the sum of several positive integers is 20. find the largest product that can be formed by these integers?

Question

.

Answer 1

The answer is 2*3*3*3*3*3*3 = 1458.

That's assuming the numbers don't have to be distinct.

If you have any set of numbers:

a1+a2+...+a_n=20

and one of the values, say a_n, is at least four, then you can get a product of greater or equal value by replacing a_n with the pair (a_n - 2) and 2. So the maximum value comes when all of the values are less than four.

If a1 = 1, you can get a greater value by combining a1 and a2, replacing with a2+a1=a2+1.

So all the values must be greater than 1 to get the maximum value.

Finally, if three of the values are 2, then you can replace those values with two threes: 2+2+2 = 3+3, but 2+2+2<3*3.

So, we now know that the maximum value is achieved when all of the values are 2 and 3, and there are at most two 2s.

But 20 = 3 + 3 + 3 + 3 + 3 + 3 + 2 = 3*6 + 2, and there is no other way to write 20 as a sum of 3s and 2s with one or two 2s.

Answer 2

2 x 3 x 4 x 5 x 6 = 720

Answer 3

20 can be made by adding:

2 + 2 + .... 10 times and the product is 2^10 = 1024

All others will be equal or less than this. For example, 5 + 5 + 5 + 5 and the product is 625, 10 + 5 + 5 (250) etc.

Answer 4

3+3+3+3+3+3+2 = 20
3x3x3x3x3x3x2 = 1458

Answer 5

720 (2*3*4*5*6)