How many integers $n$ with $70 \leq n \leq 90$ can be written as $n = ab + 3a + 7b$ for at least one ordered pair of positive integers $(a, b)$?

learnmgcat Jun 4, 2024

#1**0 **

Let's rewrite the equation $n = ab + 3a + 7b$ as $n = a(b+3) + 7b$.

We can factor the equation as $n = (a+7)(b+3) - 21$.

Since $70 \leq n \leq 90$, we have $49 \leq (a+7)(b+3) \leq 77$.

Factoring the bounds, we get $49 = 7 \times 7$ and $77 = 11 \times 7$.

So the possible values for $(a+7)(b+3)$ are $49, 56, 63, 70, 77$.

The factors of these numbers are:

$49$: $(1, 49), (7, 7)$

$56$: $(1, 56), (2, 28), (4, 14), (7, 8)$

$63$: $(1, 63), (3, 21), (7, 9)$

$70$: $(1, 70), (2, 35), (5, 14), (7, 10)$

$77$: $(1, 77), (7, 11)$

Therefore, there are $\boxed{10}$ integers $n$ with $70 \leq n \leq 90$ that can be written as $n = ab + 3a + 7b$ for at least one ordered pair of positive integers $(a, b)$.

eramsby1O1O Jun 4, 2024