In the following infinite sequence
9 98 987 9876 98765 987654 9876543 98765432 987654321 9876543219
98765432198 987654321987 ... etc
how many numbers are prime?
Cinelli
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Sequence
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Re: Sequence
None. They're all divisible by at least one of 2, 3 or 5.
Let's call your sequence X[n], starting at X[1] = 9.
The key is that you've missed out 0, so we can look for patterns in X[9n+m], where n >= 0 and 1 <= m <= 9.
All we need to do is to find divisors for each m.
Every X[3n] and X[3n+1] is divisible by 3. So that covers the cases of m=1, 3, 4, 6, 7 and 9.
Every X[9n+2, 4, 6 or 8] is divisible by 2.
Golly, that leaves only m=5. Which ends in a 5, and thus is divisible by 5.
Sorry, too easy
Let's call your sequence X[n], starting at X[1] = 9.
The key is that you've missed out 0, so we can look for patterns in X[9n+m], where n >= 0 and 1 <= m <= 9.
All we need to do is to find divisors for each m.
Every X[3n] and X[3n+1] is divisible by 3. So that covers the cases of m=1, 3, 4, 6, 7 and 9.
Every X[9n+2, 4, 6 or 8] is divisible by 2.
Golly, that leaves only m=5. Which ends in a 5, and thus is divisible by 5.
Sorry, too easy
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Re: Sequence
cinelli wrote:In the following infinite sequence
9 98 987 9876 98765 987654 9876543 98765432 987654321 9876543219
98765432198 987654321987 ... etc
how many numbers are prime?
Spoiler...
None of them - the ones ending with 9, 7, 6, 4, 3 and 1 are divisible by 3, the ones ending with 2 and 8 are divisible by 2 and the ones ending with 5 are divisible by 5, all by the standard divisibility tests. That only leaves the possibilities that primes in the sequence are 2, 3 or 5 themselves, and the sequence does not include any of them.
Gengulphus
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Re: Sequence
Well solved UncleEbenezer and Gengulphus. Perhaps it was an easy one, but I think there is a place for puzzles which aren't too taxing.
Cinelli
Cinelli
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