Solution: a = 5, b = 1, c = 2, and d = 3. (Because of the symmetry of
the equations, several rearrangements of these numbers will also satisfy the
equations.)
Got no idea where to begin with this one? There are four
unknowns but only two equations, therefore algebra won’t work. So, try some
numbers, and see what happens.
Try a = 7, b = 8, c = 9, and d = 10. Then ab = 56, a + b = 15, cd = 90, and c + d = 19. Well, that’s
no good! The products are much larger than the sums. And for even larger values
of a, b, c, and d, the difference will be more stark.
So, it seems that the solution will need to involve smaller values.
Try a = 1, b = 2, c = 3, and d = 4. Then ab = 2, a + b = 3, cd = 12, and c + d = 7. That’s not a
correct solution, either, but it does shed some light. Note that the sum a + b and the product ab
are both less than the sum c + d and the product cd. So, when you pick four numbers, the
two smallest shouldn’t be a and b. If a is the smallest value, then b
should be the largest, and c and d should be somewhere in between.
So, try rearranging. Let a
= 1, b = 4, c = 2, and d = 3. Then ab = 4, a + b = 5, cd = 6, and c + d = 5. Still not
quite there, but it’s close, isn’t it? The product ab is one less than c + d, and the product cd is one more than a + b. From that, you might realize that a
solution can be found if the value of b
is increased by one. Indeed, a = 1, b = 5, c = 2, and d = 3 is a
solution.
Interestingly, there are no others. The text below will prove
that there is only one solution, but if you want to stop here, that’d be okay.
Let a be the
smallest of the four numbers. If a > 2, then ab> 2b > a + b
= cd > c + d. But the problem
requires
that ab = c + d, so we’ve reached a
contradiction. Therefore, a < 2,
and since a must be a positive
integer, a = 1.
When a = 1, the
equation ab = c + d becomes b = c
+ d, and the equation a + b
= cd becomes 1 + b = cd,
which is equivalent to b = cd – 1. These results can be combined to
give 1 + c + d = cd, which can be
rewritten as cd – d – c
– 1 = 0, or (c – 1)(d – 1) = 2. This final equation implies
that c = 2 and d = 3, or vice versa.