GO ON BOB

Solve this cryptarithmetic (each letter represents a unique digit 0 to 9). GO + ON = BOB, given that BOB is divisible by G.

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Solution

The sum of two 2-digit numbers has to be less than 200 (because 99+99=198), so B < 2.

B is not 0, because that would require the sum of two 2-digit numbers to be less than one of those 2-digit numbers (specifically, "BOB" would be "ON" rounded down to a lower multiple of ten).

Hence B = 1. Which means "BOB" is one of 101, 121, 131, etc. (Not 111, because each letter represents a unique digit.)

Knowing that "BOB" is divisible by G, we can look up 1-digit factors of candidates for "BOB". (In the old days, mathematicians would have had tables for this. We have the Internet.)

141 has factor 3; 161 has factor 7; 171 has factors 3, 9. No other candidates for "BOB" have 1-digit factors other than 1.

Trying the candidates one at a time:

• 34 + 4_ = 141 (obviously not, as 34+49 is only 83)
• 76 + 6_ = 161 (obviously not, as 76+69 is only 145)
• 97 + 7_ = 171 (this is the only candidate where the sum is in the right range)

171 - 97 = 74, which gives 97 + 74 = 171 as the equation represented by the words.