eLitmus Cryptarithmetic Problem – 1

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