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
But try the problem first...
Answer:$216$
PRMO-2016, Problem 6
Pre College Mathematics
First hint
we don't know what will be the expression of \((9999....9)^3\). so we observe....
\(9^3\)=\(729\)
\((99)^3\)=\(970299\)
\((999)^3\)=\(997002999\)
.................
...............
Second Hint
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?
Final Step
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 \((24\times 9)\)=\(216\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
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
But try the problem first...
Answer:$216$
PRMO-2016, Problem 6
Pre College Mathematics
First hint
we don't know what will be the expression of \((9999....9)^3\). so we observe....
\(9^3\)=\(729\)
\((99)^3\)=\(970299\)
\((999)^3\)=\(997002999\)
.................
...............
Second Hint
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?
Final Step
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 \((24\times 9)\)=\(216\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
