( 1 0 0 0 2 0 0 0 )
Determine the largest 3 digit prime factor of the above number.
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Sir, I think your answer is to the problem- Prove that 661 is the largest 3 digit prime factor of ( 1 0 0 0 2 0 0 0 )
How did you initially got the number 661?
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No, since any prime number greater than 1 0 0 0 and less than 2 0 0 0 will divide ( 1 0 0 0 2 0 0 0 ) .
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I am sorry, I wanted to say you have proved that 661 is the largest 3 digit prime factor of ( 1 0 0 0 2 0 0 0 ) , but how did you get the number 661. Like you started from 1000>p>661...
(In the first comment, I missed "3 digit").
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@Vilakshan Gupta – By looking for indices, just as in my proof. Any 3 -digit prime greater than 5 0 0 has index 1 in 1 0 0 0 ! , so we are just looking for such primes with index greater than 2 in 2 0 0 0 ! . Since 6 6 1 is the largest prime less than 3 2 0 0 0 , we are done.
Mathematica
Max@Select[First /@ FactorInteger@Binomial[2000,1000],#<1000&]
returns 661
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If p is prime and 1 0 0 0 > p > 6 6 1 , then 3 p > 2 0 0 0 and so the only multiples of p between 1 and 2 0 0 0 are p and 2 p . Thus the index of p in 2 0 0 0 ! is 2 . In addition the index of p in 1 0 0 0 ! is 1 , and hence the index of p in ( 1 0 0 0 2 0 0 0 ) is 0 , meaning that p is not a factor of ( 1 0 0 0 2 0 0 0 ) . On the other hand 3 × 6 6 1 = 1 9 8 3 < 2 0 0 0 , and so the index of 6 6 1 in 2 0 0 0 ! is 3 , while the index of 6 6 1 in 1 0 0 0 ! is 1 . Thus the index of 6 6 1 in ( 1 0 0 0 2 0 0 0 ) is 1 , and hence 6 6 1 divides ( 1 0 0 0 2 0 0 0 ) , and so is the largest 3 -digit prime factor of ( 1 0 0 0 2 0 0 0 ) .