the set $s = \$ contains the first $50$ positive integers. after the multiples of 2 and the multiples of 3 are removed, how many integers remain in the set $s$?

Clearly, |S| = 50.

Count the multiples of 2 between 1 and 50:

⌊50/2⌋ = ⌊25⌋ = 25

(where ⌊x⌋ denotes the "floor of x", or the largest integer that is smaller than or equal to x; in other words, round down to the nearest integer)

Count the multiples of 3 between 1 and 50:

⌊50/3⌋ ≈ ⌊16.667⌋ = 16

Since LCM(2, 3) = 6, the sets of multiples of 2 and multiples of 3 have some overlap. Count the multiples of 6 between 1 and 50:

⌊50/6⌋ ≈ ⌊8.333⌋ = 8

Then by the inclusion/exclusion principle, we remove from S

25 + 16 - 8 = 33

elements, so that the new set S contains 50 - 33 = 17 elements.

step-by-step explanation:

the system of inequalities is

step-by-step explanation:

step 1

find the equation of the dashed line

we have the points

(0,0) and (3,1)

the line passes through the origin, so, is a direct variation

find the slope

the equation of the dashed line is

the solution of the inequality a s the shaded area above the dashed line

therefore

the equation of the inequality a is

step 2

find the equation of the solid line

we have the points

(4,0) and (0,4)

find the slope

the equation of the solid line is

the solution of the inequality b s the shaded area below the solid line

therefore

the equation of the inequality b is

step 3

the system of inequalities is