[Guest]

Determine the maximum value of a prime number x <= 999, such that Y has precisely 42 positive integer divisors (including 1 and Y), where Y = x(x+1)²

Edited January 26, 2010 by K Sengupta

[Guest]

823 - via a tuned Excel loop macro to do all the testing

[Guest]

823 - via a tuned Excel loop macro to do all the testing

Y = x(x + 1)^2

We know that x is a prime number less than 999 and Y has 42 factors. We also know that x and (x + 1) are coprime to each other.

So, number of factors of Y = 2*{no. of factors of (x + 1)^2}, hence

no. of factors of (x + 1)^2 = 21 = 3*7.

As x is prime (x + 1) can not be prime.

Hence (x + 1)^2 is in the form of a^6*b^2

=> (x + 1) = a^3*b, where a and b are prime numbers.

Since (x + 1) < 1000.

Possible values of a are 2, 3, 5, 7

When a = 7, only possible value for b = 2

=> (x + 1) = 686, x = 685(not possible as x is a prime number)

When a = 5, possible values of b = 2, 3, 7

=> (x + 1) = 250, 375, 875, x = 249, 374, 874(none is possible as none of them is prime)

When a = 3,

for b = 2

(x + 1) = 54, x = 53(possible)

All other values of b are odd numbers, so x will be even which is not possible.

When a = 2, b should be less than 125 and should be prime.

b = 113, (x + 1) = 904, x = 903(not possible)

b = 109, (x + 1) = 872, x = 871 = 13*67(not posible)

b = 107, (x + 1) = 856, x = 855(not possible)

b = 103, (x + 1) = 824, x = 823

Hence largest possible value of x = 823