Answer:
42
Explanation:
You want the largest possible value of n such that some of the numbers 1–10 have a sum of n, and the remaining numbers have a product of n.
2 numbers
The sum of the numbers 1–10 is 55. If x and y are two distinct numbers that have a product of n, then the remaining numbers will have a sum of 55 -x -y. We require the sum and product be equal, so ...
x·y = n
55 -x -y = n
Solving for y in terms of x, we find ...
x·y = 55 -x -y
x·y +x +y +1 = 56 . . . . . add 1+x+y
(x +1)(y +1) = 56
Factor pairs of 56 are ...
56 = 1·56 = 2·28 = 4·14 = 7·8
Since we require both x and y be 10 or less, the only useful pair here is 7·8. This tells us there is one possible 2-number solution: (x+1, y+1) = (7, 8), or (x, y) = (6, 7). The value of n is ...
n = 6·7 = 42
42
There is another possible solution with n = 42. We know that the digits 1, 2, and 3 have both a sum and a product of 6. So, solutions for which n = 42 are ...
{6, 7} and {1, 2, 3, 4, 5, 8, 9, 10}
{1, 2, 3, 7} and {4, 5, 6, 8, 9, 10}
More numbers
Suppose we add a number k to the list, along with x and y. Now we require ...
kxy = 55 -x -y -k
(x +1)(y +1) = (56 -k)/k
Clearly, the smallest possible value of k will give the largest product xy. For k = 1, we have
(x +1)(y +1) = (56 -1)/1 = 55 = 5·11 (x, y) = (4, 10)
For this solution, we have ...
{k, x, y} = {1, 4, 10} . . . n = 40
Larger values of k will necessarily result in smaller values of n (if there is a solution).
All solutions
An exhaustive search tells us the only two-group splits that will solve this problem are {product set} and {sum set} = ...
{6, 7} and {1, 2, 3, 4, 5, 8, 9, 10} . . . n = 42
{1, 4, 10} and {2, 3, 5, 6, 7, 8, 9} . . . n = 40
{1, 2, 3, 7} and {4, 5, 6, 8, 9, 10} . . . n = 42
The largest possible value of n is 42.
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