Question

what is the largest integer that is the divisor of (p+1)(p+3)(p+5)(p+7)(p+9) for all positive integer p.​

Answer 1

Step-by-step explanation:

1: If n is even ie if n = 2m

P(2m) = (2m+1)(2m+3)(2m+5)(2m+7)(2m+9)

Product of the constants = 1*3*5*7*9

=> p(2m) is divisible by 3,5,7,9 •••••••(1)

2: If n is odd

P(2m+1) = (2m+2)(2m+4)(2m+6)(2m+8)(2m+10)

Product of the constants

= 2 * 4 * 6 * 8 * 10 •••••••••••••(2)

=> p(2m+1) divisible by 2,4,6,8,10

Now, find GCD of (1) & (2)

GCD = 3 x 5 = 15

=> largest integer divisor of p(n) = 15

for n being odd (2k-1):

p(n)=2k(2k+2)(2k+4)(2k+6)(2k+8)

=32*k(k+1)(k+2)(k+3)(k+4)

=32*5 consecutive numbers

which will be divisible by 32*(5!)=3840

for n being even (2k)

p(n)=(2k+1)(2k+3)(2k+5)(2k+7)(2k+9)

at least 1 of the terms in the product will be divisible by 3 and at least one by 5

hence it will be divisible by 15.

Therefore, p(n) will be divisible by 15 for all positive n

So if n is odd …

So p(n)=25k(k+1)(k+2)(k+3)(k+4)

If n is even …

So p(n)=(2k+1)(2k+3)(2k+5)(2k+7)(2k+9)

So the largest divisor is 15