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

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inkyvoydBest ResponseYou've already chosen the best response.2
I 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?
 one year ago

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

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

inkyvoydBest ResponseYou've already chosen the best response.2
this 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,
 one year ago

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

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

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

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

wioBest ResponseYou've already chosen the best response.1
Hmmm, 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?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.2
yeah. 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...
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

wioBest ResponseYou've already chosen the best response.1
Now 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.
 one year ago

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

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

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

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

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

inkyvoydBest ResponseYou've already chosen the best response.2
import 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 :)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.2
Didn'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
 one year ago

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