Solved the given question by addition rules and the value of START is 10430.
Given-
The given problem is from the alphabetical puzzle. In the given problem the equation given is,
JOHN + CAN = START.
It can be written as,
J O H N
+ C A N
__________
S T A R T
__________
S must be 1, because, even if we assume all the missing digit are 9 ( the largest possible value), the sum will be smaller than than 10999 or equal to the 10998.
Similarly the value of T must be 0 because, even if we assume all the missing digit are 9 ( the largest possible value), the sum will be smaller than than 10999 or equal to the 10998.
Now as the T is two times in the solution of addition, and the value of T is 0 as we found. But T is the addition of N and N. Hence N must be 5 to make the addition 10 for the 0 value of the T.
N
+ N
____
1 0
____
J must be 9 with carried 1 to make the T 0.
A must be more than 10 to have a carry on J. A can not 1 as S has 1 digit. 2 and 3 disturbs the equation. Thus a must be 4. and for this o and c must be 6 and 7.
If a is 4 than the H and R must be 8 and 3.
The sum will look like,
9 6 8 5
+ 7 4 5
________
1 0 4 3 0
_________
Hence, solved the given question by addition rules and the value of START is 10430.
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