All Questions
9
questions
2
votes
1
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92
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Prove that $0.1111$ is not a sum of 1111 funny numbers and that every $x\in (0,1)$ is the sum of 1112 different funny numbers.
An irrational number in $(0,1)$ is funny if its first four decimal digits are the same. For example, $0.1111 + e/10^5$ is funny. Prove that $0.1111$ is not a sum of 1111 funny numbers and that every $...
- 1,225
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
2
votes
1
answer
127
views
Show that the product of the $2^{2019}$ numbers of the form $\pm 1\pm \sqrt{2}\pm\cdots \pm \sqrt{2019}$ is the square of an integer.
Show that the product of the $2^{2019}$ numbers of the form $\pm 1\pm \sqrt{2}\pm\cdots \pm \sqrt{2019}$ is the square of an integer.
I'm aware very similar problems were asked before (e.g. here and ...
- 2,031
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 ...
- 336
1
vote
4
answers
322
views
How do I solve Problem 1 of the International Olympiad of Metropolis?
The question is: Find all $n$ so that there exists $n$ consecutive numbers whose sum is a square!
My method to solve the problem: I would try to look at the values modulo $n$, and there I see there ...
user368587
2
votes
5
answers
848
views
$33^{33}$ is the sum of $33$ consecutive odd numbers. Which one is the largest? (Q25 from AMC 2012)
The number $33^{33}$ can be expressed as the sum of $33$ consecutive odd numbers. The largest of these odd numbers is
$\mathrm{A.}\ 33^{32} +32$
$\mathrm{B.}\ 33^{31} +32$
$\mathrm{C.}\ 33^{32} -32$
$\...
- 749
6
votes
1
answer
212
views
Find the remainder when the sum is divided by $1000$
Find $S \pmod{1000}$ given: $$S = \sum_{n=0}^{2015} n! + n^3 - n^2 + n - 1$$
$$S_0 = 0! + 0 - 0 + 0 -1 = 0$$
$$S_1 = 1! + 1 - 1 + 1 - 1 = 1$$
$$S_2 = 2! + 8 - 4 + 2 - 1 = 7$$
This isn't helping, so:...
- 11.2k
1
vote
1
answer
138
views
Cute convergence problem. Proving convergence of sequence regarding reciprocals of least common multiple converges.
This is the first problem of the second day of the $2014$ CIIM.
Let $\{a_n\}$ be a strictly increasing sequence of positive integers.
Prove the sequence $\sum\limits_{i=0}^n\frac{1}{[a_i,a_{i+1}]}$...
- 106k
11
votes
2
answers
203
views
Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $\sum_{k=1}^n\frac{a_k-b_k}k$
Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $$\sum_{k=1}^n\frac{a_k-b_k}k$$
By computing some partial sums, the answers are 0. It seems an inductive argument is possible.
- 4,996