Thank you all for the help! Sadly, both 5 and 14 were wrong, and the right answer was 15 lol. Here's the problem explanation that I got after getting the problem wrong:
We use Simon's Favorite Factoring Trick: observe that \(ab+2a+3b=a(b+2)+3b\), so to obtain another factor of \(b+2\), we need to add \(6\). So \(ab+2a+3b+6=a(b+2)+3(b+2)=(a+3)(b+2)\), and \(ab+2a+3b=(a+3)(b+2)-6\).
Thus we need to find how many integers \(n\) with \(76\le n\le96\) can be written as \((a+3)(b+2)\) for at least one ordered pair of positive integers \((a,b)\). Any \(n\) that works must have the property that there exists positive integers \(n_1\) and \(n_2\) with \(n_1n_2=n\) and \(n_1>n_2 \ge 3\). Otherwise, we wouldn't be able to satisfy the condition that \(a\) and \(b\) are positive integers. That is, \(n=78\) works because \(78 = 26 \cdot 3 = (23 + 3)(1 + 2)\), but \(n = 82\) doesn't because the only possible ways to decompose \(82\) as a product of two positive integers are \(41 \cdot 2\) and \(82 \cdot 1\), and in neither of those cases can both \(a\) and \(b\) be positive integers.
The rest is a brute force check: we need to eliminate prime numbers, as they definitely don't allow us to have \(a\) and \(b\) be positive integers, and we need to eliminate numbers of the form \(2p\), where \(p\) is prime. Every other number has a pair of factors \(n_1\) and \(n_2\) that satisfy \(n_1n_2=n\) and \(n_1>n_2 \ge 3\).
We complementary count: \(79, 83, 89\) are primes, and \(82, 86, 94\) are twice a prime. Our answer is thus the total number of integers \(n\) between \(76\) and \(96\) minus \(6\), or simply \(\boxed{15}\).
