# Difference between revisions of "GO ON BOB"

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+ | I believe this is original. | ||

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+ | ==Puzzle== | ||

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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. | ||

− | {{ | + | ==Help== |

+ | |||

+ | {{Hint | If you can decide what B is first, then that drastically limits the possibilities for G.}} | ||

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+ | {{Answer | <m>97 + 74 = 171</m>}} | ||

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+ | {{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). | 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). | ||

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171 - 97 <nowiki>=</nowiki> 74, which gives 97 + 74 <nowiki>=</nowiki> 171 as the equation represented by the words. }} | 171 - 97 <nowiki>=</nowiki> 74, which gives 97 + 74 <nowiki>=</nowiki> 171 as the equation represented by the words. }} | ||

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+ | ==See also== | ||

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+ | [[Cryptarithmetic]] | ||

+ | |||

+ | [[Lean meat]] | ||

[[Category: Cryptarithmetic puzzles]] | [[Category: Cryptarithmetic puzzles]] | ||

[[Category: Number theory]] | [[Category: Number theory]] |

## Current revision as of 19:34, 14 July 2013

I believe this is original.

## Puzzle

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

## Help

**Hint**

**Answer**

**Solution**

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)