Show that for any positive integer n, 5^(2n + 1) + 11^(2n + 1) + 17(2n + 1) is divisible by 33.?

using modular arithmetic

we know 5 = -1 mod 3

so 5^(2n+1) = - 1 mod 3 as (2n+1) is odd

simliarly

11^(2n+1) = -1 mod 3

17^(2n+1) = -1 mod 3

so adding
5^(2n + 1) + 11^(2n + 1) + 17^(2n + 1) = -1 -1 -1 or -3 mod 3 = 0

now a^n+b^n is divisble by (a+b) when n is odd

so 5^(2n+1) + 17^(2n+1) is divisble by 22 and hence 11
as
11^(2n+1) is divsible ny 11
so
5^(2n + 1) + 11^(2n + 1) + 17^(2n + 1) is divisble by 11

as 5^(2n + 1) + 11^(2n + 1) + 17^(2n + 1) is divisible by 3 and 11 and 3 and 11 are primes so is divisible by 3*11 or 33

hey mate,

I will employ mathematical induction
here we wish to prove
5^(2n + 1) + 11^(2n + 1) + 17(2n + 1) is divisible by 33
or
5^(2n + 1) + 11^(2n + 1) + 17^(2n + 1) = 33k , k being an integer
or
( 5^(2n + 1) + 11^(2n + 1) + 17^(2n + 1) ) mod(33) = 0

Mathematical Induction
Step (1)
Proof True for n = 1
5^(2(1) + 1) + 11^(2(1) + 1) + 17^(2(1) + 1)
= 5^3 + 11^3 + 17^(3)
Hence,

( 5^3 + 11^3 + 17^(3) ) mod 33 = 0
Thus true for n = 1

Step (2) (Inductive Step)
Assume true for n = k
( 5^(2k + 1) + 11^(2k + 1) + 17^(2k + 1) ) mod(33) = 0
or,
5^(2k + 1) + 11^(2k + 1) + 17^(2k + 1) = 33m ; m being an integer

Step (3)
Prove true for n = k + 1, i.e. Prove
( 5^(2(k+1) + 1) + 11^(2(k+1) + 1) + 17^(2(k+1) + 1) ) mod(33) = 0
( 5^(2k + 3) + 11^(2k+3) + 17^(2k+3) ) mod(33) = 0
Consider,
5^(2k + 3) + 11^(2k+3) + 17^(2k+3)
= (5^(2))(5^(2k + 1)) + (11^(2))(11^(2k + 1)) + (17^(2))(17^(2k + 1))
from the inductive step
5^(2k + 1) + 11^(2k + 1) + 17^(2k + 1) = 33m
Thus,
5^(2k + 1) = 33m - 11^(2k + 1) - 17^(2k + 1)
11^(2k + 1) = 33m - 5^(2k + 1) - 17^(2k + 1)
17^(2k + 1) = 33m - 5^(2k + 1) - 11^(2k + 1)
Thus,
(5^(2))(5^(2k + 1)) + (11^(2))(11^(2k + 1)) + (17^(2))(17^(2k + 1))
Becomes,
(5^(2))( 33m - 11^(2k + 1) - 17^(2k + 1) ) + (11^(2))(33m - 5^(2k + 1) - 17^(2k + 1)) + (17^(2))(33m - 5^(2k + 1) - 11^(2k + 1))
=( 5^(2) + 11^(2) + 17^(2) )( 33m - ( 5^(2k + 1) + 11^(2k + 1) + 17^(2k + 1) )
=( 5^(2) + 11^(2) + 17^(2) )( 33m - 33m)
=0
=0mod(33) = 0
Thus true for n = k + 1 and by Mathematical Induction is true for all natural numbers n.

Hope this helps,

David

Why do you print so carelessly? Is it 43n+83×92^(3n-1) or 43n+83×(92^3n)-1 oe is it 43(n+83)×92^(3n-1)? There are more possiblities as well.