# Thread: Prove natural number

1. ## Prove natural number

Show that $\displaystyle \frac{9^y + 6 * 2^y}{21}$ is a natural number for all y, where y is a natural number. (Without using mathematical induction)

2. Clearly, $\displaystyle 9^y+6\cdot 2^y \ \vdots \ 3$

$\displaystyle 9^y+6\cdot 2^y=9^y-2^y+7\cdot 2^y=$

$\displaystyle =(9-2)(9^{y-1}+9^{y-2}\cdot 2+\ldots+9\cdot 2^{y-2}+2^{y-1})+7\cdot 2^y=7a+7b=7c \ \vdots \ 7$

Then the number is divisible by 21.

3. Thanks but what's the a, b and c for?

4. Originally Posted by usagi_killer
Thanks but what's the a, b and c for?
What is the value of 9 - 2? Note that "9 - 2" is multiplied on some other factor, the exact value (and even expression) of which we are unclear. Since this value is unknown, we can compress the (very long) expression with "a".

Note that the other term is "7 * 2^y", where "y" is unknown, and thus "2^y" is unknown. This expression may be replaced with "b".

Then the sum is 7a + 7b = 7(a + b). Replace "a + b" with "c".