All Questions
27
questions
2
votes
0
answers
94
views
Need help with this sum involving linear congruences
A few months ago while dealing with squarefree integers in arithmetic progressions, this problem arose about finding bounded solutions to linear congruences.
The problem:
I have constants $a$ and $q$ ...
- 1,009
1
vote
0
answers
65
views
Prove that $\lim\limits_{n\to\infty} a_n=\infty$.
Let $a_n$ denote the exponent of $2$ in the numerator of $\sum_{i=1}^n \dfrac{2^i}{i}$ when written in the lowest form. For example, $a_1=1,a_2=2,a_3=2.$ Prove that $\lim\limits_{n\to\infty} a_n=\...
- 1,837
0
votes
1
answer
46
views
Prove that $\sum_{i = 1}^{N} 1+ (2i \bmod N) = N(N + 1) / 2$ for odd N.
I was able check by hand that for odd $N$ the $1+ (2i \bmod N)$ produces all values between $1$ and $N$ and for even $N$ there are repeats.
But I've no ideas on how to write a mathematical proof for ...
3
votes
0
answers
112
views
Minimum sum of digits for a polynomial
Given $n$ a natural number (greater than $0$) and $p(n)$ a polynomial, $p(n) = 2n^2 + 43n + 502$. Defining the arithmetic function $s(k) : \mathbb{N}^* \to \mathbb{N}^*$, the sum of the digits of the ...
1
vote
0
answers
89
views
A sum of modular inverses of one prime for different moduli
Note: Here, for given $m$ and $a$, I define $a^{-1}_m \in \{0, 1, 2, \ldots, m-1\} $, such that $a \cdot a^{-1}_m \equiv 1 \pmod m $.
Say, we have a prime $p$ and a whole positive number $n$ ($n <...
- 132
1
vote
1
answer
110
views
Find remainder of $\sum^{2015}_{n=1}\big(\frac{n+2}{2}\big)^{n+2}$ when divided by $23$
Find the remainder ($r$) of $\displaystyle\sum^{2015}_{n=1}\left(\frac{n+2}{2}\right)^{n+2}$ when divided by $23$.
My attempt:
$\frac{n+2}{2}=1+\frac n 2$
$A=\displaystyle \sum^{2015}_{n=1}\left(\frac{...
- 11.2k
4
votes
1
answer
110
views
If $ 1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$ where $A$ and $B$ are coprime positive integers, then $5\nmid A$ and $5\nmid B$.
Let the sum $$1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$$ where $A,B\in \mathbb{N}$ and $\gcd(A,B)=1$. Show that neither $A $ nor $B $ is divisible by $5$.
My attempt: $$\begin{...
- 750
0
votes
1
answer
180
views
How to prove $f(n)=\sum_{k=0}^{n}C_{2n+k}^{n}C_{n-1+k}^{n-1}\equiv1\pmod2$?
Prove that $$f(n)=\sum_{k=0}^{n}\binom{2n+k}{n}\binom{n-1+k}{n-1}\equiv1\pmod2,\quad\forall n\in N^{+}$$
I have tried:
By Lucas TH, we have $f(2n)\equiv f(n)\pmod 2$, but $$f(2n+1)=\sum_{k=0}^{n}\...
- 93.6k
4
votes
0
answers
146
views
A $p$-congruence involving binomial coefficients and the floor function.
Let $p$ be a prime number.
Let $k$ and $m$ be positive integers such that $k \ge m$ and $p-1$ does not divide $k-m$.
It seems to be true that $$ (-1)^{m-\lfloor
\frac{k-m}{p-1}\rfloor}\sum_{i\...
- 3,716
2
votes
2
answers
417
views
The first digit (on the left) and the last digit (on the right) of $\sum_{k=1}^{1010}k^{2020-k}$
The first digit (on the left) and the last digit (on the right) of
$$\sum_{k=1}^{1010}k^{2020-k}=1^{2019}+2^{2018}+3^{2017}+\dots+1010^{1010}$$ are to be obtained not by using computers/calculators.
...
- 5,450
3
votes
0
answers
126
views
Show that $Q(Q(Q(2005^{2005})))=7$ [duplicate]
Recently, I have found this problem:
Let $Q(n)$ denote the sum of the digits of a positive integer $n$. Prove that: $$Q(Q(Q(2005^{2005})))=7$$
This topic is related to other two question I have ...
- 6,581
3
votes
1
answer
245
views
Computing $\sum_{k=1}^n (a^k \bmod m)$
I would like to find a closed form solution for
$$\sum_{k=1}^n (a^k \bmod m)$$
$$0<a<m, n > 0$$
Note that the mod operator is within the brackets.
If a closed form solution does not exist, ...
- 133
5
votes
2
answers
277
views
Show that $\sum\limits_{i=1}^{p} i^{p}$ is divisible by $p$ for all primes $p > 2$.
Show that
$$
\sum\limits_{i=1}^{p} i^{p}
$$
is divisible by $p$ for all primes $p > 2$.
I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to ...
- 51
1
vote
1
answer
156
views
Confirmation of Proof: $\exp_a\Big\{\sum_{k=1}^n \gcd(k,n)\cos\left(\frac{2\pi k}{n}\right)\Big\}\equiv 1\pmod n$.
I stumbled upon this equation: $$\exp_a\left\{\sum_{k=1}^n \gcd(k,n)\cos\left(\frac{2\pi k}{n}\right)\right\}\equiv 1\pmod n\tag*{$\big(\forall a\in\mathbb{Z}\land n\in\mathbb{Z}^+\big)$}$$ such that $...
- 9,487
1
vote
0
answers
109
views
Is it true that $ \sum_{k=0}^{(2q+1)n+q} \binom{2\left((2q+1)n+q\right)+1}{2k} 3^k $ is divisible by $\sum_{k=0}^{q} \binom{2q+1}{2k} 3^k $
This is a continuation for this question.
Let $$S_q=\sum_{k=0}^{q} \binom{2q+1}{2k} 3^k $$
$$T_q=\sum_{k=0}^{q} \binom{2q+1}{2k+1} 3^k $$.
(i) Is it true that $S_{(2q+1)n+q}$ is divisible by $S_q$, ...
- 3,716