Problem 7: show that 4^n + 4^n+1 + 4^n+2 + 4^n+3 + 4^n+4 is divisible by 11 for any positive integer

If n is even, n

4

+4

n

is divisible by 4

∴ It is composite number

If n is odd, suppose n=2p+1, where p is a positive integer

Then n

4

+4

n

=n

4

+4.4

2p

=n

4

+4(2

p

)

4

which is of the form n

4

+4b

4

, where b is a positive integer (=2

p

)

n

4

+4b

4

=(n

4

+4b

2

+4b

4

)−4b

2

=(n

2

−2b

2

)

2

−(2b)

2

=(n

2

+2b+2b

2

)(n

2

−2b+2b

2

)

We find that n

4

+4b

4

is a composite number consequently n

4

+4

n

is composite when n is odd.

Hence n

4

+4

n

is composite for all integer values of n> 1.

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation:

600,000,000,000, or 6e14

step-by-step explanation: