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Sum of digits Problem | PRMO 2016 | Question 6

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

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