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)

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- Apr 25th 2009, 08:12 AMusagi_killerProve 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)

- Apr 25th 2009, 08:31 AMred_dog
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. - Apr 25th 2009, 08:35 AMusagi_killer
Thanks but what's the a, b and c for?

- Apr 25th 2009, 09:55 AMstapel
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".