 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  2000 9s (Posted on 2019-04-15) Find the sum of the digits of the number 999...9993. The number has 2000 repeated 9s.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution Comment 1 of 1
First calculate 999...999 ^ 2:

The base number itself is one less than 10^2000, so the square will be almost 10^4000. From the pattern set by

999^2 = 998001
9999^2 = 99980001
99999^2 = 9999800001
...

the square will have 1999 9's, an 8, 1999 zeros and a 1, in that order.

But we can go directly to the pattern of the cubes:

`   n              n^3  9              729                                                      99             970299                                                   999            997002999                                                9999           999700029999                                             99999          999970000299999`

With n consisting of 2000 9's, the cube would involve 1999 9's followed by a 7, then 1999 zeros and a 2 and finally 2000 9's.

So the desired s.o.d. is 1999*9 + 7 + 2 + 2000*9 = 36,000.

 Posted by Charlie on 2019-04-15 19:47:46 Please log in:

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