heureka

What is the smallest whole number that has

a remainder of 1 when divided by 4,

a remainder of 1 when divided by 3,

and a remainder of 2 when divided by 5

So it becomes:

x = 1 mod(4)

x = 1 mod(3)

x = 2 mod(5)

\(\begin{array}{rcll} x &\equiv& {\color{red}1} \pmod {{\color{green}4}} \\ x &\equiv& {\color{red}1} \pmod {{\color{green}3}} \\ x &\equiv& {\color{red}2} \pmod {{\color{green}5}} \\ \text{Let } m &=& 4\cdot 3\cdot 5 = 60 \\ \end{array}\)

Because 4 and 3 and 5 are relatively prim  we can go on:

\(\begin{array}{|rclcl|} \hline x &=& {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot \left[\dfrac{1}{\color{green}3\cdot 5}\pmod{4} \right] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot \left[\dfrac{1}{\color{green}4\cdot 5}\pmod{3} \right] \\ &+& {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot \left[\dfrac{1}{\color{green}4\cdot 3}\pmod{5} \right] + {\color{green}4\cdot 3\cdot 5}\cdot k \quad | \quad k \in \mathbb{Z} \\ \hline && \left[\dfrac{1}{\color{green}3\cdot 5}\pmod{4} \right] \\ && \equiv \left[ { (\color{green}3\cdot 5) }^{\varphi({\color{green}4}) -1 } \pmod {{\color{green}4}} \right] \quad | \quad \varphi({\color{green}4}) = 2 \\ && \equiv \left[ { 15 }^{2 -1} \pmod {{\color{green}4}} \right] \\ && \equiv \left[ { 15 } \pmod {{\color{green}4}} \right] \\ && \equiv \left[ { 3 } \pmod {{\color{green}4}} \right] \\ \hline && \left[\dfrac{1}{\color{green}4\cdot 5}\pmod{3} \right] \\ && \equiv \left[ { (\color{green}4\cdot 5) }^{\varphi({\color{green}3}) -1 } \pmod {{\color{green}3}} \right] \quad | \quad \varphi({\color{green}3}) = 2 \\ && \equiv \left[ { 20 }^{2 -1} \pmod {{\color{green}3}} \right] \\ && \equiv \left[ { 20 } \pmod {{\color{green}3}} \right] \\ && \equiv \left[ { 2 } \pmod {{\color{green}3}} \right] \\ \hline && \left[\dfrac{1}{\color{green}4\cdot 3}\pmod{5} \right] \\ && \equiv \left[ { (\color{green}4\cdot 3) }^{\varphi({\color{green}5}) -1 } \pmod {{\color{green}5}} \right] \quad | \quad \varphi({\color{green}5}) = 4 \\ && \equiv \left[ { 12 }^{4 -1} \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 12^3 } \pmod {{\color{green}5}} \right] \quad 12 \pmod{5} \equiv 2 \pmod{5} \\ && \equiv \left[ { 2^3 } \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 8 } \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 3 } \pmod {{\color{green}5}} \right] \\ \hline x &=& {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot [3] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot [2] + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot [3] + 60\cdot k \quad | \quad k \in \mathbb{Z} \\ x &=& 45+ 40 + 72 + 60\cdot k \\ x &=& 157+ 60\cdot k \quad | \quad 157 \pmod {60} = 37 \\ x &=& 37+ 60\cdot k \qquad k \in Z \\ \mathbf{x_{min}} &=& \mathbf{ 37} \\ \hline \end{array}\)

A permutation of the numbers (1,2,3,...,n) is a rearrangement of the numbers in which each number appears exactly once. For example, (2,5,1,4,3) is a permutation of (1,2,3,4,5). Let \pi = (x_1,x_2,x_3,---,x_n) be a permutation of the numbers (1,2,3,....,n). A fixed point of \pi is an integer k(1 ≤ k ≤ n) such that x_k=k. For example, 4 is a fixed point of the permutation $(2,5,1,4,3). How many permutations of (1,2,3,4,5,6,7) have at least one even fixed point?

I assume 1824 permutations of (1,2,3,4,5,6,7) have at least one even fixed point.

 1.) 1234567   3 (even fixed points)
 2.) 1234576   2 (even fixed points)
 3.) 1234657   2 (even fixed points)
 4.) 1234675   2 (even fixed points)
 5.) 1234765   3 (even fixed points)
 6.) 1234756   2 (even fixed points)
 7.) 1235467   2 (even fixed points)
 8.) 1235476   1 (even fixed points)
 9.) 1235647   1 (even fixed points)
 10.) 1235674   1 (even fixed points)
 11.) 1235764   2 (even fixed points)
 12.) 1235746   1 (even fixed points)
\(\cdots\)
 998.) 4731265   1 (even fixed points)
 999.) 4735162   1 (even fixed points)
 1000.) 4735261   1 (even fixed points)
 1001.) 4732561   1 (even fixed points)
 1002.) 4732165   1 (even fixed points)
 1003.) 4713562   1 (even fixed points)
 1004.) 4713265   1 (even fixed points)
 1005.) 4715362   1 (even fixed points)
 1006.) 4715263   1 (even fixed points)
\(\cdots\)
 1410.) 6217453   1 (even fixed points)
 1411.) 6274513   2 (even fixed points)
 1412.) 6274531   2 (even fixed points)
 1413.) 6274153   2 (even fixed points)
 1414.) 6274135   2 (even fixed points)
 1415.) 6274315   2 (even fixed points)
 1416.) 6274351   2 (even fixed points)
 1417.) 6275413   1 (even fixed points)
 1418.) 6275431   1 (even fixed points)
 1419.) 6275143   1 (even fixed points)
\(\cdots\)
 1626.) 7261453   1 (even fixed points)
 1627.) 7214563   3 (even fixed points)
 1628.) 7214536   2 (even fixed points)
 1629.) 7214653   2 (even fixed points)
 1630.) 7214635   2 (even fixed points)
 1631.) 7214365   3 (even fixed points)
 1632.) 7214356   2 (even fixed points)
\(\cdots\)
 1812.) 7164325   1 (even fixed points)
 1813.) 7164235   1 (even fixed points)
 1814.) 7164253   1 (even fixed points)
 1815.) 7124563   2 (even fixed points)
 1816.) 7124536   1 (even fixed points)
 1817.) 7124653   1 (even fixed points)
 1818.) 7124635   1 (even fixed points)
 1819.) 7124365   2 (even fixed points)
 1820.) 7124356   1 (even fixed points)
 1821.) 7125463   1 (even fixed points)
 1822.) 7125364   1 (even fixed points)
 1823.) 7123564   1 (even fixed points)
 1824.) 7123465   1 (even fixed points)

Let u, v, and w be vectors satisfying \(\mathbf{u}\bullet \mathbf{v} = 3,\ \mathbf{u} \bullet \mathbf{w} = 4,\ \mathbf{v} \bullet \mathbf{w} = 5 \).
Then what are \((\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w},\ (\mathbf{w} - \mathbf{u})\bullet \mathbf{v},\ (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u}\) equal to?

\(\begin{array}{|rcll|} \hline \mathbf{u}\bullet \mathbf{v} = \mathbf{v}\bullet \mathbf{u} &=& 3 \\ \mathbf{u} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{u} &=& 4 \\ \mathbf{v} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{v} &=& 5 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w} \\ &=& \mathbf{u} \bullet \mathbf{w}+2\mathbf{v} \bullet \mathbf{w} \\ &=& 4+2\cdot 5 \\ &=&\mathbf{ 14 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && (\mathbf{w} - \mathbf{u})\bullet \mathbf{v} \\ &=& \mathbf{w} \bullet \mathbf{v} - \mathbf{u}\bullet \mathbf{v} \\ &=& 5-3 \\ &=&\mathbf{ 2 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u} \\ &=& 3\mathbf{v}\bullet \mathbf{u} - 2\mathbf{w} \bullet \mathbf{u} \\ &=& 3 \cdot 3 -2\cdot 4 \\ &=& 9-8 \\ &=&\mathbf{ 1 } \\ \hline \end{array}\)

If there are vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\)\(\mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\),  and \( \mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}\).
Find coefficients a, b, and c, not all 0, such that

\(a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
and answer with \(\dfrac{a-b}{c}\).

\(\begin{array}{|lrcll|} \hline &\begin{pmatrix} 1&1&-1\\2&4&6\\1&5&15 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (2): &2a +4b+6c &=& 0 \quad | \quad : 2 \\ &a +2b+3c &=& 0 \\\\ &\begin{pmatrix} 1&1&-1\\1&2&3\\1&5&15 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) = (3) - (1) : &\begin{pmatrix} 1&1&-1\\1&2&3\\0&4&16 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) &0\cdot a +4b+16c &=& 0 \quad | \quad : 4 \\ & b+4c &=& 0 \\\\ &\begin{pmatrix} 1&1&-1\\1&2&3\\0&1&4 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (2) = (2) - (1) : &\begin{pmatrix} 1&1&-1\\0&1&4\\0&1&4 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) = (3) - (2) : &\begin{pmatrix} 1&1&-1\\0&1&4\\0&0&0 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\ \hline \end{array}\)

We set \(c=k\):

\(\begin{array}{|lrcll|} \hline (2): & 0\cdot a+1\cdot b+4 c &=& 0 \quad | \quad c=k \\ & b+4k &=& 0 \\ & \mathbf{ b } &=& \mathbf{ -4k } \\\\ (1): & 1\cdot a+1\cdot b- 1 \cdot c &=& 0 \quad | \quad c=k,\ b=-4k \\ & a-4k-k &=& 0 \\ & a-5k &=& 0 \\ & \mathbf{ a } &=& \mathbf{ 5k } \\ \hline \end{array} \)

proof:

\(\begin{array}{|rcll|} \hline a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \quad | \quad a=5k,\ b=-4k,\ c=k \\\\ \hline 5k\begin{pmatrix} 1\\2\\1 \end{pmatrix}-4k \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + k\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\\\ 5k-4k-k &=& 0 \checkmark \\ 10k-16k+6k &=& 0 \checkmark \\ 5k - 20k+15k &=& 0 \checkmark \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{a+b}{c} } \\\\ &=& \dfrac{5k-(-4k)}{k} \\\\ &=& \dfrac{5k+4k}{k} \\\\ &=& \dfrac{9k}{k} \\\\ &=& \mathbf{9} \\ \hline \end{array} \)

help

\(\begin{array}{lccccccccccc} \text{row }1:& & & & & & 1 \\ \text{row }2:& & & & & 2 & & 2 \\ \text{row }3:& & & & 3 & & 4 & & 3 \\ \text{row }4:& & & 4 & & 7 & & 7 & & 4 \\ \text{row }5:& & 5 & & 11 & & 14 & & 11 & & 5 \\ \text{row }6:& 6 & & 16 & & 25 & & 25 & & 16 & & 6 \end{array} \)

If we form a number triangle as above,

what is the sum of all the numbers in the triangle with 20 rows?

\(\text{Let row $=r$} \\ \text{Let the sum of all the numbers in the rth row $=s_r$ } \\ \text{Let the sum of the numbers in the triangle from 1 to rth row $=s_{1\ldots r}$ }\)

\(\begin{array}{|rcll|} \hline s_1 &=& 1 \\ s_2 &=& 1 +3\cdot( 1 ) \\ s_3 &=& 1 +3\cdot( 1+2 ) \\ s_4 &=& 1 +3\cdot( 1+2+4 ) \\ s_5 &=& 1 +3\cdot( 1+2+4+8 ) \\ s_6 &=& 1 +3\cdot( 1+2+4+8+16 ) \\ \ldots \\ s_r &=& 1 +3\cdot( 1+2^1+2^2+2^3+2^4+\ldots + 2^{r-2} ) \\ &=& 1+ 3\cdot(2^{r-1}-1) \\ \mathbf{s_r} &=& \mathbf{ 3\cdot 2^{r-1} -2 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_{1\ldots r} &=& \sum \limits_{i=1}^{r} \left(3\cdot 2^{i-1} -2 \right) \\ &=& 3\sum \limits_{i=1}^{r} \left(2^{i-1}\right) -\sum \limits_{i=1}^{r} 2 \\ &=& 3\sum \limits_{i=0}^{r-1} \left(2^i\right) -2\sum \limits_{i=1}^{r} 1 \\ &=& 3\sum \limits_{i=0}^{r-1} \left(2^i\right) -2r \\ &=& 3\sum \limits_{i=0}^{r} \left(2^i\right) -3\cdot2^r -2r \quad | \quad \sum \limits_{i=0}^{r} \left(2^i\right) = \left( 2^{r+1}-1 \right) \\ &=& 3\cdot \left( 2^{r+1}-1 \right) -3\cdot2^r -2r \\ &=& 3\cdot 2^{r+1}-1 -3 -3\cdot2^r -2r \\ &=& 3\cdot \left(2^{r}\cdot 2 \right) -3\cdot2^r -2r -3 \\ &=& 2^{r}\left( 6-3\right) -2r -3 \\ \mathbf{ s_{1\ldots r} } &=& \mathbf{ 3\cdot 2^{r} -2r -3 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ s_{1\ldots r} } &=& \mathbf{ 3\cdot 2^{r} -2r -3 } \quad | \quad r = 20 \\\\ s_{1\ldots 20} &=& 3\cdot 2^{20} -2\cdot(20) -3 \\ &=& 3\cdot 2^{20} -2\cdot(20) -3 \\ &=& 3\cdot 2^{20} -43 \\ &=& 3\cdot 1048576 -43 \\ &=& 3\cdot 1048576 -43 \\ &=& 3145728 -43 \\ \mathbf{ s_{1\ldots 20} } &=& \mathbf{ 3145685} \\ \hline \end{array}\)

Thank you, CPhill !

Find the sum of the numbers \(3,8,13,\ldots,253\)

arithmetic series:

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

\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)\cdot d \quad &| \quad a_n = 253,\ a_1=3,\ d=5 \\\\ 253 &=& 3+(n-1)\cdot 5 \quad &| \quad -3 \\ 250 &=& (n-1)\cdot 5 \quad &| \quad :5 \\ 50 &=&n-1 \quad &| \quad +1 \\ 51 &=&n\\ \mathbf{ n } &=& \mathbf{51} \\ \hline \end{array}\)

sum:

\(\begin{array}{|rcll|} \hline sum &=& \left(\dfrac{a_1+a_n}{2} \right) \cdot n \quad &| \quad a_n = 253,\ a_1=3,\ n=51 \\\\ sum &=& \left(\dfrac{3+253}{2} \right) \cdot 51 \\ sum &=& \left(\dfrac{256}{2} \right) \cdot 51 \\ sum &=& 128 \cdot 51 \\ \mathbf{sum} &=& \mathbf{6528} \\ \hline \end{array}\)

Thank you, sinclairdragon428 !

let a equal the number of four digit odd numbers. let b equal the number of four digit multiples of 5. find a+b.

I assume:

These are two arithmetic series.

\(\begin{array}{|rcll|} \hline 1001 + (a-1)\cdot 2 &=& 9999 \quad | \quad \small \text{first odd four digit number }=1001,\ \text{last odd four digit number }=9999\\ (a-1)\cdot 2 &=& 9999 - 1001 \\ (a-1)\cdot 2 &=& 8998 \\ a-1 &=& 4499 \\ a &=& 4499+1 \\ \mathbf{a} &=& \mathbf{4500} \\\\ 1000 + (b-1)\cdot 5 &=& 9995 \quad | \quad \small \text{first four digit number multiples of 5}=1000,\ \text{last four digit number }=9995\\ (b-1)\cdot 5 &=& 9995 - 1000 \\ (b-1)\cdot 5 &=& 8995 \\ b-1 &=& 1799 \\ b &=& 1799+1 \\ \mathbf{b} &=& \mathbf{1800} \\ \hline \end{array} \)

\(a+b = 4500+1800 = 6300\)

How many paths are there from C to B, if every step must be up or to the right?