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Grade: 10
        
when 4444^4444is written in decimal notation sum of its digit is 7 sum of digit is  A sum of digits of A is B  find sum of digitgs of B
8 months ago

Answers : (1)

Arun
13145 Points
							
remainder by 9 = digit sum of sum of first 1002 natural numbers = 1002*1003/2 mod 9
=> 501*1003 mod 9 = 6*4 mod 9 = 24 mod 9 = 6 mod 9
=> 501*1003 mod 9 = 6*4 mod 9 = 24 mod 9 = 6 mod 9

when 4444^4444 is written in decimal notation,the sum of its digits is a. Let b be the sum of the digits of a. Find the sum of the digits of b
4444^4444 = 7^4444 mod (9) = 2^4444 mod (9) = (2^3)^1481 * 2 mod (9)
=> (2^3)^1481 * 2 mod (9) = 8^1481 * 2 mod (9) = -2 mod (9) = 7 mod 9
Regards
Arun (askIITiana forum expert)
8 months ago
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