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Grade: 12
        
Find the remainder when ((13^141) +(11^141) ) is divided by 24..please help with explanation..
3 months ago

Answers : (3)

Saurabh Koranglekar
askIITians Faculty
8227 Points
							Dear student

Please begin by fermat's little theorem

Regards
3 months ago
Vikas TU
11498 Points
							
Dear student 
It is solved by using Fermats Little theorem 
Which is not a part of Your exam syllabus 
It is asked in CAT level examinatin .
Good Luck 
3 months ago
Aditya Gupta
2016 Points
							
dear student, both the answers given by vikas and saurabh are ABSOLUTELY WRONG and hence you shouldnt give any attention to their crappy excuses of using fermat theorem, to hell with fermat and his theorem!!
CORRECT SOLUTION:
we know the identity
a^(2n+1) + b^(2n+1)= (a+b)(a^2n*b^0 – a^(2n-1)*b^1 + a^(2n-2)*b^2 – ,,,,,,,, – a^1*b^(2n-1) + b^2n), where n is a natural no
simply substitute a=13 and b= 11 and n= 70, we get
(13^141) +(11^141)= (11+13)(13^140 – …... + 11^140)
= 24*k where k= (13^140 – …... + 11^140) is an integer
hence, (13^141) +(11^141) is clearly divisible by 24.
so the remainder is zero.
KINDLY APPROVE :))
3 months ago
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