The product of four different positive integers is 360. What is the maximum possible sum of these four integers?
9 answers
We can begin by prime factorizing 360: $360=2^3\cdot3^2\cdot5$. To maximize the sum of the four factors, we want to make the factors as close to each other as possible. Thus, we want to choose four factors that are as close together as possible, but whose product is still 360. Since $5$ is the largest prime factor, we can make two pairs of factors by choosing two factors from the set $\{2,3\}$ and two factors from the set $\{5\}$. The pairs are $(2,3)$ and $(2,3)$, or $(2,5)$ and $(3,3)$. For the first, the sum is $2+3+2+3=10$, and for the second, the sum is $2+5+3+3=13$. Thus, the maximum possible sum of the four factors is $\boxed{13}$.
" four different "
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
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.
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.
I like that better :)
Great, glad to hear it! Let me know if you have any more questions.
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
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
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 $\boxed{66}$.
1 answer