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10001, 100010001, .... (Posted on 2003-05-23) Difficulty: 3 of 5
Consider the following sequence:

10001, 100010001, 1000100010001, 10001000100010001, 100010001000100010001,....

Show that none of them are prime numbers.

See The Solution Submitted by Fernando    
Rating: 3.3333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re: a twist, perhaps? | Comment 7 of 12 |
(In reply to a twist, perhaps? by DJ)

My solution is complete, although my final explanation probably didn't make it very clear.

Suppose you have the equation
where you know the following:
a, b, c, and d are integers
b is prime
b is not equal to c or d

Thus, since b is prime it must be a factor of either c or d, giving an equation:
a=(c/b)d OR a=c(d/b)

In either case, since b is not equal to c or d, both terms on the right must evaluate to integers greater then 1, and we know that a cannot be prome.

This is the situation in my solution. I have derived the equation:
101(1 + 10^4 + 10^8 + ... + 10^4i) = (1 + 10^2 + 10^4 + ... + 10^2i)(10^(2i_2) +1)
In which 101 is prime, and neither of the left-hand members is equal to 101 when i>1. Therefore, by the same reasoning described above, we know that all numbers of the form
1 + 10^4 + 10^8 + ... + 10^4i
are not prime (when i=1, the equation becomes 101(10001)=101(10001), which proves nothing, but 10001=73(137), so it is also not prime).
  Posted by DJ on 2003-05-25 04:46:47

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