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Prove 11^n+2+12^2n+1 is divisible by 133 using binomial theorem

Prove 11^n+2+12^2n+1 is divisible by 133 using binomial theorem

Grade:11

2 Answers

Arun
25750 Points
6 years ago
Dear student
 
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. 
 
 
 
Regards
Arun (askIITians forum expert)
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
Since, 133 is a bigger number,Hence writing the binomial expansion in the form of its multible might be time consuming as well as difficult. Hence we proceed with Mathematical Induction to prove the result.
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 it helps.
Thanks and regards,
Kushagra

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