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

 May 16, 2023
 #1
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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).

 May 16, 2023
 #2
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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)

 May 16, 2023
 #3
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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).

 May 16, 2023
edited by Tiggsy  May 16, 2023

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