Check out some similar questions!

central2009264 · Apr 9, 2008, 08:52 AM

prove that for all integral n, An=11^(n+2)+12^(2n+1) is divisible by 133

galactus · Apr 9, 2008, 09:23 AM

This can be done with some algebra gymnastics.

Prove $11^{n+2}+12^{2n+1}=133p$ , for some integer p.

Verify n=1 is true: $11^{3}+12^{3}=3059, \;\ 3059/133=23$ ... TRUE.

Assume $P(k)=11^{k+2}+12^{2k+1}=133p$ is true.

We have to show that P(k+1) is true.

Rewrite:

$P(k+1)=11^{k+3}+12^{2k+3}=11\cdot{11^{k+2}}+144\cd ot{12^{2k+1}}$

Add and subtract $133\cdot{11^{k+2}}$ :

$11\cdot{11^{k+2}}+133\cdot{11^{k+2}}+144\cdot{12^{ 2k+1}}-133\cdot{11^{k+2}}$

$=144\cdot{11^{k+2}}+144\cdot{12^{2k+1}}-133\cdot{11^{k+2}}$

$=144(\underbrace{11^{k+2}+12^{2k+1}}_{\text{this is 133p}})-133\cdot{11^{k+2}}$

$=144(133p)-133\cdot{11^{k+2}}$

$=\underbrace{133(144p-11^{k+2})}_{\text{multiple of 133}}$

Therefore P(k+1) is true and the induction is complete. QED.

central2009264 · Apr 9, 2008, 09:33 AM

Originally Posted by central2009264

prove that for all integral n, An=11^(n+2)+12^(2n+1) is divisible by 133

Thanks!!