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 Apr 19 awarded Yearling Apr 19 awarded Yearling Feb 9 comment Repeated operations on $(a,b)$ I see, I assumed this was the same question and missed the fact that this one was restricted to the smaller number. Makes sense and I agree, it does seem that not every pair is solvable in this manner. Feb 9 comment Repeated operations on $(a,b)$ That's why there's a separate solution for $k=1$. I didn't write the solution presented there down as a proof because I'm skeptical that there isn't a shorter method, but there is certainly a solution for $k=1$ as I listed above. Feb 9 comment Repeated operations on $(a,b)$ All numbers of the form $(x,x+1)$ have a solution using 10 steps. Assume $a,b$, and $L$ doubles $a$ and adds 1 to $b$, while $R$ doubles $b$ and adds 1 to $a$, then the operations are $RRLLLLRLRR$ and the solution reaches the diagonal at $32x+68$. See a complete solution at reddit.com/r/mathriddles/comments/2v6eaj/doubling_and_adding_1 Feb 6 comment Does there exist a Benny number? $11 = 11+ s(11) - l(11) = 11 + 2 - 2$, how is that not a Benny number? Jan 27 comment Equation system modulo prime if $p=14$ then $13\equiv 5*11$, so $p$ can definitely be greater than $13$ Jan 19 comment Counting rings of order $p^3$ Ah, yes, of course. I apologize. You're absolutely correct. Jan 19 comment Counting rings of order $p^3$ I think its worth pointing out that the number listed in the series in OEIS is also 52. So in that regard MathWorld and OEIS agree. Jan 6 comment If a prime number is reversed, and then appended to itself, why is the result always a composite number? @HaraldHanche-Olsen: Any number appended in reverse to itself will necessarily have an even number of digits. Jan 2 answered Minimum moves to destroy Jan 2 comment Minimum moves to destroy I don't think you'll get an $O(n)$ algorithm, at best probably $O(n log(n))$. Dec 29 comment Does my “Prime Factor Look-and-Say” sequence always end? How would you handle the number $314928$? Dec 18 comment Is it possible to permute an unknown binary sequence so that two particular bits are equal? If we restrict the problem to input sequences in which there are at least two of each value (i.e. all sequences of 1's and 0's which have at least two 1's and at least two 0's) then it becomes possible. Dec 3 accepted Proving a property about quadratic residues Dec 3 comment Proving a property about quadratic residues Perfect, I knew I was missing something simple. Dec 3 asked Proving a property about quadratic residues Oct 7 comment Is there a discrete relationship shared between patterns and series or sequences? For your example, to define the pattern as a sequence you might use $a(n) = a(n-1) + 2010 + 200*(n \% 2)$. There's not necessarily a reason to suggest that the next number in the sequence HAS to be +2210, but absent any other available information it's a reasonable choice for continuing the pattern. Although consider if the sequence had been instead $a(n) = a(n-1)+2010+200*(f(n))$ where $f(n)$ is the fibonacci sequence starting with the unconventional $1,0,1,1,2,...$ The initial three terms would still be identical. Jul 2 awarded Curious Jun 3 answered Why do negative exponents work the way they do?