# 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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