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 one year ago
How many positive integers N<1000 are there, such that 3^NN^3 is a multiple of 5?
 one year ago
How many positive integers N<1000 are there, such that 3^NN^3 is a multiple of 5?

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inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2I rewrote it as modular arithmetic 3^N mod 10=N^3 mod 10 or 3^N mod 10 N^3 mod 10=5 But how do I solve this?

wio
 one year ago
Best ResponseYou've already chosen the best response.1If you do mod 10, the the solutions are \[ 50 =5\\ 61=5\\ 72=5\\ 83=5\\ 94=5 \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm, so I mean it eliminates what they can be... a bit.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2this problem is from brilliant.org . I have no idea what I'm supposed to do to solve it, but I know it's number theory. I tried factoring the expression into difference of cubes, bu that did not work,

wio
 one year ago
Best ResponseYou've already chosen the best response.1Are you sure that that modular arithmetic works?

wio
 one year ago
Best ResponseYou've already chosen the best response.1You're just checking that the last digit is the same...

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2yeah, but how do I even solve this problem? do you have any idea or hints? xD

wio
 one year ago
Best ResponseYou've already chosen the best response.1Honest, I'm not sure, but I do think it probably involve modular arithmetic.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm, well we definitely have to leverage the fact that if it is a multiple of five, then it's last digit is either 0 or 5, right?

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2yeah. I got that part, which is why I was able to reformulate the problem in terms of modular arithmetic, which I have no knowledge of...

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2mm. So we solve both equations for n, and subtract the intersections?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Anyway properties of mod that deal with addition?

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2uhh, I have no idea what I'm doing here lol

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay I messed up. It should be:\[ 3^nn^3 \equiv 0\pmod{10}\\ 3^nn^3 \equiv 5\pmod{10} \]

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2how does one solve these equations? if it takes too long to explain, do you know of any links where I could learn more?

wio
 one year ago
Best ResponseYou've already chosen the best response.1You'd have to learn modular arithmetic/number theory.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2can you tell me the answer? :3

wio
 one year ago
Best ResponseYou've already chosen the best response.1I'm still trying to figure it out myself.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Think about what happens to the last digit of the \(3^n\) terms... each time it is multiplied by 3.

wio
 one year ago
Best ResponseYou've already chosen the best response.1When \(n=0\), it is \(3^n = 1\pmod{10}\)

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2oh god I think I figured out how to solve this problem then

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2well you showed me how I mean

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay I programmed a bit and got these patterns: \[ n^3\\ 0^3 \equiv 0 \pmod{10}\\ 1^3 \equiv 1 \pmod{10}\\ 2^3 \equiv 8 \pmod{10}\\ 3^3 \equiv 7 \pmod{10}\\ 4^3 \equiv 4 \pmod{10}\\ 5^3 \equiv 5 \pmod{10}\\ 6^3 \equiv 6 \pmod{10}\\ 7^3 \equiv 3 \pmod{10}\\ 8^3 \equiv 2 \pmod{10}\\ 9^3 \equiv 9 \pmod{10}\\ 3^n\\ 3^0 \equiv 1 \pmod{10}\\ 3^1 \equiv 3 \pmod{10}\\ 3^2 \equiv 9 \pmod{10}\\ 3^3 \equiv 7 \pmod{10}\\ 3^4 \equiv 1 \pmod{10}\\ 3^5 \equiv 3 \pmod{10}\\ 3^6 \equiv 9 \pmod{10}\\ 3^7 \equiv 7 \pmod{10}\\ 3^8 \equiv 1 \pmod{10}\\ 3^9 \equiv 3 \pmod{10}\\ \]

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2then we just subtract the two?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Well, notice something interesting?

wio
 one year ago
Best ResponseYou've already chosen the best response.1We know that the \(n^3\) begins to cycle after 10, while the \(3^n\) begins to cycle after 4.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2ugh, the lcm of those is 20, so we'll have to check all solutions for up to 20, then multiply by 1000/20?

wio
 one year ago
Best ResponseYou've already chosen the best response.1That's what I'm thinking...

wio
 one year ago
Best ResponseYou've already chosen the best response.1Here, I'll write a program to do it up to the first 20... brb.

wio
 one year ago
Best ResponseYou've already chosen the best response.1When I go to 20....\[ \begin{array}{rlcrcl} 3:& 3^{3} && 3^4 &\equiv& 0 \pmod{10}\\ 17:& 3^{17} && 17^4 &\equiv& 0 \pmod{10}\\ \end{array} \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1When I go to 100 \[ \begin{array}{rlcrcl} 3:& 3^{3} && 3^4 &\equiv& 0 \pmod{10}\\ 17:& 3^{17} && 17^4 &\equiv& 0 \pmod{10}\\ 23:& 3^{23} && 23^4 &\equiv& 0 \pmod{10}\\ 37:& 3^{37} && 37^4 &\equiv& 0 \pmod{10}\\ 43:& 3^{43} && 43^4 &\equiv& 0 \pmod{10}\\ 57:& 3^{57} && 57^4 &\equiv& 0 \pmod{10}\\ 63:& 3^{63} && 63^4 &\equiv& 0 \pmod{10}\\ 77:& 3^{77} && 77^4 &\equiv& 0 \pmod{10}\\ 83:& 3^{83} && 83^4 &\equiv& 0 \pmod{10}\\ 97:& 3^{97} && 97^4 &\equiv& 0 \pmod{10}\\ \end{array} \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1The only problem here is that I'm not sure if Javascript can handle numbers as large as \(3^{100}\)

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2I have mathematica  but I don't know how to use it lol

wio
 one year ago
Best ResponseYou've already chosen the best response.1But we seems to consistently 1/10 to be congruent... meaning that if we do 0 to 999, we'd get 100.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Whoops.... I was never letting c go to above 10... hmmm I'm gonna have to try this...

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay, so i'm getting about 4 for every 20... that is to say 1/5 of them seem to work.

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay so here is what I did.... I kept track of \(3^n\mod{10}\) as well as \(n^3\mod{10}\)

wio
 one year ago
Best ResponseYou've already chosen the best response.1So let's just say : \[ 3^n \equiv a \pmod{10}\\ n^3 \equiv b \pmod{10} \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Now in the case that \(a>b\) you can just subtract them. In the case where \(a<b\) you have to add \(10\) to \(a\) first, then subtract.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2Yeah. Wait, did you program this bruteforce?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Here look:\[ \begin{array}{rrrrrl} n & a & b & ab & r & \text{pass/fail}\\ 0 & 1 & 0 & 1 & 1 & \text{fail}\\ 1 & 3 & 1 & 2 & 2 & \text{fail}\\ 2 & 9 & 8 & 1 & 1 & \text{fail}\\ 3 & 7 & 7 & 0 & 0 & \text{pass}\\ 4 & 1 & 4 & 3 & 7 & \text{fail}\\ 5 & 3 & 5 & 2 & 8 & \text{fail}\\ 6 & 9 & 6 & 3 & 3 & \text{fail}\\ 7 & 7 & 3 & 4 & 4 & \text{fail}\\ 8 & 1 & 2 & 1 & 9 & \text{fail}\\ 9 & 3 & 9 & 6 & 4 & \text{fail}\\ 10 & 9 & 0 & 9 & 9 & \text{fail}\\ 11 & 7 & 1 & 6 & 6 & \text{fail}\\ 12 & 1 & 8 & 7 & 3 & \text{fail}\\ 13 & 3 & 7 & 4 & 6 & \text{fail}\\ 14 & 9 & 4 & 5 & 5 & \text{pass}\\ 15 & 7 & 5 & 2 & 2 & \text{fail}\\ 16 & 1 & 6 & 5 & 5 & \text{pass}\\ 17 & 3 & 3 & 0 & 0 & \text{pass}\\ 18 & 9 & 2 & 7 & 7 & \text{fail}\\ 19 & 7 & 9 & 2 & 8 & \text{fail}\\ \end{array}\\ 4/20 = 1/5 \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1Techincally, we determined you only needed the first 20.

wio
 one year ago
Best ResponseYou've already chosen the best response.1I could have done this by hand relatively quickly, but I am very prone to mistakes. That is why I programmed it.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2I used python to get an answer of 200.

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2import numbers i=0 for n in range (1,1000): if (3**nn**3)%5==0: i=i+1 print(i) Was that the answer you got? Good thing it agreed with our methods :)

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.2Didn't actually need import numbers, but I mean funny how 5 lines of code can get a hour's worth of math work done :S

wio
 one year ago
Best ResponseYou've already chosen the best response.1I'm getting n/5 for all n that is a multiple of 20.
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