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Prove this, and get 1Frih





DecayClan
Prove that the number a=2222^5555 + 5555^2222
Can be divided by 7
(Exactly)

Find it, and win 1Frih!
(sorry, i can't give you more, thats all i have)
If i had more, iw ould give more as well Smile
matt87
I would prove this but its not worth my time for only 1 frih$
ocalhoun
true, you need to offer more FRIH$ in order to get people to do your homework. I got more than that by just commenting on it.
djared90
I agree that 1 FRIH$ is far too little to do this for you, but since I am so generous and math-loving I'll do it anyway.

To solve this, you need to use modular arithmetic.

2222^5555 can be rewritten as 3^5555 (mod 7) {using mod-7 since you are proving it is divisible by 7} and 5555^2222 can be rewritten as 4^2222 (mod 7)

Powers of 3 mod 7 are cyclic, since 3^1 = 3 (mod 7), 3^2 = 2 (mod 7), 3^3 = 6 (mod 7), 3^4 = 4 (mod 7), 3^5 = 5 (mod 7), 3^6 = 1 (mod 7), 3^7 = 3 (mod 7), and so on...So that 3^5555 (mod 7) can be rewritten as 3^5 (mod 7) = 5 (mod 7)

Likewise, powers of 4 mod 7 are also cyclic, since 4^1 = 4 (mod 7), 4^2 = 2 (mod 7), 4^3 = 1 (mod 7), 4^4 = 4 (mod 7), and so on...Thus, 4^2222 (mod 7) can be rewritten as 4^2 (mod 7) = 2 (mod 7)

Now you have 5 (mod 7) + 2 (mod 7), which equates to 7 (mod 7), and since 7 is divisible by 7, it divides 2222^5555 + 5555^2222 evenly.

Hopefully you understood all of that; otherwise, you might want to look into learning about modular arithmetic, because it really is quite interesting what you can do with it.
LandShark
wow you are amazing at math. Im going to p rint this out and give it to my math teacher
DecayClan
LoL.I made this for fun...Not my homework,i already know the answer...It's like, "who wants to try it for fun" if i really needed it, i would offer more...The 1Frih means nothing...Even the post that you say that it doesn't worth it gives you more, but thats not the point...The point is to do it because you want to, not for the 'money'...

Very well djared90!You get 1 frih!Thats the correct answer!
MrBaseball34
http://perplexus.info/show.php?pid=3108&op=sol
djared90
Well, I guess the rules that I know:

(a^3 + b^3) = (a + b)(a^2 - ab + b^2)

and

(a^2 - b^2) = (a - b)(a^2 + ab + b^2)

could be expanded to fit the rules they used, so that's also a valid explanation.

However, it would take more than "a little insight" to know to split up the original number like they did so that it comes out just right, so I still like my way with the modular arithmetic.
phx777
man, if i only reply this to tell you that your payment isnt really worth any solution, i will probably get more fri$ then i would get for the actuall solution (now lets see how much fri$ ill get for this post Razz)

yey, allmost exactly one 1 fri$, actually a lil bit over it Razz
ralphbefree
thanks for expanding my horizons into modular arithmatic... looking it up right now..
WW.peace.love.math.recycle.smiles/ralph
Vanquish
Dude, did you learn that at school?
If you did fair enough, but why would you need it in real life? I'm not flaming or anything it's just a simple question, what situations would you come across where you actualy need that formula? It still doesn't really make sence to me but i'm only 13, I still found out I got a Level 7 (Like an A) in my mock Math SATS, hopefuly I get a Level 7 in my real SATS! It's another 3 months anyway so I shall be sticking in my lessons and stop messing around, the problem is i'm distracted very easily, so if anyone else is messing around basically I mess around! Anyway back to the topic, where would you use this arithmatic in a real life situation?
matt87
I am sure it has a use somewhere, but don't exactly know where. It could be used for engineering or some other complication. I always ask this question to my teachers and they always find a way to awnswer it. Maybe try asking one of your math teachers. I am not in school right now so I don't got anyone to ask for you.
xoxmeholly
It could probably be used for like engineering, and that stuff. Maybe something out there, like space: NASA. I'm sure they use freaky math formulas. And my math teacher said that a single calculus problem would take up the entire board (it's a big board, by the way) and I'd NEVER be able to prove what on earth you'd use the formula for.

I bet this is more than 1 $FHIH

Edit: it was 2.262
Empire
I KNOW WHERE U WOULD NEED THIS!!! TO BECOME A COLLEGE MATH TEACHER!!!



damn I'm smart
Zenrei
Vanquish wrote:
Dude, did you learn that at school?
If you did fair enough, but why would you need it in real life? I'm not flaming or anything it's just a simple question, what situations would you come across where you actualy need that formula? It still doesn't really make sence to me but i'm only 13, I still found out I got a Level 7 (Like an A) in my mock Math SATS, hopefuly I get a Level 7 in my real SATS! It's another 3 months anyway so I shall be sticking in my lessons and stop messing around, the problem is i'm distracted very easily, so if anyone else is messing around basically I mess around! Anyway back to the topic, where would you use this arithmatic in a real life situation?

if you work with computers or any kind of mathimaticaly inclined work, such as relative statistics or phisics or what not... you might need this...

If you still work at mc donnalds you probably dont even need high school
its all relative Razz
NjRocket
I agree that 1 FRIH$ is far too little to do this for you, but since I am so generous and math-loving I'll do it anyway.

To solve this, you need to use modular arithmetic.

2222^5555 can be rewritten as 3^5555 (mod 7) {using mod-7 since you are proving it is divisible by 7} and 5555^2222 can be rewritten as 4^2222 (mod 7)

Powers of 3 mod 7 are cyclic, since 3^1 = 3 (mod 7), 3^2 = 2 (mod 7), 3^3 = 6 (mod 7), 3^4 = 4 (mod 7), 3^5 = 5 (mod 7), 3^6 = 1 (mod 7), 3^7 = 3 (mod 7), and so on...So that 3^5555 (mod 7) can be rewritten as 3^5 (mod 7) = 5 (mod 7)

Likewise, powers of 4 mod 7 are also cyclic, since 4^1 = 4 (mod 7), 4^2 = 2 (mod 7), 4^3 = 1 (mod 7), 4^4 = 4 (mod 7), and so on...Thus, 4^2222 (mod 7) can be rewritten as 4^2 (mod 7) = 2 (mod 7)

Now you have 5 (mod 7) + 2 (mod 7), which equates to 7 (mod 7), and since 7 is divisible by 7, it divides 2222^5555 + 5555^2222 evenly.


^^^^ That is insane lol. But if you actually take a look at it, it makes perfect sense, tried it on my calculator and got an overflow lol
Ghost Rider103
2222^5555 can be rewritten as 3^5555 (mod 7) {using mod-7 since you are proving it is divisible by 7} and 5555^2222 can be rewritten as 4^2222 (mod 7)

Powers of 3 mod 7 are cyclic, since 3^1 = 3 (mod 7), 3^2 = 2 (mod 7), 3^3 = 6 (mod 7), 3^4 = 4 (mod 7), 3^5 = 5 (mod 7), 3^6 = 1 (mod 7), 3^7 = 3 (mod 7), and so on...So that 3^5555 (mod 7) can be rewritten as 3^5 (mod 7) = 5 (mod 7)

Likewise, powers of 4 mod 7 are also cyclic, since 4^1 = 4 (mod 7), 4^2 = 2 (mod 7), 4^3 = 1 (mod 7), 4^4 = 4 (mod 7), and so on...Thus, 4^2222 (mod 7) can be rewritten as 4^2 (mod 7) = 2 (mod 7)


So wheres my one FRIH?






















Laughing
hunnyhiteshseth
Ghost Rider, NjRocket & Zenrei, you all should first see the date of the first post & frih$ of the topic starter before saying anything.

I think I needn't say more.
tamilparks
the amout is very small can you please give 100frih$
hunnyhiteshseth
Tamilparks, this topic is very old, the topic starter would have left frihost by now.
Ghost Rider103
hunnyhiteshseth wrote:
Tamilparks, this topic is very old, the topic starter would have left frihost by now.


I dont think that you are understanding us very well. lmao Laughing Laughing Laughing Laughing Laughing Laughing
hunnyhiteshseth
oh now I am understanding you very well, Laughing Laughing
mackhenry
2222^5555 can b rewritten as 3^5555 (mod 7) {using mod-7 proving it is divisible by 7} and 5555^2222 can be rewritten as 4^2222 (mod 7)

Powers of 3 mod 7 are cyclic, since 3^1 = 3 (mod 7), 3^2 = 2 (mod 7), 3^3 = 6 (mod 7), 3^4 = 4 (mod 7), 3^5 = 5 (mod 7), 3^6 = 1 (mod 7), 3^7 = 3 (mod 7), and so on...So that 3^5555 (mod 7) can be rewritten as 3^5 (mod 7) = 5 (mod 7)

Likewise, powers of 4 mod 7 are also cyclic, since 4^1 = 4 (mod 7), 4^2 = 2 (mod 7), 4^3 = 1 (mod 7), 4^4 = 4 (mod 7), and so on...Thus, 4^2222 (mod 7) can be rewritten as 4^2 (mod 7) = 2 (mod 7)

hey give my 1frih$
ya its correct wat u said hitesh
mackhenry
tamilparks wrote:
the amout is very small can you please give 100frih$

hey tamilparks its not the plce for beggers and doing nothing and asking for money wat kind of person r u Laughing
mackhenry
2222^5555 can b rewritten as 3^5555 (mod 7) {using mod-7 proving it is divisible by 7} and 5555^2222 can be rewritten as 4^2222 (mod 7)

Powers of 3 mod 7 are cyclic, since 3^1 = 3 (mod 7), 3^2 = 2 (mod 7), 3^3 = 6 (mod 7), 3^4 = 4 (mod 7), 3^5 = 5 (mod 7), 3^6 = 1 (mod 7), 3^7 = 3 (mod 7), and so on...So that 3^5555 (mod 7) can be rewritten as 3^5 (mod 7) = 5 (mod 7)

Likewise, powers of 4 mod 7 are also cyclic, since 4^1 = 4 (mod 7), 4^2 = 2 (mod 7), 4^3 = 1 (mod 7), 4^4 = 4 (mod 7), and so on...Thus, 4^2222 (mod 7) can be rewritten as 4^2 (mod 7) = 2 (mod 7)

hey give my 1frih$
hey proving it again
donate me now plz will u Pray
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