2026/058) Find the number of ordered triples of positive integers $(a,b,c)$ such that $6a+10b+15c=3000$

We are given

$6a+10b+15c=3000\cdots(1)$ 

We have rearranging the terms let us get a in terms of others 

$6a = 3000 - 10b-15c = 5(600-2b-2c)$

As the RHS is divisible by 5 and GCD(5,6) = 1 so a is divisible by 5 and so

$a=5x\cdots(2)$ for some x

Again to get b in terms of others

 $10b = 3000 - 6a-15c = 3(1000-2a-5c)$

As the RHS is divisible by 3 and GCD(3,10) = 1 so b is divisible by 3 and so

$b=3y\cdots(3)$ for some y

 Again to get c

 $15c = 3000 - 6a-10b = 2(1500-3a-5b)$

As the RHS is divisible by 2 and GCD(2,15) = 1 so c is divisible by 2 and so

$c=3z\cdots(4)$ for some z

Putting the values a,b,c from (2),(3),(4) respectively in (1)
 

$30x + 30y + 30z = 3000$

Or $x + y + z = 100$

We need to find x,y,z all 3 positive integers sum is 100

Let us assume that there are 100 stones in a line . There are 99 gaps. we can make them into 3 parts of each part containing above one by putting 2 sticks in 99 gaps. this can be done in $99 \choose 2$ ways