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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Copyright © 1996-2018 Alexander Bogomolny
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