If $3 a + 5 b = 132$ then the number of possible pairs of $a$ and $b,$ such that $a > b$ and $a$ and $b$ are positive integers is

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Solution

First find out the first pair which satisfy the given equation with minimum value of $b .$
It is $(39, 3)$

The other pairs can be calculated as for the equation, if we increase the value of $b$ by $1$ unit, then the value of $a$ must be decreased by $\frac{5}{3}$ unit.

We need values as positive integers only.
So if the value of $b$ is increased by 3 then the value of $a$ is decreased by $5$
Next pairs are $(34, 6), (29, 9), (24, 12), (19, 15)$
Next pair is $(14, 18)$ but $a > b$
So the last pair will be $(19, 15)$
So there are $5$ pairs which satisfy the given conditions.

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If 3a+5b=132 then the number of possible pairs of a and b, such that a>b and a and b are positive integers is

If $3 a + 5 b = 132$ then the number of possible pairs of $a$ and $b,$ such that $a > b$ and $a$ and $b$ are positive integers is