# 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) 12 (c) 6 (d) 9
 2. Value of S + G ? (a) 3 (b) 9 (c) 7 (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 –

1. P = {6} & U = {2, 4, 8}
2. 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.

### 2 comments on “eLitmus Cryptarithmetic Problem – 3”

• Shubham

Options do not match with the solution. None of the options are correct for 1 and t 1
• Atulya PrepInsta

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