GO ON BOB: Difference between revisions
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Solve this cryptarithmetic (each letter represents a unique digit 0 to 9). GO + ON = BOB, given that BOB is divisible by G. | Solve this cryptarithmetic (each letter represents a unique digit 0 to 9). GO + ON = BOB, given that BOB is divisible by G. | ||
{{solution | The sum of two 2-digit numbers has to be less than 200 (because 99+99<nowiki>=</nowiki>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 <nowiki>=</nowiki> 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_ <nowiki>=</nowiki> 141 (obviously not, as 34+49 is only 83) | |||
*76 + 6_ <nowiki>=</nowiki> 161 (obviously not, as 76+69 is only 145) | |||
*97 + 7_ <nowiki>=</nowiki> 171 (this is the only candidate where the sum is in the right range) | |||
171 - 97 <nowiki>=</nowiki> 74, which gives 97 + 74 <nowiki>=</nowiki> 171 as the equation represented by the words. }} | |||
[[Category: Cryptarithmetic puzzles]] | [[Category: Cryptarithmetic puzzles]] | ||
[[Category: Number theory]] | [[Category: Number theory]] | ||
Revision as of 06:35, 18 October 2010
Solve this cryptarithmetic (each letter represents a unique digit 0 to 9). GO + ON = BOB, given that BOB is divisible by G.
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)