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How many bases \(b\) are there such that 663\(_b\) is prime?

 Jun 30, 2019
 #1
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+1

Here is my "guess" (if I understand your question) !.

I don't think there is any base between 7 and 36 where 663 is prime. I couldn't find a single base between 7 and 36 where 663 is a prime number. Sorry, that is the best I could do.

 Jun 30, 2019
 #2
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+1

 

How many bases b are there such that 663b is prime? 

 

I don't think there are any.  663 is the same quantity, irrespective of what base you write it in, and that quantity is divisible.

 

.

 Jul 1, 2019
 #3
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+10

What about base 2? Rethink your answer Guest.

tommarvoloriddle  Jul 1, 2019
 #4
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Base 2 can only have digits 0 and 1.  No base up to and including 6 can have the digit 6 - that's why Guest #1 started at base 7.

Alan  Jul 1, 2019
 #5
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+1

 

tommarvoloriddle, my answer works for base 2

 

in base 10  --           663      divided by    39     equals   17        therefore not prime

in base   2  --  1010010111 divided by 100111 equals 10001    therefore not prime

 

That's what i meant by 663 is the same quantity no matter what base you write it in.  My interpretation of the question was different from that of Guest #1.  On reflection, and especially Alan's and your comments, I now realize I was looking at it the wrong way. 

.

Guest Jul 1, 2019
 #6
avatar+30021 
+3

The number 663b is   6b2 + 6b + 3  or  3(2b2 + 2b + 1)  which is the product of two integers (assuming b is an integer), hence cannot be prime for any base b.

 Jul 1, 2019

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