eLitmus Cryptarithmetic Problem – 1
J E
B B
----
J E
J E A
--------
B A D E
Assume that E = 4
Cryptarithmetic Questions asked in eLitmus 1
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
| 1. | Value of 5B? | |||
| (a) 15 | (b) 10 | (c) 5 | ||
| 2. | Value of B + A? | |||
| (a) 7 | (b) 5 | (c) 1 | (d) 8 | |
| 3. | Value of D ? | |||
| (a) 4 | (b) 3 | (c) 7 | (d) Can not be determined | |
It is highly suggested to go through these before solving - Cryptarithmetic Introduction - How to Solve Cryptarithmetic Problems
Row 1 J E Row 2 B B Row 3 ---- Row 4 J E Row 5 J E A Row 6 -------- Row 7 B A D E Assume that E = 4
Step 1
- Clearly, from the Row 2 row where
- B is written in common multiplication
- J E is multiplied with B and results in B, which results in J E.
- Thus, the B value is 1
Step 2
- In Row 4 : E + A = E.
- This, only possible when A value is 0
Row 1 J E Row 2 1 1 Row 3 ---- Row 4 J E Row 5 J E 0 Row 6 -------- Row 7 1 0 D E
Step 3
- Now in Row 5 and Row 7
- J + nothing(Row 4) gives us 10.
- Thus, there must be 1 carry from the previous step and J value must be 9
- Now, the problem looks like-
Row 1 9 E Row 2 1 1 Row 3 ---- Row 4 (carry)1 9 E Row 5 9 E 0 Row 6 -------- Row 7 1 0 D E
Step 4
- 9 + E = D
- Now, from the previous step, there can not be any carry
- As E + 0 = E, and E can have values between E = {0,9}
- Max value of E can be 8 as J = 9 (already taken)
- Thus, step 9 + E = D will have no carry from the previous step
- However, this is generating carry to the next step
- Thus, 9 + E = D + 10
- E – D = 1
- Since, E = 4 => D = 3
Login/Signup to comment
For any questions about eLitmus Cryptarithmetic questions email us at PrepInsta@gmail.com