Suppose that all positive integers which are relatively prime to $105$ are arranged into a increasing sequence: $a_1 , a_2 , a_3, . . . .$ Evaluate $a_{1000}$

By inclusion exclusion principle I found the following equation:

$n - \left( \left\lfloor \dfrac {n} {3} \right\rfloor + \left\lfloor \dfrac {n} {5} \right\rfloor + \left\lfloor \dfrac {n} {7} \right\rfloor \right) + \left( \left\lfloor \dfrac {n} {15} \right\rfloor + \left\lfloor \dfrac {n} {21} \right\rfloor + \left\lfloor \dfrac {n} {35} \right\rfloor \right) - \left\lfloor \dfrac {n} {105} \right\rfloor=1000$

I tried to solve the equation by setting $n=105k+r ; 0 \leq 0<105$. But this method is so tiring and lengthy to solve. I thought to invoke inequalities like $a-1 <\lfloor a \rfloor \leq a$, but I couldn't do as the equation contains positive as well as negative terms .Please provide me any other method to solve such equations involving floor function. Any help would be appreciated.

  • 2
    $\begingroup$ Well, I'd do it by ignoring the floor (getting something like $2187$) and searching near there. $\endgroup$ – lulu Sep 30 at 11:25
  • $\begingroup$ You already have a good formula there. All that's left to do is to do a binary search on it, and you'll have an algorithm that is pretty much as fast as possible $\endgroup$ – Sudix Sep 30 at 12:01
  • $\begingroup$ @Sudix As the left-hand side evolves in a roughly linear fashion, one can do a lot better than pure binary search by using the secant method. $\endgroup$ – Arthur Sep 30 at 12:08
  • Note that if $\gcd(a,b)=1$, then $\gcd(a+b,b)=1$

  • There are $48$ numbers which are less than $105$ which are relatively prime to $105$, since $\phi(105)=48$. Let $a_i$ be the $i$-th number which is relatively prime to $105$. It is clear that $a_{48}=104$.

  • Also the first $104$ numbers which are relatively prime are $\{1,2,4,8,\ldots,104\}$. The next $48$ numbers are $\{1+105,2+105,4+105,\ldots,104+105\}$. Thus you see that $a_{96}=209$.

Continue like this.

  • 2
    $\begingroup$ In point 3, I think you mean "The first $48$" and "The next $48$", not $104$. $\endgroup$ – Arthur Sep 30 at 11:56
  • 1
    $\begingroup$ Also, I think you meant $1,2,4,8,\ldots$, not $1,2,4,7,\ldots$. $\endgroup$ – Barry Cipra Sep 30 at 12:44

I personally think this problem is easier to approach from a slightly more brute-force angle.

How many of the $a_i$ are below $105$? How many of them are between $105$ and $210$? How many are between $210$ and $315$? How many multiples of $105$ do you have to go before you have (close to) $1000$ terms? From there it's basically trial and error.

Alternate solution, taken from the comments above.

The left-hand side of your equation is roughly linear. So you can do a linear regression from basically any two $n$-values to find the solution to your equation. For instance, inserting $n = 1$ and $n = 11$ gives $1$ and $5$ respectively.

Straight-forward linear regression from these two values (draw the line going through the points $(1, 1)$ and $(11, 5)$, and see where that line hits $y = 1000$) says the two sides of the equation will be equal at $n = 2498$. Actually inserting this value into the left-hand side of your equation we get $1142$, which is closer, but still a bit off. (Using more sensible $n$-values, like $n = 0$ and $n = 105$ will, of course, give you a much better result.)

However, one more linear regression from $n = 11$ and $n = 2498$ basically gives you the solution.


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