How many primes, written in usual base 10, have digits that are alternating 1s and 0s, beginning and ending with one?
For example (not necessarily prime):
1, 101, 10101, ...
All numbers in base-10 of alternating 1s and 0s ending in 1 are in the form (10
2n-1)/99.
As as (10
2n-1)/99 equals (10
n-1)(10
n+1)/99; and, as a prime is a positive integer that can have no other factors besides 1 and itself; either (10
n-1) or (10
n+1) must equal one of the divisors of 99 (99, 33, 11, 9, 3, 1) with the other having a prime factor and no other factors but that that composes the respective complementary divisor (1, 3, 9, 11, 33, 99).
One can consequently find that (10
n-1) must equal 99 and (10
n+1) must equal 101. Thus, the only possible number is 101, and, therefore the answer to the puzzle is
there is only 1 prime in base-10 of alternating 1s and 0s ending in 1.
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Posted by Dej Mar
on 2011-06-22 05:56:06 |
solution
