+ B Y J I U I D 2 A O 0 Y N 1 O G 7 U
Given that each alphabet in the cryptogram above represents distinct digits.
Find the number of possible solutions to the cryptogram.
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Look at the second column from left.
We know that the carry over is at most 1 .( See the theorem in 3. )
We deduce that U=0 or U=9
Case 1:U=0
+ B Y J I 0 I D 2 A O 0 Y N 1 O G 7 0
We deduce that G=3 .
+ B Y J I 0 I D 2 A O 0 Y N 1 O 3 7 0
We see that there is a carry over at the third column from right.
We deduce that N=8 or N=9
When N=8 ,
+ B Y J I 0 I D 2 A O 0 Y 8 1 O 3 7 0
O=0 , which contradicts with U=0
∴ N=9
+ B Y J I 0 I D 2 A O 0 Y 9 1 O 3 7 0
O=1
+ B Y J I 0 I D 2 A 1 0 Y 9 1 1 3 7 0
Y=2
+ B 2 J I 0 I D 2 A 1 0 2 9 1 1 3 7 0
The remaining unused digits are 4,5,6,7,8 .
(B,I,D,J,A) = (4,8,5,6,7) , (5,8,4,7,6) , (5,4,6,7,8) , (6,4,5,8,7)
There are a total number of 4 solutions when U=0 .
Case 2: U=9
+ B Y J I 9 I D 2 A O 0 Y N 1 O G 7 9
We deduce that G=2 .
+ B Y J I 9 I D 2 A O 0 Y N 1 O 2 7 9
Now, let's look at third column, O and Y are different digits so the carry over is 1 .
N+1 ≥ 1 0 ⇒ N ≥ 9
U=9 and therefore value of N does not exist.
U=9 is unsuitable.
We conclude that there are 4 solutions to the cryptogram.