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I have recently been doing problem solving in math, and I came across this problem:

Determine the number of positive multiples of $6$ or $9$ or both, less than $1000$.

I appreciate any help. Thanks!

Mathinator06
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    Can you say how many multiples of $6$ alone there are between $1$ and $1000$? – pjs36 Feb 12 '16 at 18:22
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    Maybe taking 1000/6 and removing the decimal part would give you the answer, but I am unsure. – Mathinator06 Feb 12 '16 at 18:23
  • Should there be a comma after "both"? – Bobson Dugnutt Feb 12 '16 at 18:24
  • Oh sorry, yeah I meant to put both, as in both multiples of 6 and 9 – Mathinator06 Feb 12 '16 at 18:26
  • do you guys any clue as to how to set this problem up? appreciate your help. Thanks! – Mathinator06 Feb 12 '16 at 18:30
  • You could find the number of positive multiples of $6$ less than $1000$, and the number of positive multiples of $9$ less than $1000$. Add up these numbers. You could then subtract from this sum, the number of common positive multiples of $6$ and $9$ less than $1000$. The result will be your answer (why?). (Hint: A common positive multiple of $6$ and $9$ is the same thing as a common positive multiple of $18$.) – Amitesh Datta Feb 12 '16 at 19:29

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The number of positive multiples of $6$ that are less than $1000$ is $$\left\lfloor\frac{1000}{6}\right\rfloor=166$$ Likewise, the number of positive multiples of $9$ that are less than $1000$ is $$\left\lfloor\frac{1000}{9}\right\rfloor=111$$ Adding these, we get $277$, but we have counted twice the numbers that are multiples of both $6$ and $9$. So we need to count those numbers, then subtract from the total. Numbers that are multiples of both $6$ and $9$ are numbers that are multiples of the least common multiple of $6$ and $9$. That least common multiple is $18$, so we find: $$\left\lfloor\frac{1000}{18}\right\rfloor=55$$ Then subtracting from $277$ we have the final count: $$277-55=222$$

Logophobic
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