[Guest]

N is a positive integer such that precisely 8 of its positive divisors sum to 3240. The total number of positive divisors of N may or may not be equal to 8.

Determine the minimum value of N.

Note: For the purposes of the puzzle, each of the positive integers 1 and N is deemed as a valid positive divisor of N.

[Guest]

I've not done this with any algebra - just by a bit of logical thinking. One divisor needs to be as large as possible (say =N) to keep the sum as large as possible and so minimise the product (if that makes sense). So, I've come up with the solution below.

2,3,4,5, and 6 are almost certainly divisors. 10 and 30 added to these makes 60. 3240 - 60 = 3180. N = 3180 is then a possible solution. It seems to me this must be the minimum solution, but there's no proof!

Edited October 3, 2009 by donjar

[Guest]

donjar, that's not the lowest value. For example, 1500 works. 1500+750+500+300+150+20+15+5=3240. I'm sure there is an even lower value still.

[Guest]

donjar, that's not the lowest value. For example, 1500 works. 1500+750+500+300+150+20+15+5=3240. I'm sure there is an even lower value still.

I see my error in thinking - I think. Back to the drawing board1

[Guest]

1440 also works: factors 1440, 720, 480, 360, 180, 48, 8, 4

[Guest]

I've got the answer

The minimum value of N is 1320

Edited October 3, 2009 by 3ala2

[Guest]

The minimum must be greater than 1250, since

(1 + 1/2 + 1/3 + 1/4 + ... + 1/8)x = 3240

has a solution x = 1249.58...

If there is a value of N less than 1320 it must lie somewhere between 1250 and 1320.

[plainglazed]

Probably counter to the spirit of the question/intention of the OP but...

...how about N=405; eight of those is 3240.

[superprismatic]

1260. Since 3240 = 1260 + 63 + 90 + 210 + 252 + 315 + 420 + 630

Edited October 3, 2009 by superprismatic

[Guest]

Is your answer a "hunt and peck" method, SP? Or is KS looking for a rational approach? I personally was flummoxed by this question - hardly unusual, especially on this forum.

[Guest]

An explanation for an approach that would yield SP's answer, that could very well be completely wrong:

In order to get the lowest N, you want to have as many distinct factors as possible, while keeping N in a reasonable range.

2*3*5*7*11 is 5 factors, which is equal to 2310, which is too large, so it must be 4 factors

2*3*5*7 is too small, so you want to now you need to increase, which can be done in two ways:

- make the factors bigger, e.g. turn 7 into 11

- make more of the factors, e.g. turn 2 into 4

Assuming that more factors, and smaller factors, is always better, than...

4*3*5*7 is too small

8*3*5*7 is too small

8*9*5*7 is too big

4*9*5*7 is too just right.

The trick I guess would be true prove my underlined statement

[superprismatic]

Is your answer a "hunt and peck" method, SP? Or is KS looking for a rational approach? I personally was flummoxed by this question - hardly unusual, especially on this forum.

Not quite hunt and peck but not very sophisticated either!

First, I thought that the number must be round. By round

I mean that it has a large number of factors given its

size. 30 is a round number by this definition. So, I

wrote a program to go up from some small number like 10

and print out numbers which had *lots* of factors. Well,

there were so many that I decided to modify the program

to take all subsets of 8 factors and add them up to see if

any added up to 3240. The first number that did this was

1260. In fact, I found that 1260 has so many factors that

you can find subsets of 8 which add to any number between

3239 and 3243, and is also the smallest number having these

sums for subsets of 8. I'm ashamed to say that it was

brute force. I hope KS clues us in if there is a way to

do it which is less shameful!

[Guest]

Yes, SP's answer at post #9 is indeed the answer known to me. The corresponding methodology for solving this puzzle was already provided by PM at post #7

*** For some reason, the spoiler feature does not seem to be working perfectly todaywith my browser.

Edited October 5, 2009 by K Sengupta

[Guest]

Rookie seems to have resurrected the spoiler button! Thanks for the problem, KS. I hope for more in the future, as I find them educational (and, as a bonus, annoying.)