Junior Mathematical Olympiad 2016
Solution
Q.3 Find all integer solutions of the system: 35x + 63y + 45z = l,
|x|< 9, |y|< 5, |z|< 7.
Solution. We have
l-63y = 5(7x + 9z). -5 < y < 5. Try for y = 0.ħl.ħ2.ħ3.ħ4.
Only y = 2 and y = -3 gives 1 - 63 y
as multiple of 5.
Therefore we have to solve two linear Diophantine equations:
7x + 9z = -25 and 7x + 9z = 38.
Note that g.c.d.
(7. 9) = 1. Therefore both Diophantine equations have infinitely many integer
solutions.
For first equation, guess solution with z = 5 giving x = -10. Therefore
First equation has solutions: x = -10 +
9k. z = 5 7k, k 
Z.
Now using -9 < x< 9. -7 < z < 7 we obtain 
< k < 
. 
< k < 
. k Z.
Only common
solution is k =1. Thus x = -1. v = 2. z = -2 is an integer
soln.
Similarly
for y = -3. second equation has guess solution with z
= 5 giving x = -1. therefore
Second
equation has solutions: x = -1 + 9m, z = 5 - 7m, m
Z
Now using -9 < x< 9, - 7 < z
< 7, we obtain 
< m < 
, 
< m <
, m Z.
Thus
m = 0, m = 1 are two possible values of m giving two integer solutions
x
= -1, y = -3, z = 5; x = 8, y = -3, z = -2
Hence on combining, we have three integer solutions of
the given system as:
(x,y,z) = (-1,2,-2).(-l,-3,5).and (8,-3,-2)
ALITER
Q3 Ans. - 35x + 63y + 45z = 1, where |x|<9,
|y|<5, |z|<7
⇒ 7(5x+ 9y) + 45z = 91 90
⇒ 7(5x + 9y) 91 = - 90 45z
⇒ 7(5x + 9y - 13) = - 45 (z + 2)
⇒ (5x + 9y - 13) = - 
(z + 2)
Now, see that
LHS is an integer as x and y are integers.
RHS is also an integer hence,
7| - 45 (z +
2), but 7 x 45
7| z + 2
Let, z + 2 =
7λ
⇒ z = 7 λ 2
Now, as given
in Q,
-7 < Z < 7
- 7 < 7 λ
2 < 7
-5/7 < λ < 9/7
But as Z is
an integer, therefore
λ 
{0, 1}
So, Z {-2, 5}
5x + 9y = 13, when z = -2
..(1)
And 5x + 9y = -32
. (2) when z = 5
Case I
5x + 9y = 13, when
z = -2
We can see that one
solution of this equation is (x, y) = (-1, 2)
Now, as gcd (5, 9) = 1 and -1, 2 is a particular solution to the
Diophantine equation,
x = -1 + 9t, y = 2 5t
Now, from Q,
- 9 < x < 9
⇒- 
< t < 
⇒ t {0, 1} x is an integer.
⇒ x = -1, 8
Similary, y = 2, -3
Case II
5x + 9y = -32, when z = 5
We can see that one solution of this equation (x, y) =
(-1, -3)
x = -1 + 9t, y = -3 5t
x = -1, 8 and y {-3, -8}
Solution are (x, y, z) = (-1, 2, -2) (8, -3,
-2) (-1, -3, 5) (88, -8, 5)
As |y|<5, so solutions are
(x, y, z) = (-1, 2, -2) (8,
-3, -2) (-1, -3, 5)
+++++++++++++++++++++++++