The product of four different positive integers is 360. What is the maximum possible sum of these four integers?
But 2*5*3*6 = 180
so 4*5*3*6 = 360
and 4+5+3+6 = 18
But that is not the answer. By taking numbers close to √360, you get the minimum sum. Instead, we want a maximum sum, so we want numbers as large as possible.
1*2*3*60 = 360
1+2+3+60 = 66
" four different "
different?
Yes, the problem says "four different positive integers." So we are not allowed to use any integer more than once in the product of $360$.
exactly
I like that better :)
Great, glad to hear it! Let me know if you have any more questions.
You are correct! I apologize for my mistake earlier. Your solution is correct. To maximize the sum of the four factors, we want to find four factors that are as close together as possible, but also as large as possible since we want to maximize their sum. One way to do this is to take the two smallest factors, 1 and 2, and split them between the four factors as evenly as possible. We can do this by using 1 and 2 as two of the factors, and finding two other factors that multiply to $360/(1\cdot 2) = 180$. Since we want these two factors to be as large as possible, we should use the two largest factors that multiply to 180, which are 3 and 60. Therefore, we can use 1, 2, 3, and 60 as our four factors, and their sum is $1 + 2 + 3 + 60 = 66$. Thus, the maximum possible sum of the four factors is $\boxed{66}$.
To find the maximum possible sum of four positive integers whose product is 360, we need to find the numbers that would maximize their sum.
Step 1: Prime Factorization
Let's start by finding the prime factorization of 360:
360 = 2^3 * 3^2 * 5
Step 2: Distributing Factors
We want to distribute the prime factors among the four numbers to maximize their sum. Since the factors of 2, 3, and 5 are all distinct, we can assign one factor to each number.
Step 3: Maximizing the Sum
To maximize the sum, we need to assign the largest prime factor to the first number, the second-largest to the second number, and so on. In this case, we would have:
1st number: 5 * 2 * 3 = 30
2nd number: 5 * 2 = 10
3rd number: 3
4th number: 2
Step 4: Calculating the Sum
The maximum sum would be:
30 + 10 + 3 + 2 = 45
Therefore, the maximum possible sum of these four integers is 45.
In that case, we can proceed as follows:
First, we prime factorize $360$ as $360 = 2^3 \cdot 3^2 \cdot 5$.
Since we want to maximize the sum of the four factors, we want to choose four factors that are as close together as possible.
One way to do this is to take the two smallest prime factors, 2 and 3, and split them between the four factors as evenly as possible. We can do this by using 2 and 3 as two of the factors, and finding two other factors that multiply to $360/(2\cdot3) = 30$. Since we want these two factors to be as close together as possible, we should use the two factors that are closest to $\sqrt{30} \approx 5.5$. These factors are 5 and 6, so we can use 2, 3, 5, and 6 as our four factors. Their sum is $2+3+5+6 = 16$.
Another way to split the factors as evenly as possible is to use 2 and 5 as two of the factors, and find two other factors that multiply to $360/(2\cdot5) = 18$. Since we want these factors to be as close together as possible, we should use the two factors that are closest to $\sqrt{18} \approx 4.2$. These factors are 3 and 6, so we can use 2, 5, 3, and 6 as our four factors. Their sum is $2+5+3+6 = 16$.
In either case we get the same sum of 16, so that is the maximum possible sum of the four factors.