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# Help please! Simon's Favorite Factoring Trick

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How many integers $$n$$ with $$70 \leq n \leq 90$$ can be written as $$n=ab+2a+3b$$ for at least one ordered pair of positive integers $$(a,b)$$?

This question really confused me because I don't understand what it's asking.

♥️ Thank you so much for your time and help! ♥️

Aug 13, 2024

#1
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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).

Aug 13, 2024
#2
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Thank you so much for your help, but that didn't work... I'm wondering where you got the $$-6$$ from in "n = a(b + 2) + 3(b + 2) - 6", and if that might be the problem? Could it be that you added it to one side, but not the other?

anonymousSingularity  Aug 13, 2024
#3
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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! :)

Aug 13, 2024
edited by NotThatSmart  Aug 13, 2024
#5
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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 $$\boxed{12}$$.

Aug 13, 2024
#6
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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 $$\boxed{15}$$.

Aug 13, 2024
#7
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it comes from the MathCounts 2022 National Sprint. Question #23.

Aug 20, 2024