Daily math problems for July 6, 2026 — Calculus, Logaritmi, Number theory

$f^{\mathrm{\prime }}(x)=2x-4=0\Rightarrow x=2$. Since $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$, this is a minimum.

Folosind schimbarea de bază, ${\log }_k(k+1)={\displaystyle \frac{\ln (k+1)}{\ln k}}$. Produsul telescopează: ${\displaystyle \frac{\ln 3}{\ln 2}}\cdot {\displaystyle \frac{\ln 4}{\ln 3}}\cdots {\displaystyle \frac{\ln 256}{\ln 255}}={\displaystyle \frac{\ln 256}{\ln 2}}={\log }_{2}256=8$.

$7^2=49$, $7^4={49}^2=2401\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Since $2024=4\cdot 506$, we have $7^{2024}=(7^4{)}^{506}\equiv 1^{506}=1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Last two digits: $01$.