Count the number of multiples of 3, 4, and 12 in the range 1-2005:
⌊2005/3⌋ ≈ ⌊668.333⌋ = 668
⌊2005/4⌋ = ⌊501.25⌋ = 501
⌊2005/12⌋ ≈ ⌊167.083⌋ = 167
(⌊<em>x</em>⌋ means the "floor" of <em>x</em>, i.e. the largest integer smaller than <em>x</em>, so ⌊<em>a</em>/<em>b</em>⌋ is what you get when you divide <em>a</em> by <em>b</em> and ignore the remainder)
Then using the inclusion/exclusion principle, there are
668 + 501 - 2•167 = 835
numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 <em>and</em> 4 that occur twice. Subtracting again removes them altogether.
Answer:
3
Step-by-step explanation:
as far as i can tell, if you replace all the numbers with one, your digits in the solution will look like 1111.111
D. because square root of 200 is 14.1421..
Answer:
x=9
Step-by-step explanation:
reverse the operation!
108÷12=9
It should work for most questions like this and is simple.
Yo no he podido ir a ver uow pero no me ha dado like 18
