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Multiples of 3 till 98 are 32

Multiples of 5 till 98 are 19

Multiples of 15 till 98 are 6

So 32+19-6 = 45

Tota= 98 - 45 = 53 is the answer
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I am always curious how this difficulty level is being defined, this was an easy one

total
number of multiplies of 15 - 6 (as every 15 will be checked by both inspectors, 15 has prime factors 3 and 5)
number of multiplies of 3 - 32
number of multiplies of 5 - 19

Then 32 + 19 - 6 = 45
98 - 45 = 53
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Bunuel
Two assembly line inspectors, Lauren and Steven, inspect widgets as they come off the assembly line. If Lauren inspects every fifth widget, starting with the fifth, and Steven inspects every third, starting with the third, how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector?

A. 91
B. 59
C. 53
D. 47
E. 45

We need to determine how many multiples of 5, 3, and 15 there are from 1 to 98 inclusive.

Number of multiples of 5:

(95 - 5)/5 + 1 = 19

Number of multiples of 3:

(96 - 3)/3 + 1 = 32

Number of multiples of 15:

(90 - 15)/15 + 1 = 6

We add the number of multiples of 5 and of 3 to get 19 + 32 = 51, but we must subtract the number of multiples of 15 from this total because those 6 numbers represent the double-counted multiples of 3 and 5. Thus, there are 19 + 32 - 6 = 45 multiples of 3 and/or 5 from 1 to 98, which is also the number of widgets inspected.

So, 98 - 45 = 53 widgets were not inspected.

Answer: C
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