Find the largest prime factor of 6 6 4 − 4 6 4 .
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6 6 4 − 4 6 4 = 2 6 4 ( 3 6 4 − 2 6 4 ) = 2 6 4 ( 3 3 2 + 2 3 2 ) ( 3 3 2 − 2 3 2 )
= 2 6 4 ( 3 3 2 + 2 3 2 ) ( 3 1 6 + 2 1 6 ) ( 3 1 6 − 2 1 6 )
Now, 3 3 2 + 2 3 2 = 1 1 5 3 × 1 6 0 7 1 3 3 1 1 6 9 2 9
Of the other factors in the product, the largest one is 3 1 6 + 2 1 6 = 4 3 1 1 2 2 5 7 . Regardless, of the other factors being prime or composite, it can be said that the largest prime factor in this case would be 1 6 0 7 1 3 3 1 1 6 9 2 9
How do you know that 1607133116929 can't be further factorized?
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Can this be solved efficiently without using a computer?
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Yes, with plenty of time and patience and nothing else better to do.
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@Pi Han Goh – I meant efficiently. I don't consider checking all primes less than the square root of 1607133116929 efficient enough. :D
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@Jesse Nieminen – In that case, I doubt there is an efficient approach. Primality testing is mainly left for computers to do.
Can be solved in more easier way by repeatedly factorizing it
I used a computer program to calculate the prime factors. The prime factors are 2,601,1253279,8097823. So 6 6 4 − 4 6 4 = 2 1 1 3 × 6 0 1 × 1 2 5 3 2 7 9 × 8 0 9 7 8 2 3
I don't really see an easy way to do this. And anyway, the 'correct answer' is definitely wrong.
Edit: as a matter of fact: 2 3 2 + 3 3 2 doesn't equal 1 6 0 7 1 3 3 1 1 6 9 2 9 at all...
Double edit: I'm stupid, ignore my comment
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In my comment, I say that 3 3 2 + 2 3 2 = 1 6 0 7 1 2 2 1 1 6 9 2 9 × 1 1 5 3 . You had mistook it.
Further, I think that your factorization is wrong. The reasoning is thus :
The last digit of 6 6 4 − 4 6 4 : The last digit of any power of 6 would be 6 and the last digit of any even power of 4 would also be 6 . Hence, the last digit of the difference would have to be 0 .
The last digit of the factorization : From the above logic, it is clear that at least one of the prime factors should be 5. But, that is not the case in the proposed factorization.
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You are indeed correct, my apologies. It seems python does not handle divisions with large digits very well, as my program works perfectly for numbers smaller than about 12 digits.
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It is not a very good problem.