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 September 27th, 2013, 02:40 AM #1 Newbie   Joined: Sep 2013 Posts: 6 Thanks: 0 n^2 divided by 5 will always leave remainder 1 or 4, why? Hi. Hoping for some help. When n is an integer n>/=3 and you divide n^2 by 5, the remainder will always be 1 or 4, correct? I need to understand why, but I can't seem to figure it out. Any help out there? September 27th, 2013, 02:50 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: n^2 divided by 5 will always leave remainder 1 or 4, why Or 0. You just need to consider 0-4 squared mod 5. September 27th, 2013, 05:06 AM #3 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: n^2 divided by 5 will always leave remainder 1 or 4, why Basically, what you need to understand is the n divided by 5 can leave reminders 0, 1, 2, 3 or 4. The only squares are 0 and 4, so clearly, 0^2 and 4^2 divided by 5 leaves reminder 0 and 1, respectively. September 30th, 2013, 12:06 AM #4 Newbie   Joined: Sep 2013 Posts: 6 Thanks: 0 Re: n^2 divided by 5 will always leave remainder 1 or 4, why Ah, ofc. Thank you! September 30th, 2013, 12:10 AM   #5
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Re: n^2 divided by 5 will always leave remainder 1 or 4, why

Quote:
 Originally Posted by mathbalarka Basically, what you need to understand is the n divided by 5 can leave reminders 0, 1, 2, 3 or 4. The only squares are 0 and 4, so clearly, 0^2 and 4^2 divided by 5 leaves reminder 0 and 1, respectively.
But wait, now I'm confused again. What about the reminder 4...? September 30th, 2013, 12:33 AM   #6
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Re: n^2 divided by 5 will always leave remainder 1 or 4, why

Quote:
 Originally Posted by getwolfgang But wait, now I'm confused again. What about the reminder 4...?
Yes. If n^2 is divisible by 5, then the reminder left after dividing n by 5 can be 0 or 4, since they are the only squares in the list {0, 1, 2, 3, 4}.

So, we have : reminder after dividing n by 5 is 0 or 4
Hence : reminder after dividing n^2 by 5 is 0^2(=0) or 4^2 (=1)

Get it? September 30th, 2013, 12:52 AM   #7
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Re: n^2 divided by 5 will always leave remainder 1 or 4, why

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by getwolfgang But wait, now I'm confused again. What about the reminder 4...?
Yes. If n^2 is divisible by 5, then the reminder left after dividing n by 5 can be 0 or 4, since they are the only squares in the list {0, 1, 2, 3, 4}.

So, we have : reminder after dividing n by 5 is 0 or 4
Hence : reminder after dividing n^2 by 5 is 0^2(=0) or 4^2 (=1)

Get it?
Isn't 1 a square? (1^2 = 1) and would give the reminder 4, so the answer would be that the possible reminders are 0, 1 and 4?
(If I'm completely misunderstanding you, then I appologize, I'm really tired and my brain isn't working as fast as I would like.) September 30th, 2013, 12:58 AM   #8
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Re: n^2 divided by 5 will always leave remainder 1 or 4, why

Quote:
 Originally Posted by getwolfgang Isn't 1 a square? (1^2 = 1) and would give the reminder 4, so the answer would be that the possible reminders are 0, 1 and 4?
Ah, right. n divided by 5 leaves reminder 0, 1 and 4, but that wouldn't change the answer would it? Squaring the reminders gives 0 and 1. September 30th, 2013, 01:15 AM #9 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: n^2 divided by 5 will always leave remainder 1 or 4, why The critical thing is what happens to the numbers 0-4 when you square these and reduce modulo 5. Any number, n, greater than 4 is equal to 0-4 (mod 5) and n^2 (mod 5) is the same as (n (mod 5))^2 (mod 5). So: 0^2 = 0 (mod 5) 1^2 = 1 (mod 5) 2^2 = 4 (mod 5) 3^2 = 4 (mod 5) 4^2 = 1 (mod 5) So, any number squared will be 0, 1 or 4 (mod 5). September 30th, 2013, 04:16 AM #10 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: n^2 divided by 5 will always leave remainder 1 or 4, why If OP had understood modular arithmetic he wouldn't have asked this question at all. Tags divided, leave, remainder ,

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27^2 is divided by 7 what will be reminder

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