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Sorry if this is a stupid question. I am asked to find the number of positive integers in the range $[1, 1000]$ that are divisible by $3$ and $11$ but not $9$.

Here's how I $\text{tried}$ to do it.
Let $M_3, M_9, M_{11}$ denote respectively the number of integers divisible by $3$, $9$ and $11$ respectively in the given range.
Number of integers divisible by $3$ and $11$ = $|M_3 \cap M_{11}| = |M_{33}|$ since $3$ and $11$ are prime.
Then $|M_{33}| = \lfloor\frac{1000}{33}\rfloor = 30$.
Now I have to subtract the number of integers divisible by $3$ $\text{and}$ $9$ $\text{and}$ $11$. Since every multiple of 9 is also a multiple of 3, this simplifies to $|M_9 \cap M_{11}|$. And since $9$ and $11$ are coprime, this simplifies to $|M_{99}| = \lfloor\frac{1000}{99}\rfloor = 10$.
So the final answer is $30 - 10 = 20$?

The answer key says the correct answer is $32$, which I find slightly dubious. No explanations were given.

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  • $\begingroup$ Your method is right. $\endgroup$ – André Nicolas Nov 23 '14 at 4:03
  • $\begingroup$ Do you mean to say that the correct answer is indeed 20 rather than 32? $\endgroup$ – Vizuna Nov 23 '14 at 4:11
  • $\begingroup$ Sure, as you pointed out there are only $30$ numbers divisible by $3$ and $11$. And we have to subtract a bunch. The only doubt I have is that fairly often on MSE people do not state the problem precisely. $\endgroup$ – André Nicolas Nov 23 '14 at 4:16
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I will present a solution which is entirely made up of a series of completely logical statements.

How many numbers between 1 and 1000 are divisible by 11, well 90 because 90 * 11 = 990 which is the largest multiple of 11 that is less than 1000.

So how many of these numbers are divisible by 3? well every 3rd number within the set of numbers that are divisible by 11 therefore there are 90/3 = 30

So there are 30 numbers that are divisible by 11 and 3.

Out of these numbers how many are divisible by 9? Since 3*3=9, every 3rd multiple in the set of numbers that are divisible by 3 and 11 are divisible by 9.

30/3 = 10

Therefore 10 numbers within the set of 30 numbers are divisible by 9.

30-10 = 20

Which is exactly your answer and your method of solving and with this you are correct.

I hope this was helpful.

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    $\begingroup$ If anyone else was initially unconvinced as to why we can just divide to find the number of divisors, consider a simpler example. How many numbers between 1 and 10 (inclusive) are divisible by 2? Now consider this sketch: imgur.com/a/qinKqWf. $\endgroup$ – gwg Dec 14 '18 at 19:51
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So,the LCM would be 3*11=33 (not divisible by 9) The formula is- Highest no. = Lowest no. + (n-1)*d where n is the no. of numbers which is divisible by d(in this case 33). Highest no. is the highest number in that range which is divisible whereas lowest no. is the lowest number in that range which is divisible.the highest no. = 990,lowest no. = 33 990 = 33 + (n-1)*33 990 = 33 + 33n -33 990 = 33n n = 990/33 = 30

And since every third multiple of 3 is divisible by 9, 30*2/3 = 20 Hope this helps

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  • $\begingroup$ The LCM of what? $\endgroup$ – egreg Mar 30 at 12:05

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