Help please! Simon's Favorite Factoring Trick

Analyzing the Equation and the Range

Understanding the Equation:

We can factor the equation as follows:

n = ab + 2a + 3b

n = a(b + 2) + 3b

n = a(b + 2) + 3(b + 2) - 6

n = (a + 3)(b + 2) - 6

Implications of the Equation:

This means that n + 6 must be a product of two integers greater than 2.

Considering the Range:

We're looking for n between 70 and 90, inclusive.

So, n + 6 will be between 76 and 96, inclusive.

Finding Suitable Pairs

We need to find pairs of integers greater than 2 whose product lies between 76 and 96.

Lower Bound: The smallest product we can form with integers greater than 2 is 3 * 4 = 12. This is far below 76.

Upper Bound: We can start checking larger products.

5 * 16 = 80

6 * 13 = 78

7 * 12 = 84

8 * 11 = 88

9 * 10 = 90

Counting Valid Values of n:

For each of the products above, we can find a corresponding n by subtracting 6.

Therefore, there are 5 possible values of n between 70 and 90 that can be expressed in the given form.

Answer: There are 5 integers n with 70 ≤ n ≤ 90 that can be written as n = ab + 2a + 3b for at least one ordered pair of positive integers (a, b).

Here, let me give this problem a shot. 

First, let's try to simplify this problem a bit. We can use Simon's Favorite Factoring Trick. 

Let's factor the right side a bit for n. We get

$n = ab + 2a + 3b \\ n = a(b + 2) + 3b$

Now, since we want b to be factored as well, let's add 6 to both sides. This will come in handy later. 

We get

$n + 6 = a(b + 2) + 3(b + 2) \\ n +6 = (a + 3)(b + 2)$

Now, this problem has transformed for numbers that match the conditions

$76\leq (a+3)(b+2) \leq 96$

The only numbers that don;t satsify these conditions are numbers with only two factors and one is less than 3. Thus, we can calculate each number. Here, I did it for you! :)

Thus, since there are a total of 21 numbers and 7 are not valid, we have 21 - 7 = 14 as our final answer. 

So 14 should be the answer. I'm not confident on the answer, I might have errored a bit. However, I think it's CLOSE to the right answer. Maybe 12 or 15, but 14 should be right. 

Thanks! :)

To solve for the number of integers $n$ within the interval $70 \leq n \leq 90$ that can be expressed in the form

$$n = ab + 2a + 3b,$$

we can rearrange the equation for $n$:

$$n = ab + 2a + 3b = a(b + 2) + 3b.$$

Now, we can let $m = b + 2$. Then, we can rewrite $b$ in terms of $m$ as $b = m - 2$. Substituting this back into our formula gives:

$$n = a(m) + 3(m - 2) = am + 3m - 6 = (a + 3)m - 6.$$

Now, we can rearrange this to isolate $m$:

$$n + 6 = (a + 3)m \implies m = \frac{n + 6}{a + 3}.$$

Since $m$ is a positive integer, $n + 6$ must be divisible by $a + 3$. We can explore which integers give permissible $m$ values by determining values of $n + 6$:

- The smallest $n$ is $70$, so $n + 6 = 76$.

- The largest $n$ is $90$, so $n + 6 = 96$.

Consequently, we need to examine the integers from $76$ to $96$:

$$76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96.$$

Next, we can determine the possible values of $a + 3$. Since $a$ is a positive integer, $a \geq 1$ implies $a + 3 \geq 4$.

Therefore, $m$ can take any integer value for $a + 3 \in \{4, 5, 6, \ldots\}$.

To find integers $n$, we need $n + 6$ to be divisible by each $a + 3$:

- For $k = 4$: $n + 6 = 76$, $n \equiv 0 \mod 4$ gives $n = 70, 74, 78, 82, 86, 90$.

- For $k = 5$: $n + 6 = 76 \equiv 1 \mod 5$ means $n \equiv -1 \mod 5$, giving $n = 74, 79, 84, 89$.

- For $k = 6$: $n + 6 = 76 \equiv 4 \mod 6$ means $n \equiv 2 \mod 6$, yielding $n = 72, 78, 84, 90$.

- For $k = 7$: $n + 6 = 76 \equiv 6 \mod 7$ which gives $n \equiv 1 \mod 7$ as $n = 70, 77, 84, 91$.

- For $k = 8$: $n + 6 = 76 \equiv 4 \mod 8$ gives $n \equiv 2 \mod 8$, yielding $n = 70, 78, 86$.

- For $k = 9$: $n + 6 = 76 \equiv 4 \mod 9$ gives $n \equiv 5 \mod 9$, yielding $n = 74, 83, 92$.

Now, we can collect all the possible $n$:

From $k = 4$: $70, 74, 78, 82, 86, 90$

From $k = 5$: $74, 79, 84, 89$

From $k = 6$: $72, 78, 84, 90$

From $k = 7$: $70, 77, 84, 91$

From $k = 8$: $70, 78, 86$

From $k = 9$: $74, 83, 92$

Next, let's find unique $n$ from all these lists:

- Compiling unique values from the sets we found: $70, 72, 74, 77, 78, 79, 82, 83, 84, 86, 89, 90, 91, 92$.

Thus, the unique values of $n$ between $70 \leq n \leq 90$ are:

$$70, 72, 74, 77, 78, 79, 82, 83, 84, 86, 89, 90.$$

Counting these gives us:

$$70, 72, 74, 77, 78, 79, 82, 83, 84, 86, 89, 90 \Rightarrow \text{ 12 integers. }$$

Therefore, the number of integers $n$ that can be expressed in the desired form is $\boxed{12}$.

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 $\boxed{15}$.

it comes from the MathCounts 2022 National Sprint. Question #23.