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Sum of digits Problem | PRMO 2016 | Question 6

Try this beautiful problem from Number theory based on sum of digits from PRMO 2016. You may use sequential hints to solve the problem.

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


But try the problem first…

Answer:$216$

Source
Suggested Reading

PRMO-2016, Problem 6

Pre College Mathematics

Try with Hints


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

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