# 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

Solution ### 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: 114 = 2 × 3 × 19. We are supposed to get 114 as the product of a one-digit and a two-digit numbers. This is possible to achieve in two ways: 114 = 2 × 57,76,57,83,42,19,38 and 114 = 3 ×38,76,57,83,42,19,38. Only the latter is in the form A × AB. Thus we conclude that A = 3 and B = 8,5,6,7,8,9.

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 = AB × B. However,

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 A = 3 and B = 8.

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 AB × BA: 3154 = 38 × 83.

There is an altogether different approach. Since AB × A = 114 but AB × B = 304, we conclude that A < B. How may the product B × A end with 4? In three ways:

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 3154 = 38 × 83, which thus is the only solution to the problem.

### References

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