Daily math problems for August 23, 2026 — Combinatorics, Trigonometry, Sequences & more

$\bullet$ Expand the binomial coefficient using $C_n^k=\frac{n!}{k!(n-k)!}$:

$$C_{49}^6=\frac{49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44}{720}$$

$\bullet$ The product evaluates to:

$$C_{49}^6=13983816$$

$\bullet$ Take the reciprocal:

$$\frac{1}{C_{49}^6}=\frac{1}{13983816}$$

$\bullet$ Substitute $u=\cos x$:

$$2u^2-3u+1=0$$

$\bullet$ Factor:

$$(2u-1)(u-1)=0$$

$\bullet$ So $u=\frac{1}{2}$ or $u=1$. From $\cos x=1$:

$$x=0$$

$\bullet$ From $\cos x=\frac{1}{2}$ in $[0,2\pi )$:

$$x=\frac{\pi }{3},\frac{5\pi }{3}$$

$b_5=b_1\cdot q^4=2\cdot 3^4=2\cdot 81=162$.

$y^2-6y+8=0\Rightarrow y\in \{2,4\}$. Then $x^2=2\Rightarrow x=\pm \sqrt{2}$ and $x^2=4\Rightarrow x=\pm 2$. Four distinct real solutions.

$2x^2+x+1=2(x^2-x+2)\Rightarrow 2x^2+x+1=2x^2-2x+4\Rightarrow 3x=3\Rightarrow x=1$.

Verify the domain: at $x=1$, $2x^2+x+1=4>0$ and $x^2-x+2=2>0$, so both logs are defined. The solution set is $\{1\}$.

$3x+4y=25$.

Associative ✓, identity $1$ ✓. But $2$ has no integer multiplicative inverse. Hence a monoid but not a group.

$\sin {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{\sqrt{3}}{2}}$ and $\cos {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\sqrt{3}}{2}}$. Sum: $\sqrt{3}$.

$\cos A={\displaystyle \frac{25+9-49}{2\cdot 5\cdot 3}}={\displaystyle \frac{-15}{30}}=-{\displaystyle \frac{1}{2}}$. Since $A\in (0,\pi )$, $A={\displaystyle \frac{2\pi }{3}}$.

Since $b^2=ac$, the product $abc=b\cdot ac=b\cdot b^2=b^3=64$, so $b=4$.