How many integers n with 70≤n≤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! ♥️
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).
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?
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! :)
To solve for the number of integers n within the interval 70≤n≤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⟹m=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, a≥1 implies a+3≥4.
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, n≡0mod4 gives n=70,74,78,82,86,90.
- For k=5: n+6=76≡1mod5 means n≡−1mod5, giving n=74,79,84,89.
- For k=6: n+6=76≡4mod6 means n≡2mod6, yielding n=72,78,84,90.
- For k=7: n+6=76≡6mod7 which gives n≡1mod7 as n=70,77,84,91.
- For k=8: n+6=76≡4mod8 gives n≡2mod8, yielding n=70,78,86.
- For k=9: n+6=76≡4mod9 gives n≡5mod9, 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≤n≤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⇒ 12 integers.
Therefore, the number of integers n that can be expressed in the desired form is 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≤n≤96 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>n2≥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⋅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⋅2 and 82⋅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 n1 and n2 that satisfy n1n2=n and n1>n2≥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 15.