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$$\sum_{j=1}^{546}\left[ \frac{(5j)}{(1093)}\right ]$$

Where the brackets are the floor function. I'm not even sure how to start this question besides just figuring out for how many numbers the floor of the function equates to 0, 1, 2, etc by trial and error.

Ralph B
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    You are just adding a bunch of constants over some subset of indicies. For $j$ up to 218 (now for you, how did I get 218?), the floor function gives a term of 0. So from j from 1 to 218, you are just adding 0. Now, for the next set of indices, you will get 1... how do you know where to stop and where the sum changes to terms involving 2? – Eleven-Eleven Apr 17 '19 at 16:53
  • By the way, I don't see anything wrong with that approach... sometimes trial and error is a solid way to get a handle on what is happening with each summand. – Eleven-Eleven Apr 17 '19 at 17:05

1 Answers1

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First, note that

$$0<\frac{5j}{1093}<1$$

leads to

$$0<5j<1093$$

which in turn leads to

$$0<j<218.6$$

This tells us that all $j$ from $1$ to $218$ will produce a summand of $0$

Now, do the same thing for

$$1<\frac{5j}{1093}<2$$

and find those $j$ that will give a summand of 1 for each $j$... rinse and repeat until you've gotten to your upper bound of 546...

Eleven-Eleven
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