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For $n \leq 1000$ I am interested in the integers who are divisible by 4 but not by 3 and 16. Say $a_i$ is the property that an integer is divisible by $i$. Inclusion-Exclusion gives us: \begin{align} N(a_{3}'a_{4}a_{16}') = N - N(a_3) - N(a_4') - N(a_{16}) + N(a_3a_4') + N(a_3a_{16}) + N(a_4'a_{16}) - N(a_{3}'a_{4}a_{16}') \end{align} I can determine those first 4 terms correctly:

• N = 1000

• $N(a_3)= \lfloor \frac{1000}{3} \rfloor = 333$

• $N(a_4')= 1000 - \lfloor \frac{1000}{4} \rfloor = 750 $

• $N(a_{16})= \lfloor \frac{1000}{16} \rfloor = 62$

Now it is becoming harder to working things out. We can say for $N(a_3a_4')$ that those are all the numbers that are divisible by 3 but not by 4. We can again apply inclusion-exlusion on this term but it don't seem to help. How can I work this out?

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    $\begingroup$ How can an integer be divisible by 16 but not by 4? $\endgroup$ – paul garrett May 21 '15 at 20:39
  • $\begingroup$ It's going to take a while to find an integer divisible by $\;16\;$ but not by $\;4\;$ ... $\endgroup$ – Timbuc May 21 '15 at 20:41
  • $\begingroup$ A divisor of a divisor is a divisor. $\endgroup$ – Bernard May 21 '15 at 20:41
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    $\begingroup$ You mean "divisible by 4, but not by both 3 and 16"? $\endgroup$ – Matt Gutting May 21 '15 at 20:52
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    $\begingroup$ @MarkBennet Well, when I'll believe in some infinity then I'll be able, perhaps, to respond you. In the meantime I accept infinity as a definite point in some contexts (compactification, elliptic curves with projective geometry, etc.) , very different from usual numbers and thus with no parallel notion of "divisibility". $\endgroup$ – Timbuc May 21 '15 at 21:37
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$250$ numbers are divisible by $4$. Now use inclusion/exclusion principle:

$$250-\left\lfloor\frac{250}{3}\right\rfloor-\left\lfloor\frac{250}{4}\right\rfloor+\left\lfloor\frac{250}{3\times4}\right\rfloor=125$$

Please note that we use $4$ instead of $16$, because those $250$ numbers are not consecutive - they already contain only multiples of $4$, so we need to exclude only those that are multiples of yet "another $4$" (the credit for this fix goes to @AndréNicolas, as implied in the comment below).

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  • $\begingroup$ @AndréNicolas: I agree, this answer is wrong. I will fix it or delete. Thank you for pointing that out. $\endgroup$ – barak manos May 21 '15 at 21:25
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas May 21 '15 at 21:43
  • $\begingroup$ confirmed by awk 'BEGIN{for(i=4; i<=1000; i+=4){if (i%3!=0&&i%16!=0) {print i}}}' | wc -l $\endgroup$ – Lykos May 21 '15 at 21:44

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