How many multiples of either 12 or 18 are there beteen 2000 and 3000 not countin

Question

How many multiples of either 12 or 18 are there beteen 2000 and 3000 not counting the ones that are also multiples of 10?

This means between 2000 and 3000 (inclusive):

1. how many multiples are there of EITHER 12 OR 18. cant count doubles of the ones they share

2. Cant count multiples of 12 and 18 that are also multiples of 10.

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For the first part i currently have this:

(Multiples of 12 up 5000) - (Multiples of 12 up to 1999)

+

(Multiples of 18 up to 5000) - (Multiples of 18 up to 1999)

-

(Multiples of 36 up to 5000)- (Multiples of 36 up to 1999)

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I dont however know how to deal with taking away the multiples of 12 and 18 that are also multiples of 10.

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The above is just my current idea, im looking for something much more clean and mathematical.

Explanation / Answer

The number of multiples = (Largest Number - Smallest number)/divisor + 1 (we add 1, because the smallest and the largest are both factors of the divisor and while subtracting, we are eliminiating 1 of the values, so we add 1. This can be illustrated by the following example: from 0 till 20, how many multiples of 4 are there? physical counting shows that the multiples are 4,8,12,16 and 20 = 5. The smallest is 4 and the largest is 20. Lets do it by my method, but not add 1. we would get (20-4)/4 = 16/4 = 4 multiples. so we need to add 1 always)

For multiples of 12, the largest number divisible by 12 is 3000. The smallest number is 2004.

Therefore the number of multiples is (3000-2004)/12 + 1 = 83 + 1 = 84

Now multiples of 12 which are also fmultiples of 10 are 60,120,180,240....So basically we find out how many factors of 60 are there from 2000 to 3000 (both inclusive). The largest number divisible by 60 is 3000. The smallest number is 2040. Therefore multiples of 60 = (3000-2040)/60 + 1 = 16 + 1 = 17

Therefore multiples of 12 which exclude multiples of 10 = 84 - 17 = 67

Do the same for 18.

The Largest number less than or equal to 3000. Divide 3000 by 18 and get the remainder which is 12. The rule is : To find a number lesser than the dividend we do Dividend - Remainder = 3000 - 12 = 2988.

The Smallest number greater than or equal to 2000. Divide 2000 by 18 and get the remainder which is 2. The rule is : To find a number greater than the dividend we do Dividend - Remainder + Divisor = 2000 - 2 + 18 = 2016

Therefore number of multiples of 18 = (2988 - 2016)/18 + 1 = 54 + 1 = 55

Now multiples of 18 which are also fmultiples of 10 are 90,180,270,360...So basically we find out how many factors of 90 are there from 2000 to 3000 (both inclusive). The largest number divisible by 90 is 2970. The smallest number is 2070. Therefore multiples of 90 = (2970-2070)/90 + 1 = 10 + 1 = 11

Therefore multiples of 18 which exclude multiples of 10 = 55 - 11 = 44

Therefore multiples of 12 or 18 = 67 + 44 = 111

You can use the same method for Part 1. There you dont have to worry about multiples of 10, but you have to eliminate the common multiples of 12 and 18. Take LCM of 12 and 18, which is 36

So the largest number, less than 3000 which is divisible by 36 = 2988 and the smallest number greater than 2000 is 2016. Therefore the number of multiples of 36 = (2988 - 2016)/12 + 1 = 27 + 1 = 28

Therefore multiples of 12 or 18 = 84 + 55 - 28 = 111

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