How many ordered pairs of positive integers (a, b) have the property that their product is exactly 20 times their sum?

Guest May 16, 2023

#1**0 **

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).

Guest May 16, 2023

#2**0 **

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)

Guest May 16, 2023

#3**0 **

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).

Tiggsy May 16, 2023