What is the sum of the digits of 2^1000
check my algoriathm, let y=2^1000 then logy = 1000log2 = 301 and y=10^logy=10^301
since (1,0) r da only digits of 10^n 4all n=1,2,3,4,..... The sum of digits equals 1 , but it is not the answer ...Why?
Well you seem to want 2^1000 to be equal to 10^301 which is clearly false since 5 divides the latter but not the former. The issue is that 1000log2 is not exactly equal to 301, you probably left off the decimal places which is the cause of your problem when you equate 10^logy to 2^1000.
Do you want the actual sum of all the digits in 2^1000 or do you just want the single digit reduced sum (the single digit eventually obtained from repeated summing of digits).
If its the single digit reduced sum that you want then just work modulo 9 to get the answer (which is 7 btw).
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