MHB Count of Multiples of 2 or 7 in 999

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The discussion focuses on calculating the count of multiples of 2 or 7 for natural numbers less than or equal to 999. The total count of even numbers is determined to be 499, while there are 142 multiples of 7. However, the initial sum of 641 is incorrect as it includes 71 multiples of 14 twice, necessitating their subtraction. The final count of unique multiples of 2 or 7 is confirmed to be 570.

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Find the count of multiples of 2;or 7, for all natural numbers less than or equal to 999.
 
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There are integer part of 999/2= 499 even numbers les than or equal to 999. There are integer part of 999/7= 142 multiples of 7 less than or equal to 999. But 499+ 142= [FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]1141 is too larger because it counts multiples of 14 twice. We need to subtract integer part of 999/14= 71 to account for that.
 
the answer provided is 570 which is the difference between 641 and 71. that difference between the total count of multiples of 2 and 7.
 
HallsofIvy said:
There are integer part of 999/2= 499 even numbers les than or equal to 999. There are integer part of 999/7= 142 multiples of 7 less than or equal to 999. But 499+ 142= 1141 is too larger because it counts multiples of 14 twice. We need to subtract integer part of 999/14= 71 to account for that.

Mathematicians often have trouble with simple arithmetic! :rolleyes:
 
so can you once again explain how we arrived at the value 71 and why we had to subtract it from 641?
 
multiples of 2 less than 999 ...

2(1), 2(2), 2(3), ... , 2(499)

multiples of 7 less than 999 ...

7(1), 7(2), 7(3), ... , 7(142)

multiples of 14 less than 999 which are common to both lists above ...

14(1), 14(2), 14(3), ... , 14(71)
number of multiples of 2 or 7 that are less than 999 =

(multiples of 2)+(multiples of 7)-(number of values that are multiples of both 2 and 7) =

499+142-71
 
Thanks for clarifying.
 

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