Question

Find n 11^n+2+12^2n+1 is divisible by 133 for any natural number n.Minor details should be mention.Spammers be away.

Answer 1

hey friend n can take any value of natural number..

here is the proof..


PART 1: 
First, prove it for the base case of n = 1: 

11^(1+2) + 12^(2+1) 
= 11^3 + 12^3 
= 1331 + 1728 
= 133(10) + 1 + 1728 
= 133(10) + 1729 
= 133(10) + 133(13) 
= 133(26) 

PART 2: 
Assume it is true for a natural number k. Prove it is true for the number k+1: 

True for k: 
11^(k+2) + 12^(2k+1) = 133(m), where m is an integer. 

For k+1: 
11^(k+1+2) + 12^(2(k+1)+1) 
= 11^(k+2) * 11 + 12^(2k + 2 + 1) 
= 11^(k+2) * 11 + 12^(2k + 1) * 12² 
= 11 * 11^(k+2) + (11 + 133) * 12^(2k+1) 
= 11( 11^(k+2) + 12^(2k+1) ) + 133 * 12^(2k+1) 
= 11( 133m ) + 133 * 12^(2k+1) 
= 133 ( 11m + 12^(2k + 1)) 

The stuff in the parentheses will be an integer, so the result is a multiple of 133. 

Therefore by induction a number of that form is divisible by 133

hope this is helpful.. :-)

Answer 2

let us prove it for n = 1:  

11^(1+2) + 12^(2+1)  

= 1331 + 1728  

= 133(10) + 1 + 1728  

= 133(10) + 1729  

= 133(10) + 133(13)  

= 133(26)  

let it is true for k:  

True for k:  

11^(k+2) + 12^(2k+1) = 133(m), where m is an integer.  

Now for k+1:  

11^(k+1+2) + 12^(2(k+1)+1)  

= 11^(k+2) * 11 + 12^(2k + 2 + 1)  

= 11^(k+2) * 11 + 12^(2k + 1) * 12²  

= 11 * 11^(k+2) + (11 + 133) * 12^(2k+1)  

= 11( 11^(k+2) + 12^(2k+1) ) + 133 * 12^(2k+1)  

= 11( 133m ) + 133 * 12^(2k+1)  

= 133 ( 11m + 12^(2k + 1))