Friday, 11 April 2014

proof verification - Prove by induction that 232n+312n+46 is divisible by 48 for all integers nge0




The base case, n=0:
230+310+46=48

and 48mod48=0.



Inductive Hypothesis:
Let's assume it is true for n=k.
Then p(k)=232k+312k+46


p(k+1)=232(k+1)+312(k+1)+46=529(232k)+961(312k)+46



However, I am not sure how to proceed from here in order to get an expression that is divisible by 48.

Thanks in advance for any hints.


Answer



Note that your p(k) is written wrongly.



529(232k+312k)+432(312k)+46=529(232k+312k)+9(48)(312k)+46=529(232k+312k+46)+9(48)(312k)+46(1529)=529(232k+312k+46)+9(48)(312k)46(528)=529(232k+312k+46)+9(48)(312k)46(48)(11)



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