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Ahaan S. Rungta
  • Member for 8 years, 11 months
  • Last seen more than a month ago
19 votes
3 answers
675 views

Making $121$ with five $0$s

18 votes
1 answer
2k views

Can a double-factorial be a perfect square?

17 votes
3 answers
2k views

FoxTrot Bill Amend Problems

16 votes
3 answers
1k views

Approximating $100!$

8 votes
3 answers
353 views

$ \exists a, b \in \mathbb{Z} $ such that $ a^2 + b^2 = 5^k $

7 votes
1 answer
269 views

Prove that $ \sum_{1 \le t \le n, \ (t, n) = 1} t = \dfrac {n\phi(n)}{2} $

6 votes
2 answers
2k views

Walk on Earth: Math Puzzle

5 votes
2 answers
117 views

$P(P(\cdots(P(x))))$ and its integer solutions

5 votes
4 answers
797 views

Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins

5 votes
2 answers
351 views

Show: $ f(a) = a,\ f(b) = b \implies \int_a^b \left[ f(x) + f^{-1}(x) \right] \, \mathrm{d}x = b^2 - a^2 $

4 votes
1 answer
5k views

Rigorous Statements: "It suffices to show that [...]" and Variations

4 votes
1 answer
330 views

Online Math Open Contest 2 Problem 50

4 votes
3 answers
148 views

$ 1987 \mid \left( n^n + (n+1)^n \right) $

4 votes
3 answers
194 views

Find $ \int \frac {1-x^2}{1+3x^2+x^4} \, \mathrm{d}x $

3 votes
3 answers
215 views

Computing the infinite sum, $\sum_{n=0}^\infty \frac {5^n}{25^n + 1} $

2 votes
2 answers
120 views

Prove that $\big( n^2 - 1 \big) \mid \big( 2^{n!} - 1 \big) $

2 votes
1 answer
321 views

Compute $ \lim\limits_{n\to\infty}\sqrt[n]{\log\left|1+\left(\frac{1}{n\cdot\log n}\right)^k\right|}$.

2 votes
2 answers
186 views

Convex Quadrilateral: $ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D $

2 votes
2 answers
138 views

Equality of integrals: $ \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $ [duplicate]

1 vote
3 answers
113 views

Maximizing $ \frac {y + z + yz}{\left( 1 + y + z \right)^2} $

1 vote
1 answer
332 views

Showing a topology is not metrizable

1 vote
0 answers
174 views

Counterexample for conjugate rules in $\mathbb{Z}[\sqrt[4]{2}]$

1 vote
2 answers
123 views

Marble ring: no two blacks are adjacent

0 votes
1 answer
122 views

Special permutations of $ \{ 1, 2, \cdots, 3k + 1 \} $