Determine the maximum value of a prime number x ≤ 999, such that Y has precisely 42 distinct positive integer divisors (including 1 and Y), where:
Y = x(x+1)2.
now for a number whose prime factorization is
p1^k1*p2^k2*...*pn^kn the number of divisors is given by
(k1+1)*(k2+1)*...*(pn+1)
no with x a prime number and
Y=x(x+1)^2
assume that x+1 has prime factorization of
p1^k1*p2^k2*...*pn^kn
then the number of divisors of Y is given by
2*(2k1+1)*(2k2+1)*...*(2pn+1) = 42
(2k1+1)*(2k2+1)*...*(2pn+1)=21
now 21=3*7 is the only way to factor 21 into factors greater than 2 thus we have that n=2 and k1=1 and k2=3 thus we have
x+1=p1*p2^3
now with the exception of x=2, for all prime values of x, x+1 is even. Since we are looking for the maximal value of x then we can temporarily ignore x=2 unless we are unable to find a solution with x>2.
if x>2 then x+1 is even and thus either p1=2 or p2=2
if p1=2 then we have
x+1=2p2^3
x=2p2^3-1
since x<=999 then we have
2p2^3-1<=999
2p2^3<=1000
p2^3<=500
p2<=7
so p2=3,5, or 7
if p2=3 then we have x=2*27-1=53 which is prime
if p2=5 then we have x=2*125-1=249 which is not prime
if p2=7 then we have x=2*343-1=685 which is not prime
thus it only works for p2=3 which gives x=53
if instead p2=2 then we have
x+1=8p1
x=8p1-1
now we want x<=999
thus
8p1-1<=999
8p1<=1000
p1<=125
now there are only 30 primes less than 125, working backwards from the last one we find the largest possible value for x if p2=2 is x=823 when p1=103
this is greater than what p1=2 gave of x=53
thus the maximal value for x<=999 is x=823
and to confirm
y=823*824^2=558797248
824=2^3*103
thus
y=2^6*103^2*823
thus the 42 divisors of y are
1, 2, 4, 8, 16, 32, 64, 103, 206, 412, 823, 824, 1646, 1648, 3292,
3296, 6584, 6592, 10609, 13168, 21218, 26336, 42436, 52672, 84769, 84872, 169538, 169744, 339076, 339488, 678152, 678976, 1356304, 2712608, 5425216, 8731207, 17462414, 34924828, 69849656, 139699312, 279398624, 558797248
Edited on June 17, 2010, 4:15 pm
|
|
Posted by Daniel
on 2010-06-17 13:03:40 |
analytical solution
