I'm wondering if some power of $2$ can be written in base $10$ as $2013$ followed by other digits.  Formally, does there exist $k,q,r \in \mathbb N$ such that $$2^k=2013 \cdot 10^q+r \,\,\,; \,\,\,r&lt;10^q $$  I'm not sure if it's true or not. I would go for a 'no', but I can't prove it.  Thanks for your help.
31344
Prove that for any finite sequence of decimal digits, there exists an $n$ such that the decimal expansion of $2^n$ begins with these digits.
48820
I stumbled across this problem  Find $\,10^{\large 5^{102}}$ modulo $35$, i.e. the remainder left after it is divided by $35$  Beginning, we try to find a simplification for $10$ to get: $$10 \equiv 3 \text{ mod } 7\\ 10^2 \equiv 2 \text{ mod } 7 \\ 10^3 \equiv 6 \text{ mod } 7$$ As these problems are meant to be done without a calculator, calculating this further is cumbersome. The solution, however, states that since $35 = 5 \cdot 7$, then we only need to find $10^{5^{102}} \text{ mod } 7$. I can see (not immediately) the logic behind this. Basically, since $10^k$ is always divisible by $5$ for any sensical $k$, then: $$10^k - r = 5(7)k$$ But then it's not immediately obvious how/why the fact that $5$ divides $10^k$ helps in this case. My question is, is in general, if we have some mod system with $a^k \equiv r \text{ mod } m$ where $m$ can be decomposed into a product of numbers $a \times b \times c \ \times ...$, we only need to find the mod of those numbers where $a, b, c.....$ doesn't divides $a$? (And if this is the case why?) If this is not the case, then why/how is the solution justified in this specific instance?
41968
I think i have proved that any number of the form $2^n$ can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not.  Let $g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))$, hence g is the sum of k consecutive positive integers.  $g=kx+\sum_{d=1}^{k-1}d$  Hence, $g=kx+\frac{k(k-1)}{2}$  Rearranging for x we find, $x=\frac{g}{k}-\frac{(k-1)}{2}$  So, for certain values of $k$, we can be sure that a positive integer value of $x$ can be found, and hence $g$ can be written as the sum of $k$ consecutive positive integers.There are two cases for $k$ to considor:  Case 1)  $k$ is odd. If $x$ exists, it must be the case $k$ | $g$, as for odd values of $k$, $\frac{(k-1)}{2}$ will alwyas be an integer, and hence $x$ will.  Case 2)  $k$ is even. $k$ does not divide $g$, but $\frac{g}{k}$ must be a multiple of $\frac{1}{2}$, and hence equals $\frac{h}{2}$, for some odd $h$.  Letting $j=2^n$, assume that we can write $j$ as the sum of $k$ consecutive positive integers.  If $k$ is odd, then if a $k$ exists, it must divide $j$, by Case 1. But, $j$ has no odd factors as it is a power of two, hence $k$ could not be odd.  If $k$ is even, by Case 2 it mut be the case $k$ does not divide $j$. Also, $\frac{j}{k}=\frac{h}{2}$, for some odd $h$.  But, this implies $2j$ has an odd factor, $h$. but, as $2j$ is a power of $2$, it can't have odd factors. So, $k$ can't be odd.  Hence, $k$ is neither odd or even, and this is a contradiction to our assumption. Hence no $x$ exists, and hence and number of the form $2^n$ can't be written as the sum of $k$ consecutive positive integers.  Any feedback on this proof would be appreciated.
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Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?  a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$
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I am trying to show that if $n$ is an odd natural number, then $8$ divides $n^{2}-1.$ I was able to prove that because I know that if $n$ is an odd natural number, then $n^{2}$ can be written as $8k+1$ for some $k\in \mathbb{N}.$ I would like to show this question by using the Euclidian division. Then I wrote $n^{2}-1=8k+r$, where $0\leq r  &lt; 8.$ When $r=2$ we get $n^{2}-1=8k+2.$ Since $n$ is odd, then $n^{2}-1$ is even and I got stuck. Is there way to fix that?
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Find The Last 3 digits of the number $2003^{2002^{2001}}$  BY number theory or otherwise,   Also i would like to ask is there a property observed in the numbers of the form $k^n$, where for some $k, n$ is varied then the digits of $k^n$ are periodic,  for example,  $2^n$,  its last digit is periodic with period 4, its second last digit is periodic $4\cdot 5 = 20$ its third last digit is periodic with periodic with period $20\cdot 5 =100$  I have observed this property with other numbers as well, though period might vary,for different values of $k$.
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Find all $n\in\mathbb N^+$ such that the sum of the digits of $5^n$ is equal to $2^n$.   I've solved this, but I think the proof is a bit weird and I wonder if there's a better one.    My proof:  We can see that $n\neq 1$, $n\neq 2$, $n=3$, $n\neq 4$ and $n\neq 5$.  Let $n&gt;5$. Notice that $5^6$ has $5$ digits, while $\frac{2^6}{9}=7\frac{1}{9}$, so $5^6$ has to have at least $8$ digits in order for the sum of the digits to be able to at least reach $2^6$.  Everytime $1$ is added to the exponent of $5$, the amount of digits $5^n$ gets is at most $1$, while $\frac{2^{n+1}}{9}-\frac{2^n}{9}=\frac{2^n}{9}&gt;3$, so the lowest amount of digits it has to have to reach $2^n$ becomes higher by at least $4$.  When $n=6$, the amount of digits $5^n$ has is already lower than the amount it has to have in order to reach $2^n$, and I've just proved that the amount it has to have in order to reach $2^n$ grows faster than the amount of digits $5^n$ has, therefore $5^n$ will never be able to reach $2^n$ when $n&gt;5$.  Hence the only solution is $n=3$.
73300
Possible Duplicate:          This is not a practical problem. I am asking out of curiosity.  Any links/references are most welcome.  Say, we write the digits of $\pi$ in base $10$.  Does this sequence of digits contain every possible finite length digit sequence?  What about $e$, $\sqrt{2}$ or some other commonly known irrational numbers?  Is this property of numbers independent of base?  If a number has this property when written in base $10$, will it also have it in base $2$, $3$ and all other bases?
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Prove that $n&lt;3^n$ where $n \in \mathbb N$,  when $ n=1 $, I have proved it's true. And assumed when $n = p$ , $p&lt;3^p$ is true. Can any body help me in showing that it is true for $n =p+1$
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For which odd positive integer $n$ , is it true that $-1$ is not a positive power of $2$ modulo $n$ i.e. $[-1] \ne [2^k] , \forall k &gt;0$ in $\mathbb Z_n$ ? Is there any ( at least sufficient ) characterization for such odd $n$ ?   ( I have only found $n=7$ is such a number ) . Please help . Thanks in advance
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As a follow-up to , the next two exercises state:       Use the given generators and relations to show that if $x$ is any element of $D_{2n}$ which is not a power of $r$, then:         a) $rx = xr^{-1}$, and      b) $x$ has order 2, and hence $D_{2n}$ is generated by $s$ and $sr$, each with order 2.       Here, $D_{2n}$ has the "usual" presentation  $$D_{2n} = \left\langle r, s \mid r^n = s^2 = 1, rs = sr^{-1} \right\rangle.$$  For part a), since $x$ is not a power of $r$, I have written $ x = sr^k$ for some $0 \leq k \leq n - 1$.  So then $rx = r(sr^k) = (rs)r^k = (sr^{-1})r^k = sr^{k - 1}$, but this doesn't seem to be moving me in the right direction.  Should I be using induction on the power of $r$ in the element $x$ ?  Then for part b), I simply wrote $(sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = s^2 = 1$, but I doubt that I have sufficiently justified my work.  From the comments to the previous question, I get the feeling that this is a very simple verification, but I'm having trouble.  Thanks in advance for your help!
105962
How can I check if $$12^{2013} + 7^{2013}$$ is divisible by $19$?  Also, how can I format my questions to allow for squares instead of doing the ^ symbol.
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I first noticed this is true for the integers of the sequence $9, 99, 999, 9999,\dots$, since for some term $a_n=10^n-1$ in the sequence and $p$ a prime other than $2$ or $5$, we have $a_n\equiv 0 (mod\ p)\Leftrightarrow 10^n\equiv 1 (mod\ p)$ and we can find infinitely many $n=k(p-1)$ so that $10^n=(10^{p-1})^k\equiv 1^k\equiv 1 (mod\ p)$ by Fermat's little theorem.   I'm curious, is there a similar way to show this is true for repunits $1, 11, 111,\dots$?
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I just want to know if my proof to this question is correct. First, I proved it was true for $n = 4$. $$4!&gt;2^4$$ $$24&gt;16$$ Then, I assumed that it was true for $n=k$. $$k!&gt;2^k$$ Afterwards, I tried to prove it for $n=k+1$. $$(k+1)!&gt;2^{k+1}$$ $$(k+1)k!&gt;(2)(2^k)$$ Since $k!&gt;2^k$, am I allowed to simply do $$(k+1)(2^k)&gt;(2)(2^k)$$ $$(k+1)&gt;2$$ and this is true for all $k\geq3$. Would the last two steps be illegal because I only assumed that $k!&gt;2^k$ is true?
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Prove/Disprove: Find $a\in S_4$ that is not a power of $b$ such that $ab=ba$ while $b=(123)$   What I did:  I think this statement is false.  So we need to find $aba^{-1}=b$, we know that $id\in S_4$ will surely $id*b*id^{-1}=b$ but we get $id^k=b$, I wonder if there are more elements such that $ab=ba$, how to find them?   Any help will be appreciated.
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Prove that $2^{10}+5^{12}$ is composite   I need to solve this using only high school mathematics. Any ideas?
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I just found this identity but without any proof, could you just give me an hint how I could prove it? $$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$  I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me