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charle [14.2K]
3 years ago
13

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 $s$?
Mathematics
1 answer:
Sophie [7]3 years ago
5 0

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 <u>down</u> 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.

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