Fractional decomposition

Since a and b are coprime integers, b must be a multiple of 31. There are 32 multiples of 31 between 1 and 1000, so the answer is 32.

Multiplying by $b$ , $a=\dfrac{13b}{31}$ Since $a$ is an integer, $b$ must be a multiple of 31. Let $b=31m$ . Therefore, $a=13m$ . As long as $31m<1000$ , or $m<33$ , $(a,b)=(13m,31m)$ satisfies the equation. Therefore, the answer is $\boxed{32}$ .

13/31*32/32=416/962=a/b

1<a<1000; 1<b<1000

therefore, there are 32 pairs

1000/31 = 32.25

So there are 32 sets of ordered pair

Multiples of 31 !

13/31 is an irreducible fraction. If we know that the ratio of a to b is 13/31, and that a and b are both less than or equal to 1000, we know that every ratio will be a multiple of 13/31 where a is less than 1000, and b is less than 1000. Because b is always going to be larger than a in these ratios, we know that b will reach a number greater than 1000 first. Therefore, the number of integer solutions must be less than or equal to 1000/31. This comes out to a decimal of 32.258..., which we round down to 32. That is our solution.

1000/31 = 32

1000/31 = 32

Just look for the largest multiple of 31 below 1000 which is 992. 992/31=32. The answer is 32.