Darn 2 AM Number Theory

by shiningsunnyday, Sep 1, 2016, 6:06 PM

2016 AMC 12B P22 wrote:
For a certain positive integer $n$ less than $1000$, the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$, a repeating decimal of period of $6$, and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$, a repeating decimal of period $4$. Find $n.$

Solution

Tidbit
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 2, 2016, 1:53 AM
number theory
Divisibility
modular arithmetic

Comment

5 Comments

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Wait you posted this at 11:06 A.M.
11-2 = 9
9-6 = 3
Illuminati confirmed

Er no
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 2, 2016, 11:44 AM

by zephyrcrush78, Sep 1, 2016, 9:10 PM

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Quote:
These days I'm mainly just doing problems I feel are interesting - and not forcing myself to concentrate excessively and for long durations of time.

Then how are you up at 2 doing math? I can barely hold past 9:30pm.

Lol I'm the night-owl type, definitely not the early bird.
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 2, 2016, 11:44 AM

by MathAwesome123, Sep 2, 2016, 1:04 AM

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what is (mod 1)0

Come on it's not easy typing at 2
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 2, 2016, 11:45 AM

by skipiano, Sep 2, 2016, 1:05 AM

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welp

this was AMC so you would just trial and error each of the choices duh

The choices were which range the answer was in - 0 to 200, 200 to 400, 400 to 600, 600 to 800 or 800 to 1000
This post has been edited 1 time. Last edited by shiningsunnyday, Sep 2, 2016, 11:45 AM

by cjquines0, Sep 2, 2016, 11:00 AM

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Quote:
otherwise we would have $10^3 \equiv 1 \pmod n,$ violating the order.

Why can't we have $10^3 \equiv 1 \pmod n,$?

by smo, Dec 28, 2016, 2:26 PM

To share with readers my favorite problem I came across today :) (Shout for contrib.)

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