I'd never seen this kind problem before, and don't know where to start. Any help is appreciated. Thank you very much!     A palindrome is a number that is the same forwards and backwards. For example, $212$ and $21466412$ are palindromes. Consider an arbitrary palindrome with n digits where n is even. Find a formula, in terms of $n$, of how many $n$-digit palindromes exist. All equivalent answers will be accepted.
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A palindrome is a set of characters which equal the same thing forwards and backwards, for example; abccba. For a set of a given amount of  characters (we can use 9 as an example) how would you calculate the amount of palindromes of characters mathematically? This doesn't mean the words that can be formed but rather characters which are the same at start characters and end characters. What formula can be used to calculate this for simple lengths like 9 with lowercase letters?
86936
Sorry if this has been asked before, but I have a query about the notion that the decimal expansion of $\pi$ contains every possible string of numbers (please note that I am only a "casual" maths enthusiast). If it does, then would the expansion not contain $\pi$ itself? (I.e. 3.1415926...31415926...) That would make $\pi$ a repeating decimal which could, in theory, be represented as an exact fraction. If we constrict the argument to say that $\pi$ only contains every finite sequence of numbers, then wouldn't that be contradictory (we would see 3, 31, 314, 3141, 31415... so why not all the way? We could always add another digit to create a longer finite string ad infinitum)?  Thank you in advance.
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I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.
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Sometimes, more in some languages than in others, a word is spelled using one or more numbers instead of letters. Is there a term for this?  The only example I can think of in English is se7en (seven).   Therefore,     The number se7en was not spelled entirely with letters; it was written in   ______.
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How many $4$ digit even numbers have all $4$ digits distinct ?   This question has been asked if we replace even with positive.   I was solving this Problem like: For first position $1-9$ (No $0$) for next $0-9$ (leaving out the digit which has been choose) for next $0-9$ (leaving out $2$ digits), but on last digit I get stuck. From what I should choose. Will it be $0, 2, 4, 6, 8$, because number needs to be even? How should I solve it?
33275
Can someone reverse this formula? Sorry if it's too basic, I'm an old guy and my math is rusty :-D  $s =\frac{n(n+1)}{2}$  If it ain't clear, by reverse I mean obtaining the value of n by providing s  Thanks
14842
Well if one looks at the formula of circular permutations $Pc = (n-1)!$  But as we come to that formula, I need a concrete example and an explanation.  Another thing, I have seen that when working with a bracelet, the formula changes. A concrete example please along with an explanation  Finally how the formula is worked when there are repeated elements that are permuted.  Thank you
114610
I have the following recurrence that I've been pounding on:  $$ a(0)=1\\ a(1)=1\\ a(2)=2\\ a(2n)=a(n)+a(n+1)+n\ \ \ (\forall n&gt;1)\\ a(2n+1)=a(n)+a(n-1)+1\ \ \ (\forall n\ge 1) $$  I don't have much background in solving these things, so I've been googling around looking for different techniques. Unfortunately, I haven't been able to correctly apply them yet. So far I've tried:   Characteristic equations Generating functions Plugging it into Wolfram Alpha Telescoping Observation   I'm sure that one or more of these is the way to go, but my biggest problem right now is figuring out how to deal with the idea that there are different equations for odd and even values of $n$. I'm not necessarily looking for a solution (though it would be gratefully accepted), but any nudges in the right direction would be much appreciated.    To follow up, it turned out that the speed problem described in my comment, below was related to the Decimal implementation in Python 2.7. Switching to Python 3.3 (and Java) made everything many orders of magnitude better.
14717
The problem is: Find $C_n$, the number of all ternary strings (length $n≥0$) that don't have consecutive 1's and 2's, using combinatorics.   Any tips or advices?
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One common math puzzle I've seen around asks for how many zeros are in the product of "100!"  Usually, the solution everyone gives goes something like try to match pairs of 5s and 2s that factor out of the numbers, which ends up being 24 zeroes (you can factor a 5 out of 20 of the numbers, and factor 2 5s out of 4 of the numbers; you can factor more than  24 2s out).   This however as far as I know gives the number of trailing zeroes at the end of the number, but does not account for the zeroes that are within the number. My question is, is this answer correct anyways? Can there be zeroes that aren't trailing that are inside? Why or why not and if there can be can we somehow figure out how many are within the product?   Thanks
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Let's say $A_n$ is the number of binary string that has length $n$ and does not contain the substring 101. Calculate $A_n$ for $n=1,2\cdots8.$ Find a recurrence relation for $A_n$. What does the solution of that recurrence look like?  These are the solutions that I have found for calculations for $A_n$. $1, 4, 7, 12, 20, 32, 48, 96$.  I've calculated this by hand. But how I do find the recurrence? I see that 4, 7, 12, 30 are Fibonacci - 1, but not after or before that.  But I'm not sure how to do this or if this is even correct.
113283
I have a problem understand this math problem:     Write a number on the board. This number is either multiplied by $2$ or raised to a square. If the number $"1"$ is written at the beginning, at least how many steps should be taken to reach $2^{2018}?$   A) $15$  B) $16$  C) $17$  D) $18$  E) $12$  I can't solve this problem. Because I don't understand the question. Now, I need to understand the question. Then maybe I can.  Is there a problem in the question? The question unclear for me...Can you explain me, what is the meaning of the question?
72410
Now I have proved the following two problems:      (1) Prove that a number and the sum of its digits have the same remainder upon division by 9.    (2) Given an integer, consider the difference between the sum of the digits in odd positions (counting from the right) and those in even positions. Show that the difference has the same remainder as the number itself when divided by 11.    Then, I need help on the following problem:      Partition the digits of a number into groups of two and three, respectively, starting from the right. Treat these as two- and three-digit numbers, respectively, and proceed similarly as in the two problems above. What divisibility rules arise?    In fact, I did not quite get what this problem asks for and not clear about this process. Could anyone give me some hints on this, please?  Thanks in advance.
4492
That was an interview question that I was unable to answer:  How to check that a string is a palindrome using regular expressions?  p.s. There is already a question "" and it gives a lot of answers in different languages, but no answer that uses regular expressions.
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I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$. Appreciate your help :)
16594
Can you find the smallest positive number such that if you shuffle the digits of the number in a particular order, the shuffled number becomes twice the original number.  Source:   I understand the answer is $125874 =&gt; 251748$  $251748$ is twice the $125874$ and have same digits $1,2,4,5,7$ &amp; $8$  but how to solve this non programmatic ?
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Hello everybody! The problem you see above is a combinatorial problem that I could not solve. :(  Since we are to choose distinct numbers from $\{1,2,3 \ldots K\}$, the numbers must be in increasing order because they are all positive. I think that the difficulty of this problem is in the fact that the Special Sequence can be of any length. We can go about choosing say $n$ elements in $\binom{K}{n}$ ways but again how do we control that? Talking about a recursive solution, I think that is the way to go but I can't get how to do that (that is get a recursive formula for that).  Any help in solving this problem would be much appreciated. Thanks.