## Outline Mathematics

Number Theory

# AB × BA = 3154. Find A and B.

In the multiplication problem below, A and B stand for different digits. Find A and B.

A | B | ||

× | B | A | |

------------ | |||

1 | 1 | 4 | |

3 | 0 | 4 | |

------------ | |||

3 | 1 | 5 | 4 |

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Copyright © 1996-2018 Alexander Bogomolny

### Solution

In the multiplication problem below, A and B stand for different digits. Find A and B.

A | B | ||

× | B | A | |

------------ | |||

1 | 1 | 4 | |

3 | 0 | 4 | |

------------ | |||

3 | 1 | 5 | 4 |

Well, the first product AB ×A = 114, which only has three prime factors:

It appears that the problem is redundant. Of the three given numbers, only one was needed to solve the problem. Let's see if we could also use the other two to solve it. First

304 = 2 × 2 × 2 × 2 × 38,76,57,83,42,19,38

so that it can be presented as a product of a one-digit and a two-digit numbers in two ways:

304 = 4 ×76,76,57,83,42,19,38 = 8 × 57,76,57,83,42,19,38.

Only the latter is in the form B ×AB. Therefore, again

Finally, we may start with 3154 = 2 ×19 × 83,76,57,83,42,19,38. Note that 83 is a prime. There is only one way to write 3154 as the product

There is an altogether different approach. Since AB × A = 114 but AB × B = 304, we conclude that

8 × 3,2,3,4,5,6,7,8,9, 7 × 2,2,3,4,5,6,7,8,9 and 6 × 4,2,3,4,5,6,7,8,9.

These actually give all possible solutions. Now simply verify that 27 ×72 ≠ 3154 ≠ 46 ×64. On the other hand, we do have

### References

- G. Lenchner,
*Math Olympiad Contest Problems For Elementary and Middle Schools*, Glenwood Publications, NY, 1997

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