Asked by Madhura
How may ordered pair of positive integers satisfy the condition p+q+12=pq
Answers
Answered by
Steve
I can only think of 2 and 14 which work
Answered by
Reiny
and here is why ...
pq - q = p+12
q(p - 1) = (p+12)
q = (p+12)/(p-1)
= 1 + 13/(p-1)
clearly p > 1 , to have positive values
for q to be an integer, 13/(p-1) has to be an integer, that is,
p has to be either 2 or 14
so (2,14) or (14,2)
e.g.
if p = 2, q = 14 , Steve's answer
if p=3, q= 15/2 = 7 + 1/2 , not an integer
if p = 4, q=16/3 = 5 + 1/3 , not an integer
if p=5, q = 17/4 = 4 + 1/4 , not an integer
if p=6, q= 18/5 = 3 + 1/5 , not an integer
..
if p= 20 , q = 32/19 , not an integer
...
etc
pq - q = p+12
q(p - 1) = (p+12)
q = (p+12)/(p-1)
= 1 + 13/(p-1)
clearly p > 1 , to have positive values
for q to be an integer, 13/(p-1) has to be an integer, that is,
p has to be either 2 or 14
so (2,14) or (14,2)
e.g.
if p = 2, q = 14 , Steve's answer
if p=3, q= 15/2 = 7 + 1/2 , not an integer
if p = 4, q=16/3 = 5 + 1/3 , not an integer
if p=5, q = 17/4 = 4 + 1/4 , not an integer
if p=6, q= 18/5 = 3 + 1/5 , not an integer
..
if p= 20 , q = 32/19 , not an integer
...
etc
Answered by
Madhura
Thankyou
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