Daily math problems for July 9, 2026 — Combinatorics, Calculus, Algebra & more

$\left(\genfrac{}{}{0px}{}{5}{3}\right)={\displaystyle \frac{5!}{3!\cdot 2!}}=10$.

$I_n=x^n{e}^x{\mid }_0^1-n{\int }_0^1{x}^{n-1}{e}^{x}dx=e-n{I}_{n-1}$.

A homomorphism preserves the operation: $\phi (a\cdot b)=\phi (a)\ast \phi (b)$.

$2\sin x\cos x-\sin x=0\Rightarrow \sin x(2\cos x-1)=0$. $\sin x=0$: $x\in \{0,\pi \}$. $\cos x=1\mathrm{/}2$: $x\in \{\pi \mathrm{/}3,5\pi \mathrm{/}3\}$. Total: $4$.

$S_n={\displaystyle \frac{n(3n+7)}{2}}$. Compute: $S_{17}={\displaystyle \frac{17\cdot 58}{2}}=493<500$ and $S_{18}={\displaystyle \frac{18\cdot 61}{2}}=549\ge 500$. Hence the smallest $n$ is $18$.

$\bullet$ Expand along the first row:

$$\det A=m(m^2-1)-1\cdot (m-0)+0$$

$\bullet$ Simplify:

$$\det A=m^3-2m=m(m^2-2)$$

$\bullet$ Set to zero:

$$m(m^2-2)=0$$

$\bullet$ Solutions:

$$m\in \{-\sqrt{2},0,\sqrt{2}\}$$

$A=\left(\begin{array}{cc}2& 3\\ 1& -1\end{array}\right)$ — coefficients of $x$ and $y$ in each equation.

$n\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{n^2})\sim n\cdot \frac{1}{n^2}=\frac{1}{n}\to 0$, so the limit equals $e^0=1$.

$f^{\mathrm{\prime }}(x)=3x^2-3a$, so $f^{\mathrm{\prime }}(2)=12-3a=0\Rightarrow a=4$. Check: $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$, so $f^{\mathrm{\prime }\mathrm{\prime }}(2)=12>0$, confirming a local minimum.

By Stirling, $\sqrt[n]{n!}\sim {\displaystyle \frac{n}{e}}(2\pi n{)}^{1\mathrm{/}(2n)}$. The factor $(2\pi n{)}^{1\mathrm{/}(2n)}\to 1$, so ${\displaystyle \frac{\sqrt[n]{n!}}{n}}\to {\displaystyle \frac{1}{e}}$.