Not an olympiad question
Find the number of integral solutions of
3 5 x + 6 3 y + 4 5 z = 1
for ∣ x ∣ < 9 , ∣ y ∣ < 5 , and ∣ z ∣ < 7 .
The answer is 3.
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2 solutions
this question came in jmo 2016.
it came today in jmo i got 3 solutions. {x,y,z}= {-1,2,-2} and {8,-3,-2} and {-1,-3,5}
Can you write a solution that is helpful to those who cannot solve it? Else I'm inclined to delete this solution to encourage others to contribute a relevant answer.
I also got three solutions
3 5 x + 6 3 y + 4 5 z = 1
5 ( 7 x + 9 z ) = 1 − 6 3 y
So, 5 divides 1 − 6 3 y ∣ 1 − 3 y
As − 5 ≤ y ≤ 5 only values of y=-3,2 satisfies
Case(1) take y=-3 and the equation becomes 7 x + 9 z = 3 8
7 ( x − 5 ) = 3 ( 1 − 3 z ) So 7 ∣ ( 1 − 3 z ) .
As z varies from (-7,7) .only values of z =-2,5 which gives two solution (8,-3,-2) and (-1,-3,5).
Case(2) take y=2 and the equation becomes : 7 x + 9 z = − 2 5
7 x + 9 z = − 2 8 + 3
7 ( x + 4 ) = 3 ( 1 − 3 z )
Here same as before 7 ∣ 1 − 3 z so values of z=-2,5.
But putting values of z will give only one solution because other solution does not meet condition.
So the only solution is (-1,2,-2) . So overall there are three solutions in the given range.
L e t ( 6 3 y − 1 ) + 5 ( 7 x + 9 z ) = 0 . We note that y has the smallest range. For -5<y<5, so 63y can contribute the following. ± { 6 3 , 1 2 6 , 1 8 9 , 2 5 2 } . ∴ ( 6 3 y − 1 ) = { + 6 2 , + 1 2 5 , + 1 8 8 , + 2 5 1 , − 6 3 , − 1 2 6 , − 1 9 0 , − 1 5 3 } f o r y = − 4 , − 3 , . . . 3 , 4 . But for (63-1) to be integer , it must be a multiple of 5. So +125 and -190 are only useful. T h a t m e a n s ( 7 x + 9 z ) m u s t b e = 5 − 1 2 5 o r 5 1 9 0 , t h a t i s − 2 5 o r + 3 8 . We get (7x+9z)= -25 with x=-1 and z=-2 . We get (7x+9z)= 38 with x=-1 and z=5 ...... and also with x=8 and z=-2 .
So there are only three integer solutions.