heureka

The larger of two numbers is 33 more than the smaller.
When added together, the sum of the larger number and five times the smaller number is 129.
What are the two numbers?

\(\text{Let large number $=l$} \\ \text{Let small number $=s$}\)

\(\begin{array}{|lrcll|} \hline (1) & l &=& s +33 \\ (2) & l+5s &=& 129 \\ & s+33 + 5s &=& 129 \\ & 6s &=& 129 - 33 \\ & 6s &=& 96 \quad |\quad : 6 \\ & \mathbf{s} &=& \mathbf{16} \\\\ & l &=& s+33\\ & l &=& 16+33 \\ & \mathbf{l} &=& \mathbf{49} \\ \hline \end{array}\)

The two numbers are 16 and 49

Compute 

\((1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ) \dotsm (1 + \tan 45^\circ)\)

Formula:

\(\begin{array}{|rcll|} \hline \tan(x-y) &=& \dfrac{\tan x-\tan y}{1+\tan x \tan y }\\\\ \tan(45^\circ) &=& 1 \\ \hline \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{(1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ) \dotsm (1 + \tan 22^\circ)(1 + \tan 23^\circ) \dotsm (1 + \tan 43^\circ)(1 + \tan 44^\circ)(1 + \tan 45^\circ)} \\\\ &=& (1 + \tan 1^\circ)(1 + \tan 44^\circ) \times (1 + \tan 2^\circ)(1 + \tan 43^\circ) \times (1 + \tan 3^\circ)(1 + \tan 42^\circ) \\ && \vdots \\ && \times (1 + \tan 22^\circ)(1 + \tan 23^\circ) \times (1 + \tan 45^\circ) \\\\ &=& (1 + \tan 1^\circ)\Big(1 + \tan (45^\circ-1^\circ)\Big) \times (1 + \tan 2^\circ)\Big(1 + \tan (45^\circ-2^\circ)\Big) \times (1 + \tan 3^\circ)\Big(1 + \tan (45^\circ-3^\circ)\Big) \\ && \vdots \\ && \times (1 + \tan 22^\circ)\Big(1 + \tan (45^\circ-22^\circ)\Big) \times (1 + \tan 45^\circ) \\\\ &=& (1 + \tan 1^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 1^\circ} {1+\tan 45^\circ\tan 1^\circ} \Big) \times (1 + \tan 2^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 2^\circ} {1+\tan 45^\circ\tan 2^\circ} \Big) \times (1 + \tan 3^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 3^\circ} {1+\tan 45^\circ\tan 3^\circ} \Big) \\ && \vdots \\ && \times (1 + \tan 22^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 22^\circ} {1+\tan 45^\circ\tan 22^\circ}\Big) \times (1 + \tan 45^\circ) \quad | \quad \mathbf{\tan 45^\circ = 1} \\\\ &=& (1 + \tan 1^\circ)\Big(1 + \dfrac{1-\tan 1^\circ} {1+\tan 1^\circ} \Big) \times (1 + \tan 2^\circ)\Big(1 + \dfrac{1-\tan 2^\circ} {1+\tan 2^\circ} \Big) \times (1 + \tan 3^\circ)\Big(1 + \dfrac{1-\tan 3^\circ} {1+\tan 3^\circ} \Big) \\ && \vdots \\ && \times (1 + \tan 22^\circ)\Big(1 + \dfrac{1-\tan 22^\circ} {1+\tan 22^\circ}\Big) \times (1 + 1) \\\\ &=& (1 + \tan 1^\circ)*\dfrac{(1+\tan 1^\circ+1-\tan 1^\circ)} {(1+\tan 1^\circ)} \times (1 + \tan 2^\circ)*\dfrac{(1+\tan 2^\circ+1-\tan 2^\circ)} {(1+\tan 2^\circ)} \times (1 + \tan 3^\circ)*\dfrac{(1+\tan 3^\circ+1-\tan 3^\circ)} {(1+\tan 3^\circ)} \\ && \vdots \\ && \times (1 + \tan 22^\circ)*\dfrac{(1+\tan 22^\circ+1-\tan 22^\circ)} {(1+\tan 22^\circ)} \times 2 \\\\ &=& 2\times 2\times 2 \\ && \vdots \\ && \times 2 \times 2 \\\\ &=& 2^{22}\times 2 \\\\ &=& 2^{23} \\\\ &=& \mathbf{8388608} \\ \hline \end{array} }\)

Given that \(x+y=\dfrac7{10}\) and \(x-y=\dfrac5{14}\),

find the value of \(x^2-y^2\), expressed as a common fraction.

\(\begin{array}{|rcll|} \hline x^2-y^2 &=& (x+y)(x-y) \\\\ x^2-y^2 &=& \dfrac7{10}*\dfrac5{14} \\\\ \mathbf{x^2-y^2} &=& \mathbf{\dfrac14} \\ \hline \end{array}\)

compute

\(\sin^{6}(3.25^\circ) + 3\sin^{2}(3.25^\circ)\cos^{2}(3.25^\circ) + \cos^{6}(3.25^\circ)\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sin^{6}(3.25^\circ) + 3\sin^{2}(3.25^\circ)\cos^{2}(3.25^\circ) + \cos^{6}(3.25^\circ) \quad | \quad \cos^{2}(3.25^\circ)=1- \sin^{2}(3.25^\circ)} \\ &=& \sin^{6}(3.25^\circ) + 3\sin^{2}(3.25^\circ)\Big(1- \sin^{2}(3.25^\circ)\Big) + \Big(1- \sin^{2}(3.25^\circ)\Big)^3 \\ &=& \sin^{6}(3.25^\circ) + 3\sin^{2}(3.25^\circ)- 3\sin^{4}(3.25^\circ) + \Big(1- \sin^{2}(3.25^\circ)\Big)^3 \\ &=& \sin^{6}(3.25^\circ) + 3\sin^{2}(3.25^\circ)- 3\sin^{4}(3.25^\circ) + 1- 3\sin^{2}(3.25^\circ)+ 3\sin^{4}(3.25^\circ)-\sin^{6}(3.25^\circ) \\ &=& \mathbf{1} \\ \hline \end{array} \)

Please help ASAP

\(\begin{array}{|rcll|} \hline AB=AC \\ \hline \angle ABC &=& \angle ACB \quad | \quad \angle ABC =2x \\ 2x &=& \angle ACB \\ \mathbf{\angle ACB} &=& \mathbf{2x} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD=BC \\ \hline \angle BDC &=& \angle ACB \quad | \quad \angle ACB =2x \\ \mathbf{\angle BDC} &=& \mathbf{2x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \angle ADB &=& 180^\circ - \angle BDC \quad | \quad \mathbf{\angle BDC=2x} \\ \mathbf{\angle ADB} &=& \mathbf{180^\circ - 2x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ \text{In $\triangle$ BCD }: } \\ \hline 2x+2x+x &=& 180^\circ \\ 5x &=& 180^\circ \\\\ x &=& \dfrac{180^\circ}{5} \\\\ \mathbf{x} &=& \mathbf{36^\circ} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ \text{In $\triangle$ ADB }: } \\ \hline A+ (180^\circ-2x) +x &=& 180^\circ \\ A+ 180^\circ-2x +x &=& 180^\circ \\ A+ 180^\circ-x &=& 180^\circ \\ A -x &=& 0 \\ A &=& x \quad | \quad \mathbf{x=36^\circ} \\ \mathbf{A} &=& \mathbf{36^\circ} \\ \hline \end{array} \)

Thank you, Melody!

Simpify \(\dfrac{\sin 25^\circ \sin 35^\circ \sin 85^\circ}{\sin 75^\circ}\)

Formula:

\(\begin{array}{|rcll|} \hline \sin(x)\sin(y) &=& \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\\\ \sin(x)\cos(y) &=& \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\\\ \cos(60^\circ) &=& \dfrac12 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{\sin(25^\circ) \sin(35^\circ) \sin(85^\circ)}{\sin(75^\circ)} } \\\\ && \qquad \boxed{ \sin(x)\sin(y) = \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(35^\circ-25^\circ)-\cos(35^\circ+25^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\cos(60^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\dfrac{1}{2} \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4} } \\\\ &=& \dfrac{\Big(\dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4}\Big) \sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{2}*\cos(10^\circ)\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ && \qquad \boxed{ \sin(x)\cos(y) = \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(85^\circ-10^\circ)+\sin(85^\circ+10^\circ) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(95^\circ) \Big) \quad | \quad \sin(95^\circ)=\sin(180^\circ-95^\circ)=\sin(85^\circ) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big) } \\\\ &=& \dfrac{\dfrac{1}{2}*\dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)+\dfrac{1}{4}*\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)}{\sin(75^\circ)} \\\\ &=& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}\)

What is the smallest integer n, greater than 1, such that \(n^{-1}\pmod{130}\) and \(n^{-1}\pmod{231}\) are both defined?

Modular multiplicative inverse is defined if \(\gcd(130,n)=\gcd(231,n)=1\)

\(\begin{array}{|rcll|} \hline \text{factor}~ 130 &=& 2*5*13 \\ \text{factor}~ 231 &=& 3*7*11 \\ \hline \end{array}\)

The first prim number = 2. This number is a prime factor in 130, so \(\gcd(130,2) \ne 1\)
The 2nd prim number = 3. This number is a prime factor in 231, so \(\gcd(231,3) \ne 1\)
The 3rd prim number = 5. This number is a prime factor in 130, so \(\gcd(130,5) \ne 1\)
The 4th prim number = 7. This number is a prime factor in 231, so \(\gcd(231,7) \ne 1\)
The 5th prim number = 11. This number is a prime factor in 231, so \(\gcd(231,11) \ne 1 \)
The 6th prim number = 13. This number is a prime factor in 130, so \(\gcd(130,13) \ne 1\)
The 7th prim number = 17. This number is not a prime factor in 130 and not a prime factor in 231
, \(\gcd(130,17) = \gcd(213,17) = 1\)

The smallest integer n is 17

check:

\(\begin{array}{|rcll|} \hline 17^{-1} \pmod{130} &\equiv& 23 \\ 17*23 & \equiv& 1 \pmod{130}~ \checkmark \\\\ 17^{-1} \pmod{231} &\equiv& 68 \\ 17*68 & \equiv& 1 \pmod{231}~ \checkmark \\ \hline \end{array}\)

You have a triangle with 45 identical lines in this way as you can see on the picture.

What is gamma?

\(\begin{array}{|r|r|} \hline & angle \\ \text{lines} & =\dfrac{\text{lines}-1}{2} \\ \hline 5 & 2\gamma \\ 7 & 3\gamma \\ 9 & 4\gamma \\ 11 & 5\gamma \\ 13 & 6\gamma \\ 15 & 7\gamma \\ \ldots & \ldots \\ 45 & \dfrac{\text{45}-1}{2}=22\gamma\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2*22\gamma+\gamma &=& 180^\circ \\ 45\gamma &=& 180^\circ \\ \gamma &=& \dfrac{180^\circ}{45} \\ \mathbf{\gamma} &=& \mathbf{ 4} \\ \hline \end{array}\)

Thank you, Dragan!