Try this beautiful problem from Number system based on sum of digits.

## Sum of digits | PRMO | Problem 6

Find the sum of digits in decimal form of the number \((9999….9)^3\) (There are 12 nines)

- $200$
- $216$
- $230$

**Key Concepts**

Number system

Digits

counting

## Check the Answer

Answer:$216$

PRMO-2016, Problem 6

Pre College Mathematics

## Try with Hints

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 \((24\times 9)\)=\(216\)

## Other useful links

- https://www.cheenta.com/area-of-triangle-problem-amc-10a-2009-problem-10/
- https://www.youtube.com/watch?v=V01neV8qmh4