eLitmus Cryptarithmetic Problem – 4
Cryptarithmetic Questions asked in eLitmus 4
Ques. For the following Cryptarithmetic find the answers to the below questions? S I R
x M L T
------------
I U R S
R M J R
R E L U
-----------------
R S S J S S
| 1. | Which of the following set contains only even numbers? | |||
| (a) I, R, M | (b) S, I, R | (c) S, R, L | (d) M, L, T | |
| 2. | Which of the following set contains odd numbers? | |||
| (a) T, M, I | (b) S, I, R | (c) J, T, L | (d) R, M, L | |
| 3. | Value of S? | |||
| (a) 2 | (b) 3 | (c) 5 | (d) 4 | |
It is highly suggested to go through these before solving
Row1 S I R Row2 x M L T Row3 ------------ Row4 I U R S Row5 R M J R Row6 R E L U Row7 ----------------- Row8 R S S J S S
Step 1
- Now, R x L = _L,
- By Rule 3 we know that it is of form X x Y = _X
- Thus, the possible values are –
L = {6} and R= {2, 4, 8}
Or, R = {5} and L = {3, 7, 9}
Let us take the first case where (L, R) = (6, 2) and the new problem looks like :
Row1 S I 2 Row2 x M 6 T Row3 ------------ Row4 I U 2 S Row5 2 M J 2 Row6 2 E 6 U Row7 ----------------- Row8 2 S S J S S
Step 2
- Now, S = 2 + 2 = 4 + (no carry)
- Since, there will be no carry from previous step
Putting all the values, the problem looks like –
Row1 4 I 2 Row2 x M 6 T Row3 ------------ Row4 I U 2 4 Row5 2 M J 2 Row6 2 E 6 U Row7 ----------------- Row8 2 4 4 J 4 4
Step 3
- We can find value of T = 7
- As 2 x T = 4
- Possible values of T = {2, 7} to give 4
- Value 2 is already taken by R so T = 7
Replacing with new values
Row1 4 I 2 Row2 x M 6 7 Row3 ------------ Row4 I U 2 4 Row5 2 M J 2 Row6 2 E 6 U Row7 ----------------- Row8 2 4 4 J 4 4
Step 4
- Now, I x 7 = _2
- Also, there is one carry from previous step : 2 x 7 = _4 (1 carry)
- Thus, I x 7 results in a number ending from 1
- This is only possible if I = 3
As : I x 7 = 21 + 1 carry = 22 this satisfied I x 7 = _2 condition
Row1 4 3 2 Row2 x M 6 7 Row3 ------------ Row4 3 U 2 4 Row5 2 M J 2 Row6 2 E 6 U Row7 ----------------- Row8 2 4 4 J 4 4
Step 5
- If we look at the step 3 + M + 6 = _4
- The max carry this step can generate to next step is 1
- As for all possible values of M = {0,9} it still will result in 1 carry to next step
- As 3 + 9 + 6 = 18 (max value possible)
Thus,
- 2 + E + (1 carry) = 4
- Thus, E = 1
Row1 4 3 2 Row2 x M 6 7 Row3 ------------ Row4 3 U 2 4 Row5 2 M J 2 Row6 2 E 6 U Row7 ----------------- Row8 2 4 4 J 4 4
Step 6
- Creating a sub problem of the question –
Row1 4 3 2 Row2 x M Row3 ------------ Row4 2 1 6 U
Now,
- M x 4 = 21
- This is only possible when, M = 5 and obv. 1 carry from previous step
- Thus M = 5
Now, you can easily find the values of U and J, which would be 0 and 9
Thus,
U = 0, E = 1, R = 2, I = 3, S = 4, M = 5, L = 6, T = 7, J = 9
4 3 2
x 5 6 7
----------
3 0 2 4
2 5 9 2
2 1 6 0
----------------
2 4 4 9 4 4 
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