heureka

If $a\equiv 18\pmod{42}$ and $b\equiv 73\pmod{42}$, then for what integer $n$ in the set $\{100,101,102,\ldots,140,141\}$ is it true that $$a-b\equiv n\pmod{42}~?$$

Help please!

\(\begin{array}{|rcll|} \hline a &\equiv& 18\pmod{42} \\ b &\equiv& 73\pmod{42} \\ \hline a-b &\equiv& 18- 73 \pmod{42} \\ a-b &\equiv& -55 \pmod{42} \\ a-b &\equiv& -55+42+42 \pmod{42} \\ a-b &\equiv& 29 \pmod{42} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a-b &\equiv& 29 \pmod{42} \\ a-b &=& 29 + z\cdot 42,\ z \in \mathbb{Z} \\ && \begin{array}{|r|c|c|} \hline & \mathbf{n} = (a-b)\pmod{42} & \\ z & =29 + z\cdot 42 & n \text{ in set} \\ \hline 0 & n=29 & \\ \hline 1 & n=71 & \\ \hline 2 & n=113 & \checkmark \\ \hline 3 & n=155 & \\ \hline \ldots \\ \hline \end{array} \\ \hline \end{array}\)

It is true for \(n = 113 \)

Find the remainder when \(3 \times 13 \times 23 \times 33 \times \ldots \times 183 \times 193\) is divided by \(5\).

\(\begin{array}{|rcll|} \hline 3 \pmod{5} &\equiv& 3 \\ 13 \pmod{5} &\equiv& 3 \\ 23 \pmod{5} &\equiv& 3 \\ 33 \pmod{5} &\equiv& 3 \\ \ldots \\ 183 \pmod{5} &\equiv& 3 \\ 193 \pmod{5} &\equiv& 3 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{3 \times 13 \times 23 \times 33 \times \ldots \times 183 \times 193 \pmod{5}} \\ &\equiv& 3 \times 3 \times 3 \times 3 \times \ldots \times 3 \times 3 \pmod{5} \\ &\equiv& 3^{20} \pmod{5} \quad | \quad 3^2\pmod{5} \equiv 9 \pmod{5} \equiv -1 \pmod{5} \\ &\equiv& \left(3^2\right)^{10} \pmod{5} \\ &\equiv& \left(-1\right)^{10} \pmod{5} \\ & \mathbf{\equiv}& \mathbf{1 \pmod{5}} \\ \hline \end{array}\)

The remainder is 1

Danke asinus!

In convex quadrilateral abcd,ab=bc=13 , cd=da=24, and angle D= 60 degrees. 

Points X andY  are the midpoints of BC and  DA respectively. Compute XY^2 (the square of the length of  XY).

\(\text{Let $\angle DAC =\angle ACD = 60^\circ $} \\ \text{Let $\angle ACY = \dfrac{\angle ACD}{2} = 30^\circ $} \\ \text{Let $ AD=AC=CD =24 $} \\ \text{Let $ CY = u $} \)

\(\mathbf{u=\ ?}\)

\(\begin{array}{|rcll|} \hline u^2+12^2 &=& 24^2 \\ u^2 &=& 24^2-12^2 \\ \mathbf{u^2} &\mathbf{=}& \mathbf{432} \\ \hline \end{array}\)

cos-Rule \(\mathbf{\cos(B)=\ ?}\):

\(\begin{array}{|rcll|} \hline 24^2 &=& 13^2+13^2-2\cdot 13 \cdot 13 \cdot \cos(B) \\ \ldots \\ \mathbf{\cos(B)} &\mathbf{=}& \mathbf{1-\dfrac{24^2}{2\cdot 13^2}} \\ \hline \end{array}\)

\(\mathbf{\cos\left(\dfrac{B}{2}\right)=\ ?}: \)

\(\begin{array}{|rcll|} \hline \cos(B) &=& 2\cos^2\left(\dfrac{B}{2}\right) - 1 \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 1+ \cos(B) \quad | \quad \mathbf{\cos(B) = 1-\dfrac{24^2}{2\cdot 13^2} } \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 1+ 1-\dfrac{24^2}{2\cdot 13^2} \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 2-\dfrac{24^2}{2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& 1-\dfrac{24^2}{2^2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& \dfrac{2^2\cdot 13^2-24^2}{2^2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& \dfrac{10^2}{2^2\cdot 13^2} \\ \cos\left(\dfrac{B}{2}\right) &=& \dfrac{10}{2\cdot 13} \\ \mathbf{\cos\left(\dfrac{B}{2}\right)} &\mathbf{=}& \mathbf{\dfrac{5}{13}} \\ \hline \end{array}\)

\(\mathbf{\sin(\alpha)=\ ?}\):

\(\begin{array}{|rcll|} \hline 180^\circ &=& B + 2\alpha \\ \ldots \\ \alpha &=& 90^\circ -\dfrac{B}{2} \\ \sin(\alpha) &=& \sin\left( 90^\circ -\dfrac{B}{2} \right) \\ \sin(\alpha) &=& \cos\left(\dfrac{B}{2} \right) \\ \mathbf{\sin(\alpha)} &\mathbf{=}& \mathbf{\dfrac{5}{13}} \\\\ \cos(\alpha) &=& \sqrt{1-\sin^2(\alpha)} \\ &=& \sqrt{1-\dfrac{5^2}{13^2}} \\ &=& \sqrt{ \dfrac{13^2-5^2}{13^2}} \\ &=& \sqrt{ \dfrac{12^2}{13^2}} \\ \mathbf{\cos(\alpha)} &\mathbf{=}& \mathbf{\dfrac{12}{13}} \\ \hline \end{array}\)

cos-Rule \(\mathbf{\overline{XY}^2=\ ?}\):

\(\begin{array}{|rcll|} \hline \overline{XY}^2 &=& 6.5^2+u^2-2\cdot 6.5 \cdot u \cdot \cos(\alpha+30^\circ) \\ &=& 6.5^2+432-13 \cdot u \cdot \cos(\alpha+30^\circ) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \cos(\alpha+30^\circ) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \cos(\alpha) \cos(30^\circ)-\sin(\alpha)\sin(30^\circ ) \Big) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \dfrac{12}{13}\cdot \dfrac{\sqrt{3}} {2} -\dfrac{5}{13}\cdot \dfrac{1} {2} \Big) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \dfrac {12 \sqrt{3}-5}{2\cdot 13} \Big) \\ &=& 474.25- \sqrt{432} \cdot \Big( \dfrac {12 \sqrt{3}-5}{2 } \Big) \\ &=& 474.25- \sqrt{\dfrac{432}{4}} \cdot \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{108} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{4\cdot 27} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{2^2\cdot 3^2\cdot 3} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- 6\sqrt{ 3} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- 6\sqrt{ 3} \cdot 12 \sqrt{3} -5\cdot 6\sqrt{ 3} \\ &=& 474.25- 72\ \cdot 3 +30\sqrt{ 3} \\ &=& 474.25- 216 +30\sqrt{ 3} \\ &=& 258.25 + 30\sqrt{ 3} \\ &=& 258.25 + 51.9615242271 \\ \mathbf{\overline{XY}^2} &\mathbf{=} & \mathbf{310.211524227} \\ \hline \end{array}\)

Suppose that ABCD is a trapezoid in which AD || BC. Given AC perpendicular CD, AC bisects angle BAD, and the area of ABCD is 42, then compute the area of triangle ACD.

On the xy-plane, the origin is labeled with an M.

The points (1,0), (-1,0), (0,1), and (0,-1) are labeled with A's.

The points (2,0), (1,1), (0,2), (-1, 1), (-2, 0), (-1, -1), (0, -2), and (1, -1) are labeled with T's.

The points (3,0), (2,1), (1,2), (0, 3), (-1, 2), (-2, 1), (-3, 0), (-2,-1), (-1,-2), (0, -3), (1, -2), and (2, -1) are labeled with H's.

If you are only allowed to move up, down, left, and right, starting from the origin,

how many distinct paths can be followed to spell the word MATH?

\(\begin{array}{|lclcl|} \hline \text{Let } d &-& \text{ down } &=& 0 \\ \text{Let } r &-& \text{ right } &=& 1 \\ \text{Let } u &-& \text{ up } &=& 2 \\ \text{Let } l &-& \text{ left } &=& 3 \\ \hline \end{array}\)

From M to "down-A", all distinct paths

are all three digit numbers to base 4:

\(\begin{array}{|r|r|l|} \hline base_4 &\text{path} & \text{spell the word MATH} \\ \hline 000 & ddd & \checkmark \\ 001 & ddr & \checkmark \\ 002 & ddu & \\ 003 & ddl & \checkmark \\ 010 & drd & \checkmark \\ 011 & drr & \checkmark \\ 012 & dru & \\ 013 & drl & \\ 020 & dud & \\ 021 & dur & \\ 022 & duu & \\ 023 & dul & \\ 030 & dld & \checkmark \\ 031 & dlr & \\ 032 & dlu & \\ 033 & dll & \checkmark \\ \hline && 7 \text{ distinct paths to spell the word MATH } \\ \hline \end{array} \)

For reasons of symmetry:

\(\begin{array}{|lcll|} \hline \text{From M to "down-A" there are } 7 \text{ distinct paths to spell the word MATH } & base_4 \quad 000 \to 033 \\ \text{From M to "right-A" there are } 7 \text{ distinct paths to spell the word MATH }& base_4 \quad 100 \to 133 \\ \text{From M to "up-A" there are 7 } \text{ distinct paths to spell the word MATH } & base_4 \quad 200 \to 233 \\ \text{From M to "left-A" there are } 7 \text{ distinct paths to spell the word MATH } & base_4 \quad 300 \to 333 \\ \hline \end{array} \)

\(4\times 7 = 28\) distinct paths can be followed to spell the word MATH

Zeigen Sie durch Induktion über n und mit Hilfe des binomischen Lehrsatzes die Pascal'sche Identität:

Ich vermute die Aufgabe lautet folgendermaßen:

\(\displaystyle \sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix}s_{n}(p)=(n+1)^{q+1}-1 \\ \text{Es sei im folgenden } \displaystyle s_{n}(p)= \sum \limits_{k=1}^{n} k^p\)

Beweise mit vollständiger Induktion:

\(\displaystyle \sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix}s_{n}(p)=(n+1)^{q+1}-1,\ \text{ für alle } n \ge 1\)

Induktionsanfang:

\(\begin{array}{|lll|} \hline n=1 & \text{linke Seite:} & \displaystyle \sum\limits_{p=0}^{q} \begin{pmatrix}q+1\\p\end{pmatrix}s_{1}(p) \qquad \boxed{s_1(p) = \displaystyle\sum \limits_{k=1}^{1} k^p = 1^p = 1 } \\ & &= \displaystyle \sum\limits_{p=0}^{q} \begin{pmatrix}q+1\\p\end{pmatrix} \\ & &= \displaystyle \begin{pmatrix}q+1\\0\end{pmatrix} + \begin{pmatrix}q+1\\1\end{pmatrix} + \ldots + \begin{pmatrix}q+1\\q\end{pmatrix} \\ & &= \displaystyle \begin{pmatrix}q+1\\0\end{pmatrix} + \begin{pmatrix}q+1\\1\end{pmatrix} + \ldots + \begin{pmatrix}q+1\\q\end{pmatrix} + \begin{pmatrix}q+1\\q+1\end{pmatrix} - \begin{pmatrix}q+1\\q+1\end{pmatrix} \\ & &= \left(1+1\right)^{q+1} - \begin{pmatrix}q+1\\q+1\end{pmatrix} \\ & &= \left(1+1\right)^{q+1} - 1 \\ & &\mathbf{= 2^{q+1}-1} \\\\ & \text{rechte Seite:} & \left(1+1\right)^{q+1} - 1 \\ & &\mathbf{= 2^{q+1}-1} \\ \hline \end{array} \)

\(\text{Für $\mathbf{n=1}$ sind beide Seiten gleich, und die Aussage ist wahr!}\)

Die Induktionsannahme (I.A.) lautet:

\(\begin{array}{|rcll|} \hline \displaystyle \sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix}s_{n}(p)=(n+1)^{q+1}-1 \\ \hline \end{array}\)

Der Induktionsschluss von \(\mathbf{n}\) nach \(\mathbf{n+1}\):

\(\begin{array}{|rcll|} \hline \displaystyle \sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix}s_{n+1}(p) &=& \Big((n+1)+1\Big)^{q+1}-1 \\ &=& \left(n+2\right)^{q+1}-1 \\ \hline \end{array} \)

\(\bf{\text{linke Seite Vorbereitung :}} \\ \begin{array}{|rcl|} \hline \displaystyle s_{n+1}(p) &=& \displaystyle \sum \limits_{k=1}^{n+1} k^p \\ &=& \displaystyle \sum \limits_{k=1}^{n} k^p +(n+1)^p \\ &=& \displaystyle s_{n}(p) +(n+1)^p \\ \hline \end{array} \)

\(\small{ \bf{\text{linke Seite:}} \\ \begin{array}{|rcll|} \hline \mathbf{\displaystyle \sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix}s_{n+1}(p) } \\ &=& \displaystyle\sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix} \Big( s_{n}(p) +(n+1)^p \Big) \\\\ &=& \displaystyle\sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix} s_{n}(p)+\sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix} (n+1)^p \\\\ &\overset{I.A.}{=} & \displaystyle (n+1)^{q+1}-1 +\sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix} (n+1)^p \\ &=& \displaystyle\sum \limits_{p=0}^{q}\begin{pmatrix}q+1\\p\end{pmatrix} (n+1)^p + (n+1)^{q+1}-1 \\ &=& \begin{pmatrix}q+1\\0\end{pmatrix}(n+1)^0 + \begin{pmatrix}q+1\\1\end{pmatrix}(n+1)^1 + \ldots + \begin{pmatrix}q+1\\q\end{pmatrix}(n+1)^q + (n+1)^{q+1}-1 \\ &=& \begin{pmatrix}q+1\\0\end{pmatrix}(n+1)^0 + \begin{pmatrix}q+1\\1\end{pmatrix}(n+1)^1 + \ldots + \begin{pmatrix}q+1\\q\end{pmatrix}(n+1)^q + \begin{pmatrix}q+1\\q+1\end{pmatrix}(n+1)^{q+1}-1 \\ &=& \Big((n+1)+1\Big)^{q+1}-1 \\ &=& \mathbf{\Big(n+2\Big)^{q+1}-1} \\ \hline \end{array} } \)

\(\bf{\text{rechte Seite:}} \\ \begin{array}{|ll|} \hline \mathbf{\Big((n+1)+1\Big)^{q+1}-1 } \\\\ &=& \mathbf{\Big(n+2\Big)^{q+1}-1} \\ \hline \end{array}\)

\(\bf{\text{Ergebnis:}} \\ \begin{array}{|ll|} \hline \displaystyle \Big(n+2\Big)^{q+1}-1 = \Big(n+2\Big)^{q+1}-1\ \checkmark \\ \hline \end{array}\)

Thank you, Melody !

There are exactly four integers n such that the equation below is an integer.

Find the largest \(n\).

\(\dfrac{(n+1)^2}{n+23}\)

\(\begin{array}{|rcll|} \hline \dfrac{(n+1)^2}{n+23} &=& \dfrac{n^2+2n+1}{n+23} \\\\ n^2+2n+1\ :\ n+23 &=& n-21 + \dfrac{484}{n+23} \quad \text{Polynomial long division} \\\\ 484 &=& 0 \pmod{n+23} \quad | \quad \dfrac{484}{n+23} \text{ must be an integer } \\\\ 484-0 &=& m\cdot (n+23), \ m \in \mathbb{Z} \\ 484 &=& m\cdot n +23m \\ m\cdot n &=& 484 - 23m \\ n &=& \dfrac{484 - 23m}{m} \\ n &=& \dfrac{484}{m} - 23 \quad | \quad n \text{ is max, when } m = 1! \\ n_{\text{max}} &=& 484 - 23 \\ \mathbf{ n_{\text{max}} } & \mathbf{=} & \mathbf{ 461 } \\ \hline \end{array}\)