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#1 2007-09-22 16:55:46

Identity
Member
Registered: 2007-04-18
Posts: 934

Some number theory problems

1. If p is a prime ≥ 5, prove p²-1 is divisible by 24.

2.

ends in what digit?

3. What is the smallest prime divisor of

Thanks wink

Last edited by Identity (2007-09-22 16:56:07)

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#2 2007-09-23 02:13:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Some number theory problems

1.

If p is a prime, then it isn't divisible by 2 or 3. Therefore, dividing it by 6 would give a remainder of either 1 or 5 and so p can be generally represented by (6n±1).

p²-1 = (6n±1)²-1 = 36n² ±12n +1 -1 = 36n²±12n = 12n(3n±1)

If n is odd, then 3n±1 is even.
If n is even, then n is even. tongue

Therefore, n(3n±1) is always even.
This means that 12n(3n±1) is always divisible by 24 and so p²-1 also is.

2.

If we're only interested in the last digit, then we can turn 2137 into 7.
Now we look for a pattern in the last digit of the first few powers of this.

7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807

But as before, we're only interested in the last digit of each of these, and so the sequence goes 7, 9, 3, 1, 7...

To generalise, 7^k = 7^(k+4), in respect of last digit. That in turn means that we can take off multiples of 4 from 753 without affecting anything. 753 = 188x4+1, and so the last digit of 2137^753 is the same as the last digit of 7^1.

3.

Using similar reasoning to the last question, it can be deduced that the last digit of 3^11 is 7, and the last digit of 5^13 is 5. Therefore, the last digit of their sum would be 2 and so their smallest prime divisor would also be 2.


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-09-23 02:53:08

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: Some number theory problems

Thanks, I think p=6n±1 was genius smile. never would have seen it coming myself.

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