An algebra problem by Hafizh Ahsan Permana
How many pairs of positive integer ( x , y ) satisfy the equation below?
x 1 + y 1 = 6 1
Note: ( 5 , 4 ) and ( 4 , 5 ) are different solutions.
The answer is 9.
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2 solutions
Number of solutions to questions of this type is the number of factors of n 2 . It can also be proved.
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Yes...but hey factors of 36- oh.yes...oh my god..i multiplied 2+1 and 2+1 to get 8 :(.....
please help we can write it as x =y/y -6 this is equal to x = 1 + 6/y-6 since we want integral solutions so y-6 should divide 6 thus we get y = 8,9,12 therefore only 3 solutions
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y = 6x/(x-6), and x = 6y/(y-6); pairs are (7,42), (8,24), (9,18), (10,15), (12,12), (15,10), (18,9), (24,8) and (42,7).
x 1 + y 1 6 ( x + y ) x y − 6 x − 6 y ( x − 6 ) ( y − 6 ) = 6 1 = x y = 0 = 3 6
Since 3 6 = 2 2 × 3 2 has ( 2 + 1 ) ( 2 + 1 ) = 9 pairs of positive factors ( a , b ) . Therefore, there are also 9 pairs of positive integer ( x , y ) = ( a + 6 , b + 6 ) satisfying the equation.
6 x + 6 y = x y x y − 6 x − 6 y = 0 ( x − 6 ) ( y − 6 ) = 3 6 now 36 can be factorized into (x-6) and (y-6 ) as 1 . ) 1 ∗ 3 6 − − − − − > 2 s o l u t i o n s 2 . ) 2 ∗ 1 8 − − − − > 2 s o l u t i o n s 3 . ) 3 ∗ 1 2 − − − − > 2 s o l u t i o n s 4 . ) 4 ∗ 9 − − − > 2 s o l u t i o n s 5 . ) 6 ∗ 6 − − > 1 s o l u t i o n s s o l u t i o n s ( x , y ) = 2 + 2 + 2 + 2 + 1 = 9