eLitmus Cryptarithmetic Problem – 2
Cryptarithmetic Questions asked in eLitmus 2
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
H O W
x C U T
----------
D D C P
D W P W
D U O A
----------------
D C E P D P| 1. | Value of T + O + P ? | |||
| (a) 3 | (b) 8 | (c) 6 | (d) 9 | |
| 2. | Value of D ? | |||
| (a) 1 | (b) 2 | (c) 3 | (d) 4 | |
| 3. | Value of 2C ? | |||
| (a) 13 | (b) 14 | (c) 17 | (d) 11 | |
It is highly suggested to go through these before solving
Row1 H O W Row2 x C U T Row3 ---------- Row4 D D C P Row5 D W P W Row6 D U O A Row7 ---------------- Row8 D C E P D P
Step 1
- W x U = _W (Rule 3)
- Case 1 :
- When W={2, 4, 8} and U={6}
- Case 2 :
- When W={5} and U={3, 7, 9}
Assuming U = 6, the problem will look like –
Row1 H O W Row2 x C 6 T Row3 ---------- Row4 D D C P Row5 D W P W Row6 D 6 O A Row7 ---------------- Row8 D C E P D P
Step 2
- D + 6 + (carry) = C
- [Carry may be 0, 1, 2 (as 3 digit addition in previous steps )]
- Thus C = {7, 8, 9}
- Start hit and trial C={7, 8, 9} and W={2, 4 ,8}
Hit and Trial 1
- C = 7 and W = 2
- Then D=9 [ As, C + W = D] If we take take D=9 then
Row1 H O 2 Row2 x 7 6 T Row3 ---------- Row4 (carry)1 9 9 7 P Row5 9 2 P 2 Row6 9 6 O A Row7 ---------------- Row8 0 5 E P 9 P
Step 3
Assuming: C = 7 and W = 2
- Now this causes conflict as –
- D = 9 from last step as 7 + 2 = 9
- D = 0 as 9 + 1 (carry) from previous step 9 + 6 = 5 (1 carry)
To values of D not possible thus thus hit and trial is wrong we must do another hit and trial
- Let’s take C=7 and W=4
Row1 H O 4 Row2 x 7 6 T Row3 ---------- Row4 1 1 7 P Row5 1 4 P 4 Row6 1 6 O A Row7 ---------------- Row8 1 7 E P 1 P
Step 4
Assuming C=7 and W=4
Then, D=1 as C + W = D
- Thus, value for A = 8
- As, 7 x 4 = _A
Rewriting the whole problem
Row1 H O 4 Row2 x 7 6 T Row3 ---------- Row4 1 1 7 P Row5 1 4 P 4 Row6 1 6 O 8 Row7 ---------------- Row8 1 7 E P 1 P
Step 5
Assuming C=7 and W=4
H O 4
x 7
----------
1 6 O 8
- 7 x H = 16
- How, is this possible?
- This is possible when
- 7 x 2 = 14 and 2 carry from previous step
- Thus, H = 2
Row1 2 O 4 Row2 x 7 6 T Row3 ---------- Row4 1 1 7 P Row5 1 4 P 4 Row6 1 6 O 8 Row7 ---------------- Row8 1 7 E P 1 P
Step 6
- 7 + 4 = 1 (1 carry to next step)
- Thus, 1 (carry) + 1 + P + 8 = P
- 10 + P = P means 1 carry to next step
- 0 + P = P, can only be possible when, P = 0
Rewriting the problem –
Row1 2 O 4 Row2 x 7 6 T Row3 ---------- Row4 1 1 7 0 Row5 1 4 0 4 Row6 1 6 O 8 Row7 ---------------- Row8 1 7 E 0 1 0
Step 7
- 4 x T = _0
- This is possible when, T = 0, T = 5
- T = 0 not possible as P = 0 in previous step
- Thus, T = 0
Also clearly,
- O = 3 as,
2 O 4
x 6
----------
1 4 0 4
- 6 x 4 = 4 (2 carry)
- 6 x O + 2 (carry) = _0
- This is only possible when O = 3
- 6 x 3 + 2 (carry) = 20
H=2, O=3, W=4, C=7, U=6, T=5, D=1, S=0, E=9
2 3 4
x 7 6 5
------------
1 1 7 0
1 4 0 4
1 6 3 8
----------------
1 7 9 0 1 0 
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In step 2, how is C = {7,8,9} ? Please help
These questions were really helpful for eLitmus exam got placed via eLitmus in 2 companies with an offer of 6 LPA.
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