I've got a copious amount of Nines

Number Theory Level 3

Let $N = \underbrace{999 \ldots 9}_{ 2013 \space 9 \text{'s} }$ . What is the last three digits of the sum of digits of $N^3$ ?

The answer is 234.

4 solutions

Priyanka Das
Dec 14, 2013

A pretty pattern is evident:-

$9^{3} = 729$

$99^{3} = 970299$

$999^{3} = 997002999$

$9999^{3} = 999700029999$

$99999^{3} = 999970000299999$

Clearly for a number with n 9's, its cube contains (n-1) 9's followed by 7 followed by (n-1) 0's followed by 2 followed by n 9's

Thus, the sum of the digits $= (n-1)*9 + 7 + 2 + n*9 = 2n*9$

For n = 2013, the sum of the digits will be $= 2*2013*9 = 36\boxed{234}$

I found a way of establishing that pattern, which is to view $99...999^3$ as

$(10^n -1)^3 = 10^{3n} - 3 \times 10^{2n} + 3 \times 10^{n} - 1$

This explains the 9999 at the start, then the 0000, and finally the 9999 at the end.

Chung Kevin - 7 years, 6 months ago

I tried the same way

Eddie The Head - 7 years, 5 months ago

Did the same way

Happy Melodies - 7 years, 5 months ago

That's a better way for sure. But patterns are beautiful to look at as well :)

Priyanka Das - 7 years, 5 months ago

Or we can see that the pattern of the sum of digits:

$9^3=729$ Sum of digits= $18$

$99^3=970299$ Sum of digits= $18\times2=36$

$999^3=997002999$ Sum of digits= $18\times3=54$ ......

Pattern: For example, look at $99^3$ . The number $99$ has two $9's$ .So the sum of digits of $99^3=18\times2=36$ .

The number $99...9$ has $2013$ $9's$ ,so the sum of digits of $99...9^3$ (2013 9's)= $18\times2013=36234$ The answer is $\boxed{234}$ .

That's how i get the answer. :D

A Former Brilliant Member - 7 years, 6 months ago

But.....do someone know that how to write 999...9 with 2013 9's as the problem? (Not like me :( ) Thanks!

A Former Brilliant Member - 7 years, 6 months ago

Do you mean this one: $\underbrace{999...999}_\text{2013 9's}?$ If that's which you mean, try this: \underbrace{999...999}_\text{2013 o 9's}

Tunk-Fey Ariawan - 7 years, 4 months ago

Exactly :D

Krishna Ar - 7 years ago

I did the same

Vishal S - 6 years ago

Roy Roque Rivera Jr
Jan 5, 2014

We have

$99^{3}=970299$ $999^{3}=997002999$ $9999^{3}=999700029999$

We can therefore expect that $N^{3}$ has 2012 9's followed by one 7 followed by 2012 zeros then one 3 then finally followed by 2013 9. Summing these digits give us $(2012)9+7+2+(2013)9=36234$ . The last three digits is $\boxed{234}$

Muhammad Poapa
Dec 15, 2013

Well, 1. Try to find out the patterns.

Here's what I got, how many 9 you have then? (Ex. 9999^{3}, you'll have the sum of the digits is 8.9 = 72. Try to 99999^{5} you'll have the sum of the digits 10.9 = 90)
So the pattern is, how many 9.2.9)
In case, there's 2013 9's so the sum of digits 4026.9 = 36234.
The last three digits, 234

Sorry for bad english

Nikhil Tyagi
Feb 1, 2014

By taking cube of 999{which is 997002999} sum of its digits is 6 9. Similarly cube(9999) has 8 9 as sum of its digits . so any number only constituting of n times 9 as its digits will have its sum of digits as 2 n 9. Therefore sum of digits 2013 2 9 comes out to be 36234 so its last digits are \boxed{234}

sorry for very bad formatting the sum of digits of 999^{3} is 6 \times 9

Nikhil Tyagi - 7 years, 4 months ago

I've got a copious amount of Nines

The answer is 234.

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