#2**0 **

sorry the problem is more sorry

How many different integers between 100 and 500 are multiples of either 6, 8,or both?

SmartMathMan Jan 9, 2018

#3**+1 **

Multiples of 8 =

[500 - 100 ] / 8 - 1 = 50 - 1 = 49 between

Multiples of 6 =

[ 500 - 100 ] / 6 - 1 = 66 between

The LCD for 6, 8 is 24....we must subtract the multiples of 24 between 100 and 500 from the above total

[ 500 - 100] / 24 = 16 between

So.....the total multiples of 6, 8 (or both) between 100 and 500 =

49 + 66 - 16 =

99

CPhill Jan 9, 2018

#4**+2 **

These are simply the number of terms in Arithmetic Series:

Number of terms =[ Last term - First term] / Common diffrence + 1

N =[496 - 104] / 8 + 1

N=[ 392] / 8 + 1

N= 49 + 1

N = 50 - multiples of 8 between 100 and 500

N =[ 498 - 102] / 6 + 1

N =[ 396 ] / 6 + 1

N= 66 + 1

N = 67 - multiples of 6 between 100 and 500

Total = 50 + 67 =117 - multiples of 6 and 8 between 100 and 500.

However, 6 and 8 have an LCM of 24.Therefore, multiples of 24 between 100 and 500 are

N = 16 - multiples of 24 between 100 and 500 which have been counted TWICE!. So the number of integers that are multiples of 6 and 8, or both, between 100 and 500 is:

117 - 16 = 101 Integers

Guest Jan 9, 2018