# eLitmus Cryptarithmetic Problem – 5

## Cryptarithmetic Questions asked in eLitmus 5 Ques. For the following Cryptarithmetic find the answers to the below questions?
```                 T  E  A
x  H  A  D
----------
L  D  T  R
H  R  S  A
E  W  D  A
-----------------
L  E  S  S  E  R```
 1. Value of  S ? (a) 6 (b) 7 (c) 8 (d) 9
 2. Which of the following follows the Pythagoras theorem ? (a) H, A, D (b) H, A, B (c) T, E, A (d) T, E, D
 3. Value of H + R + S + A ? (a) 14 (b) 15 (c) 16 (d) 17

It is highly suggested to go through these before solving

```Row1              T  E  A
Row2           x  H  A  DRow3           ----------
Row4           L  D  T  R
Row5        H  R  S  A
Row6     E  W  D  A
Row7    -----------------
Row8    L  E  S  S  E  R```

### Step 1

• A x A = _A
• This means by Rule2 that A = {5,6}
```Row1              T  E  A
Row2           x  H  A  DRow3           ----------
Row4           L  D  T  R
Row5        H  R  S  A
Row6     E  W  D  A
Row7    -----------------
Row8    L  E  S  S  E  R```

### Step 2

• Case I – when A={5} then H={3, 7, 9}
• Case II – when A={2, 4, 8} then H={6}

Taking A = 5 from Step 1 and just replacing a values, the problem now looks like –

```Row1              T  E  5 Row2           x  H  5  DRow3           ----------
Row4           L  D  T  R
Row5        H  R  S  5
Row6     E  W  D  5
Row7    -----------------
Row8    L  E  S  S  E  R```

### Step 3

• Looking at the 5 x D = _R
• Possible values for R can be either 0 or 5
• As if you multiply 5 to number you will either get 0 or 5 in units digit
• R value will be 0 as A is already 5
• R = 0

Replacing the values –

```Row1              T  E  5 Row2           x  H  5  DRow3           ----------
Row4           L  D  T  0
Row5        H  0  S  5
Row6     E  W  D  5
Row7    -----------------
Row8     L  E  S  S  E  0```

### Step 4

• Let us create a subproblem –
```Row1              T  E  5
Row2           x        5
Row3           ----------
Row4           H  0  S  5
```

Now, in this –

• T x 5 = _0
• This means there is no carry from previous step
• This means E x 5 < 10
• Only possible way it could happen is E = 1
• Thus the value for S will be –
• S = 1 x 5 + 2(carry from previous step)
• S = 7

Values till now are –

R = 0, E = 1, A = 5, S = 7 replacing and writing it again.

```Row1              T  1  5 Row2           x  H  5  DRow3           ----------
Row4           L  D  T  0
Row5        H  0  7  5
Row6     1  W  D  5
Row7    -----------------
Row8     L  1  7  7  1  0```

### Step 5

• Now, T = 6
• As T + 5 = _1 (no carry from previous step possible)
• L = 2
• L can not 1, E is already 1
• Thus, there must be 1 carry coming from previous step
• L = 1 + 1(carry) = 2
• Rewriting the problem
```Row1              6  1  5 Row2           x  H  5  DRow3           ----------
Row4           2  D  6  0
Row5        H  0  7  5
Row6     1  W  D  5
Row7    -----------------
Row8     2  1  7  7  1  0```

### Step 5

• D = 4
• As D + 7 + 5 + 1(carry from previous step) = _7
• Thus D = 4
```Row1              6  1  5 Row2           x  H  5  4Row3           ----------
Row4           2  4  6  0
Row5        H  0  7  5
Row6     1  W  4  5
Row7    -----------------
Row8     2  1  7  7  1  0```

### Step 6

• H = 3
• 6 x 5 = H0
• H + W + 0 (no carry from previous step) = 1
• Thus, W = 8
```              6  1  5
x  3  5  4
----------
2  4  6  0
3  0  7  5
1  8  4  5
----------------
2  1  7  7  1  0```

[Relax – it’s going to take some time to understand the whole concept.
If you are facing any difficulty.

Please go through the Cryptarithmetic Tutorial.]

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