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Prove that 7^51-1 is a multiple of 103.

According to the rule Ferm 7^102 is a multiple of 103.

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I think if 7^51 when divided by 103 gives remainder -1,or say 42, then 7^102 = (7^51)^2 should also give remainder (-1)^2 ,ie, 1 ... so i think you got it wrong somewhere. Correct me if i'm wrong.

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Where did you find it out that 7^51 dividing by 103 gives the remainder -1. Also remainder cannot be negative number. So by -1 you mean 102. And read firstly the theorem of Ferm. Your answer is actually absolutely wrong, sorry don't wanna be rough but it's so. I have got this problem from the book. So I am sure the problem is alright.
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I understand ... i know i'm not very good :P
No, not -1 .. i miscalculated.
I just thought of the core remainder,ie., 1 or 102
But why wouldn't it work if
7^51 = 1 mod 103
then (7^51)^2 = 1 mod 103, ie., 7^102 is not div. by 103
Sorry I don't know maths that well nor any theorems... i'm just a school student using my common sense
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Point out where I'm doing the mistake - then it'll be very helpful :)
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Mysterio. Understand this. I need to prove that 7^51-1=0 mod103. To prove it I need to prove that 7^51=1 mod103. According to the theorem I found out this 7^102=1 mod103 that means 7^102-1=0 mod103 that means (7^51-1)(7^51+1)=0 mod 103 that means either 7^51-1 or 7^51+1 (one of them) should be a multiple of 103. Now I should prove that 7^51+1 isn't a multiple of 103. It's pretty hard. Maybe there is another way :)
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Ok i get it :) You typed it a bit wrong in the question :)
You said 7^102 is a multiple of 103 .... you should have said it left remainder 1 :)
Anyway i find it pretty hard too .. i'm trying :)
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And are you preparing those types of problems for INMO ?
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No I am not. I just was solving by my interest. That's it. Pretty interesting. Math is really fun!
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Agree :)
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but I did it in the calculator and it didn't work!

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Sorry but I cannot agree with you. I also divided it by limitless calculator and I have got this:
(7^51-1)/103=122 225 779 597 397x10^27.
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Dulat: Doesn't direct calculation as you've done constitute proof?
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Haha.) Yes. But in the olympiad I will not have any calculator. So I should prove in other way. I think you understand it.)
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Learn 'BASIC' and program it. "103" is an odd number, you cannot 'halve' the multiplier and get the same answer.

Comments (1)
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It is impossible to get the answer somehow. It can be proved just by the "Small theorem of Ferm". By showing that the remainder of 7^51 is 1 in mod103 and 1-1=0 will prove that 7^51-1 has no remainder and can be divided by 103.
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