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Sum of digits Problem | PRMO 2016 | Question 6

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

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