How many integers between $10000$ and $99999$ ( inclusive) are divisible by $3$ or $5$ or $7$?
Sebastian Wright 
How many integers between $10000$ and $99999$ (inclusive) are divisible by $3$ or $5$ or $7$ ?
My Try :
Total Integers between $10000$ and $99999$ are $89999$.
$\left\lfloor\frac{89999}{3}\right\rfloor+\left\lfloor\frac{89999}{5}\right\rfloor+\left\lfloor\frac{89999}{7}\right\rfloor$ - $\left\lfloor\frac{89999}{3\times5}\right\rfloor-\left\lfloor\frac{89999}{3\times7}\right\rfloor-\left\lfloor\frac{89999}{5\times7}\right\rfloor$ + $\left\lfloor\frac{89999}{3\times5\times7}\right\rfloor$ =
$48857$
I don't have an answer for this. Am I right here ?
$\endgroup$ 13 Answers
$\begingroup$The number of integers between $3$ and $4$ inclusive divisible by $3$ is not $\lfloor\frac{(4-3+1)}{3}\rfloor=\lfloor\frac{2}{3}\rfloor=0$ but is instead $\lfloor \frac{4}{3}\rfloor - \lfloor\frac{2}{3}\rfloor = 1$.
In the same way, you should not be using $\lfloor \frac{90000}{7}\rfloor$ but instead you should be using $\lfloor\frac{99999}{7}\rfloor - \lfloor\frac{9999}{7}\rfloor$
That is to say, the number of numbers $n$ in the range $a\leq n\leq b$ which is divisible by $7$ are those numbers in the range $1\leq n\leq b$ divisible by seven which are not numbers in the range $1\leq n\leq a-1$ which are divisible by $7$.
$\endgroup$ 5 $\begingroup$One cannot simply divide the entire range by the number whose multiples you want to count; this may end up over/undercounting. Example: $\lfloor \frac{90000}{35} \rfloor = 2571$ but there are actually $2572$ multiples of $35$. Rounding off to the nearest integer(instead of floor) might always give correct answer, but I'm not sure; I use the method below.
There are $99999-10000+1 = 90000$ integers in the range in consideration, $[10000, 99999]$.
The lowest number in this range divisible by $3, 5, 7$ are $10002, 10000, 10003$ respectively. The highest number divisible by $3, 5, 7$ are $99999, 99995, 99995$ respectively.
You also have to find the highest and lowest numbers divisible by $15$, $35$, $21$ and $105$. Do that and count their multiples as shown below.
There are $\frac{99999-10002}{3}+1$ integers ($10005,10008,...99999$ plus one for not counting $10002$ initially) divisible by $3$. Another way to think of it is that the range contains the $3334^{th}$ multiple of $3$, all the way up to the $33333^{rd}$ multiple of $3$. $$ \frac{99999-10002}{3}+1 = \frac{3\cdot (33333-3334)}{3} +1 $$
You can calculate the rest as you did above using inclusion-exclusion.
$\endgroup$ $\begingroup$Another way:
$10000=3\times 3333+1\implies$ First integer $>10000$ divisible by $3$ is $10002$.
So if we count a number of integers in sequence $10002,10005,\cdots,99999$ we have our answer. the common difference is $3$ first term is $10002$ hence, $a_n=10002+(n-1)3=99999\implies n=30000$. Similarly for others too.
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