1
$\begingroup$

I put everything on a google document since I don't know how to use LaTeX or whatever. Here's the doc. I would prefer if you looked at the doc, but I've done my best to translate it onto here the best I can. Here it is.

$$\sum_{x=1}^{10} \frac yx = \text{Positive Integer}$$ If $n_1$ is the smallest $y$ value that satisfies the equation above, $n_2$ is the second smallest, etc. then what is the sum of $\sum_{z=n_1}^{n_{100}} z$?

(Example: n1 = 2520, because 2520 is the smallest number that is divisible by every number between 1 and 10.)

I simply have no idea how to even approach this problem. I just learnt about sigma (yes I’m only in algebra II), and so I started fumbling around with it and ended up with this question. I know that $\sum_{z=n_1}^{n_{100}} z$ technically means all the integers between n1 and n100 added together, but what I actually want to do is n1+n2+n3...+n100. How do I write that in sigma notation, or is that a separate part of algebra?

I gave this question a shot, so here it is. First things first, if x is divisible by y, then x*z (z being a non-zero integer) is still divisible by y. That means n2 might be 2n1, n3 might be 3n1, etc. I can generalize it by saying that nx=x(n1). So, the expression can be simplified to something along the lines of [100 over sigma over z=1 of 2520z]. So, that gives me 100(2520+252000)/2, which equates to 12726000. I highly doubt this is the correct answer, but if this question was on an exam or whatever, I would probably just put it as the answer and hope for the best."

$\endgroup$
  • 1
    $\begingroup$ The key to show that only the multiples of $2520$ are solutions is to consider that $$\sum_{j=1}^{10} \frac{1}{j}=\frac{7381}{2520}$$ and that $\gcd(7381,2520)=1$ $\endgroup$ – Peter Nov 6 '17 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.