My trip to New York was a lot of fun. Some friends from Kentucky were visiting me this weekend, and that was fun too. But in all the chaos Sunday Chess Problem ended up taking the week off. Sorry about that! It will return next week.

POTW, on the other hand, is not taking the week off. Alas, this is the last one for the semester. No more POTW until the fall. It is a great sadness, but try to muddle through.

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890

890 is correct.

The key is to notice that if n is a multiple of 10, then n^3 + 100 will be a multiple of 100. So we look for other numbers, relatively prime to 100, that can be additional divisors. We notice that for any m such that m mod 9 = 8, it will also be true that m^3 mod 9 = 8, and 100 mod 9 = 1. So we try 890, and find that 890^3 + 100 = 704969100. The digits in that number sum to 27, which is a multiple of 9, and since it ends with two zeroes, it must also be a multiple of 100. Therefore 890^3 + 100 is indeed divisible by 890 + 10, and to verify we find that 704969100 / 900 = 783299.

If we look for other numbers of the form 100m -10, we find that 890 is the largest that satisfies (100m - 10) mod m = m - 1. So it is the desired solution.

I haven't tried all of the possibilities, so there may be some I have overlooked, but among the smaller positive integers that work are 2, 5, 10, 15, 40, 90, and 290. The first two I found by trial and error; the others I found by a similar method to the above.

I guess I'm confused. Should the problem be worded as, "...evenly divisible..."?

Not being a math guy, I wrote a Windows PowerShell script to run through some possibilities:

$array = (1..100)

foreach ($n in $array) {

$ncubed = [math]::pow($n,3)

$ncubedplus = $ncubed + 100

$finaln = $ncubedplus / ($n + 10)

"$n ==> $finaln"

}

Every number can be divided by some other number, but only a handful divide evenly. If that's the actual intent, then some modulo step would be needed to find just the whole integers.

@James: It's usually implicit when the problem specifies that you are dealing with integers that "divisible" means evenly divisible.

A C code fragment to do this by brute force (all variables are of type long):

for(i = 1; i < 1000; i++){

icubeplus = i * i * i + 100;

iplus = i + 10;

if(icubeplus mod iplus == 0)

printf("%6d %14d %6d %10d\n",i,icubeplus,iplus,icubeplus/iplus);

}

The script you posted won't find the answer, because as noted above, there are at least two numbers greater than 100 that satisfy the condition (namely, 290 and 890).

Did I oversimplify this one? z^3 + 100= (z^3 + 1000) - 990. The z^3 + 1000 term is divisible by z+10 for any z. Thus the whole ecpression is divisible by z+10 iff 900 is. The largest integer divisible by 900 is of course 900. z+10=900 giving z=890 as the solution.

James, the theme of this semester's POTW is number theory. In number theory, the notion of "divisible" is very precisely defined. An integer a is divisible by another integer b if and only if an integer c exists such that a=bc. That definition precludes any cases where the quotient of the integers is not an integer. For example, 10 is divisible by 2, but not by 4 because 2x5 = 10, but no integer value for c exists such that 4c=10, only the noninteger value 5/2. It is worth noting that an integer is always divisible by itself since for any a there is an integer value for c such that a = ac, namely c=1. I made use of this fact in my POTW solution above.

Eric, referring to my solution, 440 will also be a solution. Any value of x such that 900 is divisible by x+10 will work. Factors of 900 are 2,3,5,6,9,10,12,15,18,20,25,30,36,45,50,60,75,90,100,150,180,300,450 and 900. The list of all solutions is thus -8,-7,-5,-4,-1,0,2,5,8,10,15,20,26,35,40,50,65,80,90,140,170,290,440, and 890.

Eric Lund, thanks for the reply. You're correct that my code would not have found the answer, in that I set my array too small, from 1 to 100. Playing with it more, I set the array higher, and quickly found that 890 was the maximum value.

My code would be modifed as such to find only whole integers:

$array = (1..1000000)

foreach ($n in $array) {

$ncubed = [math]::pow($n,3)

$ncubedplus = $ncubed + 100

$finaln = $ncubedplus / ($n + 10)

if ($finaln % 1 -eq 0) {

“$n ==> $finaln”

}

}

19 solutions to the problem, with 890 being the largest.

I went the other way around, by subbing m=n+10 and recasting the problem as the greatest m such that (m-10)^3 + 100 is divisible by m. The constant term in expanding the cube is -1000, to which 100 is added to get -900. This means that in order for m to divide (m-10)^3 + 100 evenly, it must be a factor of 900, the greatest being 900. Subbing back, n=900-10=890