Try this beautiful problem from Number system based on sum of digits.
Find the sum of digits in decimal form of the number \((9999....9)^3\) (There are 12 nines)
Number system
Digits
counting
Answer:$216$
PRMO-2016, Problem 6
Pre College Mathematics
we don't know what will be the expression of \((9999....9)^3\). so we observe....
\(9^3\)=\(729\)
\((99)^3\)=\(970299\)
\((999)^3\)=\(997002999\)
.................
...............
we observe that,There is a pattern such that...
In \((99)^3\)=\(970299\) there are 1-nine,1-seven,1-zero,1-two,2-nines & \((999)^3\)=\(997002999\) there are 2- nines,1-seven,2-zeros,1-two,3-nines....so in this way.....\((999....9)^3\) will be 11-nines,1-seven,11-zeros,1-two,12-nines..........
Therefore \((999....9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)
can you finish the problem?
Therefore \((999....9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\).......
total numbers of Nines are (11+12) and (7+2)=9(another one) .....so total (11+12+1)=24 nines and the sum be \((24\times 9)\)=\(216\)
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