how to always round up to the next integer

Math.Ceiling((double)list.Count() / 10);

(list.Count() + 9) / 10

Everything else here is either overkill or simply wrong (except for bestsss' answer, which is awesome). We do not want the overhead of a function call (Math.Truncate(), Math.Ceiling(), etc.) when simple math is enough.

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OP's question generalizes (pigeonhole principle) to:

How many boxes do I need to store x objects if only y objects fit into each box?

The solution:

1. derives from the realization that the last box might be partially empty, and
2. is (x + y - 1) ÷ y using integer division.

You'll recall from 3^rd grade math that integer division is what we're doing when we say 5 ÷ 2 = 2.

Floating-point division is when we say 5 ÷ 2 = 2.5, but we don't want that here.

Many programming languages support integer division. In languages derived from C, you get it automatically when you divide int types (short, int, long, etc.). The remainder/fractional part of any division operation is simply dropped, thus:

5 / 2 == 2

Replacing our original question with x = 5 and y = 2 we have:

How many boxes do I need to store 5 objects if only 2 objects fit into each box?

The answer should now be obvious: 3 boxes -- the first two boxes hold two objects each and the last box holds one.

(x + y - 1) ÷ y =
(5 + 2 - 1) ÷ 2 =
6 ÷ 2 =
3

So for the original question, x = list.Count(), y = 10, which gives the solution using no additional function calls:

(list.Count() + 9) / 10