Tricky math question, smart people answer please!!!!!?

I've seen a problem like this on a contest. Here's my stab at it:

I can prove with certainty that the divide for the Maximum Product must contain only 3's and 2's. 4 is optional, because 4*(n-4) = 2*2*(n-4), so both forms are acceptable.

Consider taking a 5 as one of the factors. Isn't 2*3 = 6 better, and will get a higher product?
In fact, for any even number n = 2k, n > 4, you have k^2 > 2k, and for any odd number n = 2k+1, n > 3, you have k*(k+1) > 2k+1. Therefore, for any possible addends > 4, it is better to split them up to achieve a greater product.

So, to get the greatest product, you need to find integers a and b such that 2a + 3b = n, and 2^a*3^b is maximized. Solving the first equation, we get a = (n - 3b)/2, and by substitution, (along with converting to the natural exponent):

2^a*3^b = e^(ln(2)*(n-3b)/2 + ln(3)*b)

Now we need to maximize the expression b*(ln(3) - 3*ln(2)/2)) + n*ln(2)/2

Well, turns out that ln(3) - 3*ln(2)/2 is a positive value, approximately 0.058892, so the larger b is, the larger our product is. Therefore, the optimal solution is obtained by taking as many 3's as possible. If the remainder is 2, just take the 2, if it is 0, you're good, but if it is 1, you need to kill a 3 and take two 2's instead.

So the Maximum Product for 1000 is 3^332*2*2, which is a 160 digit number that I am not typing out.

Hope that helps

i believe it is a fibbonacci sequence. In fibbonacci, the final 2 numbers upload to make the subsequent interior the sequence. case in point, a million,a million,2 could be a million+a million=2 etc. c+d=0, b+c=d, a+b=c, a=-d-b. in case you have gotten that far...

If you generalize that to include real numbers, so that we can write it as f(x) = x^(1000/x) then:

f(x) = e^(1000lnx/x)
f'(x) = (1000 - 1000lnx)/x² * e^(1000lnx/x) = 0
0 = [1000e^(1000lnx/x) - 1000lnxe^(1000lnx/x)] / x^2
0 = 1000e^(1000lnx/x) - 1000lnxe^(1000lnx/x) {x != 0}
e^(1000lnx/x) = lnxe^(1000lnx/x)
lnx = 1
x = e

then your best "product" is
e^(1000/e)

e lies between 2 and 3 (closer to 3) so the best integer product would be as many 3s as possible and the rest 2s, no other digits.

3^332 * 2^2 = 1.015e+159
for comparison
3^(1000/3) = 1.097e+159
and
e^(1000/e) = 5.861e+159