# Divisibility

• Oct 17th 2010, 04:54 PM
kiddopop
Divisibility
For a = 238000 = 2^4 * 5^3 * 7 * 17 and b = 299880 = 2^3 * 3^2 * 5 * 7^2 * 17

(a) Is there an integer k so that a divides b^k? If so, what is the smallest possibility for k? If not, why not.

(b) Is there an integer k so that b divides a^k? If so, what is the smallest possibility for k? If not, why not.
• Oct 17th 2010, 06:43 PM
Soroban
Hello, kiddopop!

Quote:

$\displaystyle \begin{array}{ccccc}a &=& 238,\!000 &=& 2^4\cdot5^3\cdot7\cdot17 \\ b &=& 299,\!880 &=& 2^3\cdot3^2\cdot5\cdot7^2\cdot17 \end{array}$

$\displaystyle \text{(a) Is there an integer }k\text{ so that }a\text{ divides }b^k\,?$
. . . .$\displaystyle \text{If so, what is the smallest possibility for }k\,?\:\text{ If not, why not?}$

Can this be an integer? .$\displaystyle \dfrac{b^k}{a} \;=\;\dfrac{(2^3\cdot3^2\cdot5\cdot7^2\cdot17)^k}{ 2^4\cdot5^3\cdot7\cdot17}$

We have: .$\displaystyle \displaystyle \frac{2^{3k}\cdot3^{2k}\cdot5^k\cdot7^{2k}\cdot17^ i}{2^4\cdot5^3\cdot7\cdot17} \;=\; 2^{3k-4}\cdot3^{2k}\cdot5^{k-3}\cdot7^{2k-1}\cdot17^{k-1}$

The answer is yes, for $\displaystyle k \ge 3.$

Quote:

$\displaystyle \text{(b) Is there an integer }k\text{ so that }b\text{ divides }a^k\,?$
. . . .$\displaystyle \text{ If so, what is the smallest possibility for }k\,?\:\text{ If not, why not?}$

Can this be an integer? .$\displaystyle \dfrac{a^k}{b} \;=\;\dfrac{(2^4\cdot5^3\cdot7\cdot17)^k}{2^3\cdot 3^2\cdot5\cdot7^2\cdot17}$

We have: .$\displaystyle \displaystyle \frac{2^{4k}\cdot5^{3k}\cdot7^k\cdot17^k}{2^3\cdot 3^2\cdot5\cdot7^2\cdot17} \;=\;\dfrac{2^{4k-3}\cdot5^{3k-1}\cdot7^{k-2}\cdot 17^{k-1}}{3^2}$