eLitmus Cryptarithmetic Problem – 5
Cryptarithmetic Questions asked in eLitmus 5
Important
The Cryparithmetic methods discussed below are developed by PrepInsta only and available on 2 PrepInsta owned websites only. Anyone copying the method will be legally sued as these are not open source but are PrepInsta's Proprietary methods
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 D
Row3 ---------- 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 D
Row3 ---------- 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 D
Row3 ---------- 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 D
Row3 ---------- 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 D
Row3 ---------- 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 D
Row3 ---------- 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 4
Row3 ---------- 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.]
Login/Signup to comment
For any questions about eLitmus Cryptarithmetic questions email us at PrepInsta@gmail.com