show that 27 times 23(n) + 17 times 10(2n) is divisible by 11 for all possible integers n.?

Question

the pronumerals in the brackets represent powers.

Answer 1

F(n) = 27*23^n + 17*10^(2n) is divisible by 11

Check if F(1) is divisible by 11
F(1) = 27*23 + 17*100
= 621 + 1700
= 2321
= 11*211

F(1) is divisible by 11

Assume F(m) is divisible by 11
F(m) = 27*23^(m) + 17*10^(2m)
F(m+1) = 27*23^(m+1) + 17*10^(2(m+1))

F(m+1) - F(m) = 27 * 23^m *[23 - 1] + 17*10^(2m)*[10^2 - 1]
= 27 * 23^m * 22 + 17 * 99 * 10^(2m)
= 11[27 * 23^m * 2 + 17 * 9 * 10^(2m)]

F(m+1) - F(m) is divisible by 11
If F(m) is divisible by 11, them F(m+1) will be divisible by 11

F(1) is divisible by 11
Therefore F(2) must be divisible by 11....

F(n) is divisible for all integral values of n

Answer 2

The word is "positive," not "possible"!
27 x 23^n + 17 x 10^2n = 27 x 23^n + 17 x 100^n =
27 x (2 x 11 + 1)^n + 17 x (9 x 11 + 1)^n;

Since all the cross-multiplication components of raising either of these to a power still leaves one more than a multiple of 11 in each case, we can ignore the parts that are perfect multiples of 11, which multiplied by anything else, will still be multiples of 11, so we're left with 27 x 1 + 17 x 1 = 44 = 4 x 11.

QED, in a roundabout sort of way.

Answer 3

I am assuming that 23(n) is 23 to the power n (23^n)

when n = 1, 27x23 + 17x 100 = 2321 which is divisible by 11

By math induction if n is true and n+1 is true and 1 is true the statement is true

so evaluate 27 x 23^(n+1) + 17 x 10^(2n+2)
= 27 x 23^n x 23 + 17 x 10^2n x 100
= 27 x 23^n x (100 - 77) + 17 x 10^2n x 100
= 100(27 x 23^n + 17 x 10^2n) - 77 x 27 x 23^n
this first expression is divisible by 11
and since 77 is divisible by 11 the second term is also.

So this statement is always true.