 4
G = 1
S + G = 4 + 1 = 5
Answer 3: Value of 2U
U = 2
2*U = 2*2 = 4
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Options do not match with the solution. None of the options are correct for 1 and t
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