I'm reading Hardy's wonderful book on theory of numbers. At some point he is proving that "almost all" integer numbers contain digit 9 (or any other sequence of digits, like "9345"). It's fairly obvious: number of integers up to $n$ digits is $a_n=10^n-1$. Number of integers up to $n$ digits with digit 9 missing is $b_n=9^n-1$. And because $\lim_{n\to\infty}{b_n}/{a_n}=0$ it means that $b_n \ll a_n$ for big enough $n$. In other words, the number of integers without digit 9 is "neglectable" (comparatively small).   So far so good, but in the footnote Hardy mentiones an interesting consequence without proof. Let $n$ be any number that is missing decimal digit 9. The following "quasiharmonic" sum is convergent:   $$\sum_{n=1}^{\infty}\frac 1n=\frac11+\frac12+\dots+\frac17+\frac18+\frac1{10}+\frac1{11}+\dots\frac1{18}+\frac1{20}+\frac1{21}+\dots$$  It also means that the sum of reciprocals of all numbers containing digit 9 must be divergent:  $$\frac19+\frac1{19}+\frac1{29}\dots+\frac1{89}+\frac1{90}+\frac1{91}+\frac1{92}+\dots$$  Is there a simple proof for this?
18378
I know that the harmonic series $1 + \frac12 + \frac13 + \frac14 + \cdots$ diverges. I also know that the sum of the inverse of prime numbers $\frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \cdots$ diverges too, even if really slowly since it's $O(\log \log n)$.  But I think I read that if we consider the numbers whose decimal representation does not have a certain digit (say, 7) and sum the inverse of these numbers, the sum is finite (usually between 19 and 20, it depends from the missing digit). Does anybody know the result, and some way to prove that the sum is finite?
90761
A curiosity that's been bugging me. More precisely:     Given any integers $b\geq 1$ and $n\geq 2$, there exist integers $0\leq k, l\leq b-1$ such that $b$ divides $n^l(n^k - 1)$ exactly.   The question in the title is obviously the case when n = 10. This serves as a motivating example: if we take b = 7 and n = 10, then k = 6, l = 0 works (uniquely), and if we calculate $\dfrac{n^k - 1}{b}$, we see that it turns out to be 142857 - the repeating part of the decimal expansion of 1/7. A (very sketchy, but correct!) sketch proof, which I've included for completeness:   Notice that $\dfrac{1}{99\dots 9} = 0.00\dots 01\; 00\dots 01\; 00\dots 01\dots$. Notice that $1/7$ must have either a repeating or a terminating decimal expansion: just perform the long division, and at each stage you will get remainders of 0 (so it terminates) or 1, 2, ..., 6 (so some of these will cycle in some order). It turns out the decimal expansion is repeating, and the 'repeating part' is 142857. This has length (k=)6. $\dfrac{1}{10^6 - 1} = 0.000001\;000001\dots$, so $\dfrac{142857}{10^6 - 1} = 0.142857\;142857\dots = 1/7$, and so $7\times 142857 = 10^6 - 1$. And we can do the same thing with $\dfrac{1}{10\dots 0} = 0.00\dots 01$ and terminating decimals. And the same proof of course holds in my more general setting.   But this (using the long division algorithm after spotting an unwieldy decimal expansion) feels a little artificial to me, and the statement is sufficiently general that I'm sure there must be a direct proof that I'm missing. Of course, the $n^k$ part is easy, but the $n^k-1$ part has me a little stumped. My question is: is there a direct proof of the latter part?  Thanks!
111940
Does every number not ending with zero have a multiple without zero digits at all?   We consider numbers in their usual base 10 representation. we don't consider numbers ending with $0$, because all multiples of such numbers end with $0$.   It can be proven that all powers of $2$ have such a multiple: Indeed, for every number of the form $2^n$, we can find a multiple $l$ of it whose last $n$ digits are nonzero (This can be done by repeatedly multiplying a number by $10^k+1$, as explained ), and then forget all other digits; these last $n$ digits constitute a number that is still divisible by $2^n$, since it is obtained from $l$ by subtracting a multiple of $10^n$ which is in particular a multiple of $2^n$.  The same argument shows that every power of $5$ has a multiple without zeros, but this does not seem to generalize to other numbers.  My intuition is actually that the claim is not true, since the density of numbers without any zero digits approaches $0$ for big numbers. This is because the probability of a random number with at most $n$ digits to contain only nonzero digits is $(\frac{9}{10})^n$. Thus big numbers "nearly always" contain a zero. This hints on the existence of a counterexample, but does not prove its existence.
19737
I'm trying to teach myself some number theory. In my textbook, this proof is given:  Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3. Let $n=a_0a_1\ldots a_k$ be the decimal representation of an integer $n$. Thus $n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k$ where $0\le a_i&lt;10$ The key observation is that $10\equiv1\pmod3$, i.e., $[10]=[1]$. Hence $[10^i]=[10]^i=[1]$, i.e., $10^i\equiv1\pmod 3$. The assertion is an immediate consequence of this congruence.  I don't understand the last statement. Why does it follow from that congruence?
35607
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.
45720
If $n$ is a positive integer such that the sum of all positive integers $a$ satisfying $1\leq a\leq n$ and $\gcd(a,n)=1$ is equal to $240n$ then the number of summands namely $\phi(n)$ is    120 124 240 480   My attempt:   If $n $= $p_1^{q_1}p_2^{q_2}...p_k^{q_k}$ where $p_1, p_2,...p_k$ are primes.  $240 n = \frac{n(n+1)}{2} - \frac{p_1^{q_1+1}-1}{p_1 -1}\frac{p_2^{q_2+1}-1}{p_2 -1}... \frac{p_k^{q_k+1}-1}{p_k -1}  +1 $  From here, I was not able to proceed. Guesswork tells me that the number has to be 480 because the sum is $240n$.
21336
I am trying to group numbers based on the count of digits in the square they produce, and below is the list containing the grouping that I've done so far.  1 - 3     // Numbers producing 1 digit when squared ... 4 - 9     // Numbers producing 2 digits ... 10 - 31   // Numbers producing 3 digits ... 32 - 99   // and so on ... 100 - 316   317 - 999   1000 - 3162 3163 - 9999 10000 - 31622 31623 - 99999 100000 - 316227 316228 - 999999   As one can observe, every range that produces odd digit count has a lower bound of $10^{i-1}$, while its upper bound has an observable pattern. I want to know if this can be represented by a formula or if the upper bound is a known constant such that we can use it as $C \cdot 10^{i-1}$.
1706
I am having trouble with the following proof:     For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$   My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.
12867
How many $N$ digits binary numbers can be formed where $0$ is not repeated. Note - first digit can be $0$.  I am more interested on the thought process to solve such problems, and not just the answer. If anyone can cite some resources for learning how to solve such problems would be great.
120131
The sum of the digits in $9 k$ (where $k$ is an integer) is a multiple of $9$: for example   $$9\cdot 1=9$$ $$9\cdot 7=63 \qquad  \text{and } 6+3=9\cdot 1$$ $$9\cdot 11=99 \qquad \text{and }  9+9=9\cdot 2$$  But why?
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.
69873
I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.    Question:  If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.    Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$  Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$   therefor  $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$  $9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$  Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.    The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$.  I think this is fair game here but not 100% confident.  Thanks.
59751
Original Title: Tauberian theorems and Cesàro sum  Theorem (Landau-Hardy, From Rudin's Principle of Mathematical Analysis Exercise 3.14)  $\newcommand\abs[1]{\left\lvert#1\right\rvert}$ If $\{s_n\}$ is a complex sequence, define its arithmetic means $\sigma_n$ by $$\sigma_n=\frac{s_0+s_1+\dotsb+s_n}{n+1}\qquad(n=0,1,2,\dotsc)$$ Put $a_n=s_n-s_{n-1}$ for $n\ge1$. Assume $M&lt;+\infty$ and $\abs{na_n}\le M$ for all $n$, and $\lim_{n\to\infty}\sigma_n=\sigma$, then $\lim_{n\to\infty}s_n=\sigma$.  The outline of the proof  If $m&lt;n$, then $$s_n-\sigma_n=\frac{m+1}{n-m}(\sigma_n-\sigma_m)+\frac1{n-m}\sum_{k=m+1}^n(s_n-s_k)\tag{*}$$ Notice that $\abs{s_n-s_k}\le(n-m-1)M\,/\,(m+2)$, fix $\epsilon&gt;0$ and associate with each $n$ the integer $m$ that satisfies $$m\le\frac{n-\epsilon}{1+\epsilon}&lt;m+1$$ Then $(m+1)\,/\,(n-m)\le1/\epsilon$ and $\abs{s_n-s_k}&lt;M\epsilon$. Hence $$\limsup_{n\to\infty}\,\abs{s_n-\sigma}\le M\epsilon$$  Questions and thoughts  It seems that the equation (*) comes out strangely. I wonder how to discover such kind of strange identities. So is there any observation, even deeper, to look through that equation?  Thanks!
18232
The staff of  has shown that the sum of all the integers from $0$ to $\infty$ is $-\frac1{12}$. Recently I was looking for the sum of all the (positive) integers from $0$ to $n$ and I found that: $$\sum_{i=0}^n i=\frac{n(n+1)}{2}$$  So I decided to take the limit: $$\lim_{n\to \infty}\frac{n(n+1)}{2}$$ but that tends towards $\infty$ when I expected that to be $-\frac1{12}$! Where did I got wrong? (the result is also confirmed by Wolfram Alpha)
63064
David Burton Elementary Number Theory Revised Printing question 4.3.5(a) is as follows:  Using the 9-test or 11-test, find the missing digits in the calculations below:  $(52817 * 3212146) = 169655x15282$  I can tell the number of the right isn't divisible by 9 or 11 since 5+2+8+1+7 = 23 and 3+2+1+2+1+4+6=19 are both not a multiple of 9 while 5-2+8-1+7=17 and 3-2+1-2+1-4+6=3 are both not a multiple of 11 so neither of the factors of 169655x15282 have 9 or 11 as factors.  Thus, $1+6+9+6+5+5+x+1+5+2+8+2 \not \equiv 0$ (mod 9) and $1-6+9-6+5-5+x-1+5-2+8-2 \not \equiv 0$ (mod 11)  so $x \ne 4$ and $x \ne 5$  Now, if I simply multiply 52817 by 3212146, I can see that $x = 9$ but I'm not sure how to get there from what I have.    How might I bridge this gap or is this just a typo in the text? Later editions seem to have altered the numbers involved.  Thanks
101242
Written, in Ex.8 Ch.9.1 of the book Advanced Calculus by P. M. Fitzpatrick :     Suppose that $\sum\limits_{k=1}^\infty a_k$ and $\sum\limits_{k=1}^\infty b_k$ are series of positive numbers such that $$\lim_{n \to \infty} \frac{a_n}{b_n}=l \ \ \ \text{and} \ l&gt;0.$$ Prove that the series $\sum\limits_{k=1}^\infty a_k$ converges iff the series $\sum\limits_{k=1}^\infty b_k$ converges.   Am I correct by the following (sketch of) proof? :  1- For a given $\epsilon_1$ there is $N_1$ such that $\left|\frac{a_n}{b_n} - l\right| &lt; \epsilon_1$ for all $n \ge N_1$.  2- Since the series $\sum\limits_{k=1}^\infty a_k$ converges, for a given $\epsilon_2$ there is $N_2$ such that $\left|b_{n+1}+\dots+b_{n+k}\right|&lt; \epsilon_1$ for all $n \ge N_2$ any for all natural numbers $k$.  3- Define $N = \max {\{N_1,N_2}\}$.   4- From Step (1), $a_{n+k} &lt; (\epsilon_1+l) b_{n+k}$ for all $n \ge N$ any for all natural numbers $k$. [Also, $a_i$'s and $b_i$'s are all positive]. Thus $a_{n+1}+\dots+a_{n+k} &lt; (\epsilon_1+l) (b_{n+1}+\dots+b_{n+k})&lt; \epsilon_3$. Then the convergence of the series $\sum\limits_{k=1}^\infty b_k$ implies the convergence of the series $\sum\limits_{k=1}^\infty a_k$.  5- For the reverse implication, we use the fact that $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=l \iff \lim\limits_{n \to \infty}\frac{b_n}{a_n}= \frac{1}{l}= l' &gt;0$ and repeat the process this time for a and b exchanged.   6- The Quotient Property For Sequences hold for a nonzero limit in the denominator, but since the limit in the numerator also is zero so we may use the The Quotient Property For Sequences.    Thanks.
97727
I found this very amusing comic on the internet the other day:   The last frame seems to claim that the harmonic series converges if you throw out all the terms with a $9$ in the denominator. Is this true? Where could I find a proof? Is it only true for terms with a $9$ or is it true for any number? If it's only true for $9$, why is that?