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I know that there are 50 multiples of 2, and 11 multiples of 9 in 1-100, but I'm not sure how I would use this information to draw the above conclusion.

Conversely, there are 50 integers that are NOT multiples of 2, and 89 integers that are NOT multiples of 9.

How would I prove the proposition with this?

123
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  • See http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html – lab bhattacharjee Feb 26 '16 at 16:19
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    And don't forget $99 (=9\cdot 11)$. – John Dawkins Feb 26 '16 at 16:24
  • Can you work out how many common multiples of 2 and 9 in that range? If you can, then a simple Venn diagram should lead you to the answer. – user49685 Feb 26 '16 at 16:26
  • That's where I'm having trouble. I'm not sure how to find the common multiples without just exhaustively enumerating all of them. (Which works for this example, but doesn't scale to larger numbers) – 123 Feb 26 '16 at 16:29
  • Ok, so now you list 30 multiples of 2; and 7 multiples of 9. Look at the 2 lists, can you list some common multiple of both 2 and 9? Do you realize they have some pattern? (And no, you don't need to list all, just list some, and try to look for the pattern) – user49685 Feb 26 '16 at 16:31
  • Are half of the multiples of 9, multiples of 2? – 123 Feb 26 '16 at 16:32
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    Yes, that's an excellent observation. Now look closely, all of the common multiples of both 2; and 9 are the MULTIPLES of ... (what number) ? – user49685 Feb 26 '16 at 16:34
  • Ahh, they are all multiples of (2⋅9) – 123 Feb 26 '16 at 16:39
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    Just a little note, it's usually NOT true that the common multiples of a, and b are the multiples of a⋅b. For example, the multiples of 4 are: 0; 4; 8; 12; 16; 20; 24; 28; 32; 36; 40; ... The multiples of 6 are: 0; 6; 12; 18; 24; 30; 36; 42; 48; 54; 60; ... So the common multiple of 4 and 6 are: 0; 12; 24; 36; 48; ... (they are multiples of 12, NOT 24 = 4⋅6). This 12 are call the LEAST COMMON MULTIPLE of 4, and 6, that you'll soon learn in school. Just note that the least common multiple of a and b is usually NOT a⋅b. But in your problem, yes 18 is the least common multiple of 2, and 9. – user49685 Feb 26 '16 at 16:47

1 Answers1

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You need to observe that an integer may be both a multiple of $2$ and a multiple of $9.$ As you have calculated, there are $50$ multiples of $2$ and $11$ multiples of $9.$ Every multiple of $18$ was counted twice. There are $5$ of these.

Our count becomes $100 - (50 + 11 - 5) = 100 - 56 = \boxed{44}.$

K. Jiang
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