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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
Last edited by Identity (2007-09-22 16:56:07)
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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.
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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Thanks, I think p=6n±1 was genius . never would have seen it coming myself.
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