[Guest]

Determine the probability that for a base ten positive integer N drawn at random between 2 and 101 inclusively, the number N³ + 1 is expressible as the product of three distinct positive integers that correspond to three consecutive terms of an arithmetic progression.

Edited December 27, 2009 by K Sengupta

[Guest]

the question is really asking "for how many n greater than 1 and less than 102 is n^3+1 divisible by 3?"

because an arithmetic progression can be written for the sum of 3 terms as (a-b)+a+(a+b)=3a

This is true for all n that is congruent to 2 mod 3.

Thus the odds of such an occurrence over any interval is approximately one third.

However, in this particular interval, because both the first and last terms are congruent to 2 mod 3, there are 34 such n on this interval spanning 100 terms. thus 34/100 or 34% is the appropriate answer.

[Guest]

Probability is zero.

how do you factor n^3 + 1?

(n + 1) * (n^2 - n + 1). The latter has no integer roots.

Edited December 28, 2009 by actualry

[Guest]

the question is really asking "for how many n greater than 1 and less than 102 is n^3+1 divisible by 3?"

because an arithmetic progression can be written for the sum of 3 terms as (a-b)+a+(a+b)=3a

This is true for all n that is congruent to 2 mod 3.

Thus the odds of such an occurrence over any interval is approximately one third.

However, in this particular interval, because both the first and last terms are congruent to 2 mod 3, there are 34 such n on this interval spanning 100 terms. thus 34/100 or 34% is the appropriate answer.

The true intent of the OP is:

"Determine the probability that for a base ten positive integer N drawn at random between 2 and 101 inclusively, the number N^3+1 = p*q*r, where p, q, r are three distinct positive integers such that p, q and r (in this order) corresponds to three consecutive terms of an arithmetic progression."

Edited December 28, 2009 by K Sengupta

[Guest]

Probability is zero.

how do you factor n^3 + 1?

(n + 1) * (n^2 - n + 1). The latter has no integer roots.

Since n is a positive integer drawn at random between 2 and 101 inclusively, there will always exist some values of n for which the said factorization is possible.

For example, for n = 4 we observe that n^3 + 1 = 65, which is the product of the positive integers 1, 5 and 13. However, this fails the second criteria as 1, 5, 13 does not correspond to three consecutive terms of an arithmetic sequence.

[superprismatic]

The probability is 0.05:

Here are the 5 out of the hundred:

N=3; N³+1=28 = 1*4*7

N=12; N³+1=1729 = 7*13*19

N=27; N³+1=19684 = 19*28*37

N=48; N³+1=110593 = 37*49*61

N=75; N³+1=421876 = 61*76*91

[Guest]

Since (n3+1) = (n+1) (n2-n+1) = (n+1) ({n+1}2 - 3n).

So if 3n = r2 for some integer "r" then three integers in AP are (n+1-r), (n+1), (n+1+r). So definite solutions atleast are 3, 12, 27, 48,75. There may be other solutions also but for given range these are the only solutions.

[Guest]

the question asks N^3 + 1= (X-1)(X)(X+1) = X^3 - X

so just go to the polynomial solver and enter

1X^3 + 0X^2 - 1X - {9,28,65 ....}

for all N^3 + 1 | N=(2,101)

if the solver gives an interger as an answer than that is an answer.

count the number of answers and divide by 99

[Guest]

it seems to be impossible because for N=101 N^3+1~=100*101*102 but not quite. As N increases the answers get closer together but never match. I say it is 0% probability for the range (2,101) or any number higher for that matter.

Edited January 3, 2010 by gkibarricade

[Guest]

sorry, ignore my previous posts i misunderstood the question

[Guest]

If and only if N is 3 times a perfect square, then N^3+1 will be a multiple of three terms in an arithmetic progression.