Seek help on divisibility?

Question

show that 27 x 23^n + 17 x 10^2n is divisble by 11 for all positive integers n.

Appreciate if anyone can assit. Thanks

Answer 1

Using the MOD method...

27 = 22 + 5 = 2 mod 5
23 = 22 + 1 = 2 mod 1
17 = 11 + 6 = 1 mod 6
10^2 = 100 = 99 + 1 = 9 mod 1

Just replace the numbers in the expression with the mods.

27 x 23^n + 17 x (10^2)^n
= 5 * 1^n + 6 * 1^n
= 5 + 6
= 11

11 is divisible by 11, so the expression is divisible by 11.

Answer 2

This is just an exercise in modular arithmetic. The remainder of an expression of sums, products, powers etcetera is the same as the sums, products, powers etcetera of the remainders. Reducing things to their remainders relative to 11, we proceed:

27 x 23^n + 17 x 100^n
= 5 x 1^n + 6 x 1^n
= 5 x 1 + 6 x 1
= 11 = 0.

Indeed, we see that 27 x 23^n + 17 x 10^2m is divisible by 11 for all pairs of positive integers n and m. Perhaps the question used 'n' twice to try to mislead you into thinking the matched exponents should be used to simplify the expression - but even if they could, it is completely unnecessary, as you can see.

Answer 3

This is just an exercise in modular arithmetic. The remainder of an expression of sums, products, powers etcetera is the same as the sums, products, powers etcetera of the remainders. Reducing things to their remainders relative to 11, we proceed:

27 x 23^n + 17 x 100^n
= 5 x 1^n + 6 x 1^n
= 5 x 1 + 6 x 1
= 11 = 0.

Indeed, we see that 27 x 23^n + 17 x 10^2m is divisible by 11 for all pairs of positive integers n and m. Perhaps the question used 'n' twice to try to mislead you into thinking the matched exponents should be used to simplify the expression - but even if they could, it is completely unnecessary, as you can see.

Answer 4

Try with n=0:

27 * 23^0 + 17*10^(2 * 0) = 27*1 + 17*1 = 44 = 11*4

Assume for any non-negative n, and prove for n+1:

There is a positive integer k, such that:
11k = 27*23^n + 17 * 10^2n

27*23^(n + 1) + 17*10^[2(n + 1)] =
= 27*23^(n + 1) + 17 * 10^(2n + 2) =
= 27*23^n * 23 + 17 * 10^2n * 100 =
= 27*23^n * 23 + 17 * 10^2n * (23 + 77) =
= 23*(27*23^n + 17*10^2n) + 77*10^2n =
= 23*11k + 77*10^2n

77*10^2n is divisible by 11.

QED

Answer 5

Use the proving method to proof it then.

First, start by choosing any positive integer of you choice and sub it into the equation. (Choose your lucky number, which can be 1 or 9999999999. :P )
Prove that the above equation is definitively correct first.

Second, is the reasoning part.
(27 x 23^n) + (17 x 10^2n) = 11a, where a is an integer
a = ((27 x 23^n) + (17 x 10^2n))/11 --------- (1)

Therfore is follows that, in the next series
(27 x 23^(n + 1)) + (17 x 10^(2n + 1)) = 11(a + 1)
Sub in the equation (1) into the above equation, and prove that the right hand side equates to the left hand side, and your job is done. Do the simple workings yourself then.

Have fun!!!