Problem 5
How many ordered triplets
(
a
,
b
,
c
)
of non-zero positive integers satisfy
5 a + 6 b + 3 c = 5 0 ?
The answer is 13.
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2 solutions
Since b and c contributes in multiples of 3,
5
0
−
5
a
=
3
∗
X
.....X is an integer.
So " a "can only be 1, 4, 7...corresponding numbers left to be absorbed by b, and c.... are
4
5
,
3
0
,
1
5
.
There has to be at least one c. Thus 6b can absorb
4
2
,
2
7
a
n
d
1
2
.
There can be.... 42/6 =7 " b "in 42 , .... 27/6 = 4 in 27,.... 12/6= 2 in 12.
7
+
4
+
2
=
1
3
since they must be non zero, the problem is equivalent to finding the number of non-negative triples of 5a+6b+3c= (50-5-6-3) = 36.
But then, this is the coefficient of x 3 6 in the power series of 1 / [ ( 1 − x 3 ) ( 1 − x 5 ) ( 1 − x 6 ) ] , that is, 13.
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Nice thinking about 36. How did you think of this power series ?
Can you please elaborate on this?
Nice solution.
5a+6b+3c=50 ===> 5a+3(2b+c)=50
Let 2b+c=y ==> 5a+3y=50
solve for 5a+3y=50 from a=1 (Since given that a,b,c are positive non zero integers)
We get (a,y)={(1,15),(4,10),(7,5)}
Now similarly Solve for values of b&c from y=2b+c={15,10,5}
We get (b,c)={(1,13)(2,11)(3,9)(4,7)(5,5)(6,3)(7,1) (1,8)(2,6)(3,4)(4,2) (1,3)(2,1)} fro y=15,10,5
==>Total number of distinct solutions=13