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When the number of 1s is odd, the number is divisible by the repunit you get by taking out all of the zeroes; e.g. 10101 is divisible by 111, 101010101 is divisible by 11111, and so on. This is because 1+102+104+⋯+102n=(1+10+102+⋯+10n)(1−10+102−⋯+10n). Note that the last sign in that alternating factor is a + only if n is odd.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Umm.. Yes, I can :)
Let's prove the contrapositive. In other words, we're going to prove that if n is not a prime number, then 9109−1 will never be a prime number.
Let n be equal to ab where a and b are two positive integers greater than 1.
Now notice that
910ab−1=910a−1×(1+10a+102a+103a+⋯10a(b−1))
which is definitely not a prime number.
A fun fact: primes in the form of 910n−1 are called repunit primes.
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This is true for any base.A repunit is of the form b−1bn−1 in base b.The above result holds in any base that a repunit is a prime if n is a prime.
Here's a related problem:
How many numbers of the form 1010…0101 are prime?
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Some initial thoughts:
101 is prime.
10101 is a multiple of 3. (This holds for all other cases where the number of 1s are a multiple of 3)
1010101 is the product of at least two factors, namely 101 and 10001. This holds for all greater cases where the number of 1s are even.
Haven't thought of when the number of 1s are odd though.
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When the number of 1s is odd, the number is divisible by the repunit you get by taking out all of the zeroes; e.g. 10101 is divisible by 111, 101010101 is divisible by 11111, and so on. This is because 1+102+104+⋯+102n=(1+10+102+⋯+10n)(1−10+102−⋯+10n). Note that the last sign in that alternating factor is a + only if n is odd.
only 101 is a prime
Yeah the one way of doing this. I was not sure how to put proof related questions on this forum. But this is good level 4 question otherwise :)