Daily math problems for August 4, 2026 — Calculus, Algebra, Absolute value & more

$2^n$ grows without bound: $\lim 2^n=\mathrm{\infty }$.

$3\ast 5=3+5-2=6$.

$x-3=\pm 5$ gives $x=8$ or $x=-2$. Sum $=8+(-2)=6$.

${\displaystyle \frac{1-\cos x}{x^2}}={\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\cdot \text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2\to {\displaystyle \frac{1}{2}}$.

$\bullet$ Compute the modulus of the base:

$$\mathrm{\mid }1+i\sqrt{3}\mathrm{\mid }=\sqrt{1+3}=2$$

$\bullet$ Modulus distributes over powers:

$$\mathrm{\mid }(1+i\sqrt{3}{)}^{2024}\mathrm{\mid }=2^{2024}$$

$\bullet$ The denominator is real and positive, so:

$$\mid {\displaystyle \frac{(1+i\sqrt{3}{)}^{2024}}{2^{2023}}}\mid ={\displaystyle \frac{2^{2024}}{2^{2023}}}=2$$

${\sum }_{k=0}^6(-1{)}^k\left(\genfrac{}{}{0px}{}{6}{k}\right)=(1-1{)}^6=0$.

$\tan {\displaystyle \frac{\pi }{3}}=\sqrt{3}$, and ${\displaystyle \frac{\pi }{3}}\in \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$, so $\arctan \sqrt{3}={\displaystyle \frac{\pi }{3}}$.

${\displaystyle \frac{n(n-1)}{2}}=45\Rightarrow n(n-1)=90\Rightarrow n=10$.

Let $L$ be the trail length. Define

- $u(t)$: the monk's distance from the base on Monday at time $t\in [7,17]$ - $d(t)$: her distance from the base on Tuesday at the same clock time

Both functions are continuous on $[7,17]$ (a hiker doesn't teleport).

Now consider $f(t)=u(t)-d(t)$. By the problem statement:

- $u(7)=0$ and $d(7)=L$, so $f(7)=-L<0$. - $u(17)=L$ and $d(17)=0$, so $f(17)=+L>0$.

Since $f$ is continuous and changes sign on $[7,17]$, the Intermediate Value Theorem guarantees some $t^{\ast }\in (7,17)$ with $f(t^{\ast })=0$, i.e. $u(t^{\ast })=d(t^{\ast })$. At that clock time, the monk stands at the same point on the trail on both days. $\mathrm{■}$

The physical intuition ("imagine two monks: one ascending Monday, one descending Tuesday at the same time — they must meet") collapses into a one-line IVT argument once you let $f$ do the work.

By induction, $A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$, so $(A^n{)}_{12}=n$.