How many ordered pairs of positive integers (a, b) have the property that their product is exactly 20 times their sum?
We have that ab=20(a+b). Solving for b, we get b=20a/(20-a). We want b to be an integer, so a−20 must divide 20. The divisors of 20 are 1, 2, 4, 5, 10, and 20. For each divisor d, we can take a=20d, and then b=20. Thus, there are 6 ordered pairs (a,b).
You have to specify a ceiling as to how high you want to check.
Checking pairs up to 100, you get the the following:
a b
1 = (70, 28)
2 = (60, 30)
3 = (45, 36)
4 = (40, 40)
5 = (36, 45)
6 = (30, 60)
7 = (28, 70)
+397 We have ab = 20(a + b) = 20a + 20b,.......(*)
so
ab - 20a = 20b,
a(b - 20) = 20 b,
a = 20b/(b - 20).
a is to be positive so b must be greater than 20, and since the roles of a and b can be reversed a must also be greater than 20.
Let a = 20 + m, and b = 20 + n, then, substituting into (*),
(20 + m)(20 + n) = 20(20 + m + 20 + n),
400 + 20m + 20n + mn = 800 + 20m + 20n.
mn = 400.
So we are looking for two positive integers having a product of 400.
(1) m = 1, n = 400 from which a = 21, b = 420.
(2) m = 2, n = 200 ............... a = 22, b = 220.
(3) m = 4, n = 100 ............... a = 24, b = 120.
etc.
There are five others, (fifteen in all, counting reversals).
