If $n$ is an integer, $1 \leq n \leq 2010,$ how many fractions $\frac{n^2}{2010}$ yield repeating decimals?

I have no idea where to start. Can someone help me?

Guest Nov 15, 2018

#1**+1 **

a=1; b=listfor(n, 1, 3, (n^2/2010))

(0.0004975124378109452736318407960199004975124378109452736318407960199004975124378109452736318407960199005,

0.001990049751243781094527363184079601990049751243781094527363184079601990049751243781094527363184079602,

0.004477611940298507462686567164179104477611940298507462686567164179104477611940298507462686567164179104)

P.S. I ran them on my computer and ALL of them repeat!!. And their "period" of repeats is exactly the same for ALL of them - that is, they ALL repeat every 33 DIGITS. Look at the 3 examples above(1^2/2010, 2^2/2010, 3^2/2010), which I have calculated to 100 decimal places and observe that they ALL repeat every 33 digits. What is the "mathematical explanation" for it, that I honestly don't know? Maybe somebody like heureka, Alan and other Mods might have an explanation. It might have something to do with 2010 itself. Try factoring it and this is what you get: 2010 = 2 * 3 * 5 * 67. Notice that it has **4 distinct prime factors** multiplied together. I believe that is the reason it repeats every 33 digits, because its largest factor is 67 and 33 is 67/2 nearly!!

Guest Nov 15, 2018

edited by
Guest
Nov 15, 2018

edited by Guest Nov 15, 2018

edited by Guest Nov 15, 2018

#2**+1 **

Good work guest. It is good to see you trying to work this out.

It would be good if you becamse a member becasue then it is more freindly and there are other little perks as well.

((2*3)*5)*67 = 2010

The prime factors of 2010 are 2,3,5 and 67

A fraction will only terminate if the prime factors of the denominator are powers of 2, 5 or 10.

\(\text{So for }1 \leq n \leq 2010\\ \frac{n^2}{2010}\)

will TERMINATE only if the 3 and the 67 can cancel out with a 3 and a 67 in the numerator.

3*67=210

2010/201=10

Only 10 will terminate.

So 2000 will not terminate.

Melody Nov 17, 2018