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 \(\boxed{15}\).