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# Understand the problem

The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$?

##### Source of the problem
American Mathematics Competition
Number Theory

7/10

##### Suggested Book

Elementary Number Theory by David M, Burton

So, well have a long look at the problem. With a little bit of thought, you might even crack this without proceeding any further !

Think you could do with some help ? Okay let’s get ourselves a headstart. As you might have guessed, sometimes the most fruitful thing to be done is to observe what’s going on in these kind of problems. The intuition is clear, there are too many instances of ‘8‘-s for someone to account for them manually.  So, yes, that’s something we can expect. Try working out with a few simple cases like, k=1, k=2, and so on…

Okay, so let’s see what’s happening for a few small values of k…
k=1
8 * 8 = 64 k=2
8 * 88 = 704 k=3
8 * 888 = 7104 k=4
8 * 8888 = 71104 …
So, Wait ! Look closely…Do you see something ?

A really nice pattern is evolving. If you can see it, and yet find it a bit difficult to articulate it mathematically, don’t worry. See for k=4, the product gives us the result 71104. That means, we have the starting digit to be 7. And the ending suffix is 04. What varies are the ones. See, for k=4, we have ( k-2 = 2 ) ones. That’s it ! The generalization should be fairly simple for us to do now… For every k >=2, the product result consists of a 7 to start with, exactly k-2 1’s follow, and we conclude with single occurrences of 0 and 4 each. Now, think…can you take this till the end ?

Once we’ve seen the pattern, it’s easy to get this done.  What we have noted, we need that to sum up to 1000.  In simple mathematical terms, 7 + (k-2).1 + 0 + 4 = 1000
11 + k = 1002
k = 991

So, we need ‘8’ to come 991 times in the multiplicand, so that the digits sum up to 1000. So, that seals the deal !

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