All Questions
22
questions
1
vote
1
answer
129
views
Proving $\sum_{1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $
Show that $$\sum_{i=1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $$ where $ \lceil\cdot\rceil$ denotes the ceiling function.
My method: one way would be observe each part of the ...
2
votes
2
answers
153
views
$\sum_{k=0}^\infty[\frac{n+2^k}{2^{k+1}}] = ?$ (IMO 1968)
For every $ n \in \mathbb{N} $ evaluate the sum $ \displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right]$ ($[x]$ denotes the greatest integer not exceeding $x$)
This was IMO 1968, 6th ...
2
votes
4
answers
62
views
For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
By dividing the numbers between $1$ and $5^m$ as intervals of $5^k$, I was getting the following ...
3
votes
2
answers
689
views
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
How many values of $n$ are there for which $n!$ ends in $1998$ zeros?
My Attempt:
Number of zeros at end of $n!$ is
$$\left\lfloor \frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^2}\right\rfloor+\...
2
votes
1
answer
447
views
Sum of all divisors of the first $n$ positive integers.
Yesterday I was going through Möbius Function notes, and found that
writing $n = p_{1}^{\alpha_1}p_{2}^{\alpha_2}\cdots p_{r}^{\alpha_r}$,
the sum of all divisors can be written as.
$$
e(n) = \prod_{...
3
votes
2
answers
227
views
Find n in sum that results in a number $aaa$
Lets say that we have the sum $1+2+3+\ldots+n$ where $n$ is a positive natural number and that this sum should equal a three digit number in which all the digits are the same, for example $111, 222,$ ...
0
votes
1
answer
100
views
Is this a valid definition of the rationals?
$$\mathbb{Q}=\left\{\sum_{n=1}^k f(n)\mid k,n\in\mathbb{N}\land f\text{ is a finite composition of $+$, $-$, $\div$, $\times$}\right\}$$
Reasoning:
Any real number can be described by a (sometimes ...
3
votes
2
answers
199
views
Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$
Calculate $ \biggr\lfloor \frac{1}{4^{\frac{1}{3}}} + \frac{1}{5^{\frac{1}{3}}} + \frac{1}{6^{\frac{1}{3}}} + ... + \frac{1}{1000000^{\frac{1}{3}}} \biggr\rfloor$
I am just clueless. I just ...
2
votes
2
answers
83
views
Suppose that $1+2+...+n=\overline{aaa}$. Which of the following items CERTAINLY divides $n$? $5,6,7,8,11$
Suppose that $1+2+...+n=\overline{aaa}$. Which of the following items CERTAINLY divides $n$?
$5,6,7,8,11$
I converted the given relation into the following:
$$n(n+1)=2*3*37*a$$
Now I think ...
3
votes
2
answers
102
views
How many integers $n$ for $3<n<100$ are such that $1+2+3+\cdots+(n-1)=k^2$, with $k \in \mathbb{N^*}$?
I know that the sum $1+2+3+\cdots+(n-1)$ equals $\frac{(n-1)\cdot n}{2}$.
I wrote the equation in the two following forms:
$$(n-1)\cdot \frac{n}{2}=k^2$$
$$(n-1)\cdot n=2k^2$$
And I tried to find ...
4
votes
1
answer
500
views
Does anyone know how to reduce this sum of sums into something simpler in order to find a special value? [duplicate]
to clarify the difference between this and the supposed duplicate, these two questions talk about completely different functions with completely different purposes
I was given this from a friend. ...
0
votes
2
answers
124
views
Proof of Number Theoretic Function $\sigma$ [closed]
If $N$ is a positive integer then,$$\sum\limits_{n=1}^{N}\sigma(n)=\sum\limits_{n=1}^{N}n\lfloor\dfrac{N}{n}\rfloor$$, where $\lfloor.\rfloor$ denotes greatest integer function and $\sigma(n)$ denotes ...
3
votes
4
answers
186
views
Summation or Integral representation ${e^{2}\above 1.5pt \ln(2)}=10.66015459\ldots$
How can I construct a summation or integral representation of $${e^2\above 1.5pt \ln(2)}.$$ Naively I would write $$\Bigg(\sum_{n=0}^{\infty}{2^n \above 1.5pt n!} \Bigg)\Bigg(\sum_{n=1}^\infty {(-1)^{...
11
votes
1
answer
306
views
Proving if it is possible to write 1 as the sum of the reciprocals of x odd integers
Let $x$ be an even number. Is it possible to write 1 as the sum of the reciprocals of $x$ odd integers? Write a proof supporting your answer.
I tried a lot of these, and I think it is no because I ...
14
votes
4
answers
462
views
Show that the numerator of $1+\frac12 +\frac13 +\cdots +\frac1{96}$ is divisible by $97$
Let $\frac{x}{y}=1+\frac12 +\frac13 +\cdots +\frac1{96}$ where $\text{gcd}(x,y)=1$. Show that $97\;|\;x$.
I try adding these together, but seems very long boring and don't think it is the right way ...