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How many integers n with 70n90 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
avatar+1776 
0

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
avatar+9 
0

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
avatar+1953 
+1

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+3bn=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(a+3)(b+2)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
avatar+1262 
0

To solve for the number of integers n within the interval 70n90 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=m2. Substituting this back into our formula gives:

n=a(m)+3(m2)=am+3m6=(a+3)m6.

Now, we can rearrange this to isolate m:

n+6=(a+3)mm=n+6a+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, a1 implies a+34.

 

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

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

- For k=4: n+6=76, n0mod4 gives n=70,74,78,82,86,90.

- For k=5: n+6=761mod5 means n1mod5, giving n=74,79,84,89.

- For k=6: n+6=764mod6 means n2mod6, yielding n=72,78,84,90.

- For k=7: n+6=766mod7 which gives n1mod7 as n=70,77,84,91.

- For k=8: n+6=764mod8 gives n2mod8, yielding n=70,78,86.

- For k=9: n+6=764mod9 gives n5mod9, 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 70n90 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 12 integers. 

Therefore, the number of integers n that can be expressed in the desired form is 12.

 Aug 13, 2024
 #6
avatar+9 
0

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 76n96 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 n1 and n2 with n1n2=n and n1>n23. Otherwise, we wouldn't be able to satisfy the condition that a and b are positive integers. That is, n=78 works because 78=263=(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 412 and 821, 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 n1 and n2 that satisfy n1n2=n and n1>n23.

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 15.

 Aug 13, 2024
 #7
avatar+49 
0

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

 Aug 20, 2024

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