heureka

Factor x^2-6x-9

$$\small{\text{$ x^2-6x-9 = \left[ x-3(1+\sqrt{2}) \right] \left[ x-3(1-\sqrt{2}) \right] $}}$$

tcdf(-50, 1.25, 20)

Wolfram Alpha:

How come 30^99 + 61^100 is divisible by 31

$$\small{\text{$30 \equiv - 1 \pmod {31}$}}\\ \small{\text{ and $61 \equiv - 1 \pmod {31}$}}\\\\ \small{\text{$(-1)^{99} + (-1)^{100} \stackrel{?}\equiv 0 \pmod{31}$}}\\\\ \small{\text{$-1 + 1 \equiv 0 \pmod{31}$}}\\\\$$

$$\small{\text{$ \begin{array}{| r|l|| r|l|| r|l|| r|l|| r|l|| r|l|| r|l|| r|l|| r|l|| r|l| } \hline \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} & \small{\text{first}} & \small{\text{following}} \\ \small{\text{digit}} & \small{\text{digits}} & \small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} &\small{\text{digit}} & \small{\text{digits}} \\ \hline a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd & a & \textcolor[rgb]{1,0,0}{b}cd \\ & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 & & b+c+d < 10 \\ \hline 0 & \textcolor[rgb]{1,0,0}{0}00 & 1 & \textcolor[rgb]{1,0,0}{1}00 & 2 & \textcolor[rgb]{1,0,0}{2}00 & 3 & \textcolor[rgb]{1,0,0}{3}00 & 4 & \textcolor[rgb]{1,0,0}{4}00 & 5 & \textcolor[rgb]{1,0,0}{5}00 & 6 & \textcolor[rgb]{1,0,0}{6}00 & 7 & \textcolor[rgb]{1,0,0}{7}00 & 8 & \textcolor[rgb]{1,0,0}{8}00 & 9 & \textcolor[rgb]{1,0,0}{9}00 \\ 1& 001 & 2& 101 & 3& 201 & 4& 301 & 5& 401 & 6& 501 & 7& 601 & 8& 701 & 9& 801&&\\ 2& 002 & 3& 102 & 4& 202 & 5& 302 & 5& 402 & 7& 502 & 8& 602 & 9& 702 &&&&\\ 3& 003 & 4& 103 & 5& 203 & 6& 303 & 5& 403 & 8& 503 & 9& 603 &&&&&&\\ 4& 004 & 5& 104 & 6& 204 & 7& 304 & 5& 404 & 9& 504 &&&&&&&&\\ 5& 005 & 6& 105 & 7& 205 & 8& 305 & 5& 405 &&&&&&&&&&\\ 6& 006 & 7& 106 & 8& 206 & 9& 306&&&&&&&&&&&&\\ 7& 007 & 8& 107 & 9& 207&&&&&&&&&&&& &&\\ 8& 008 & 9& 108&&&&&&&&&&&&&& &&\\ 9& 009 & &&&&&&&&&&&&&&&&&\\ \hline 1& 010 & 2& 110 & 3& 210 & 4& 310 & 5& 410 & 6& 510 & 7& 610 & 8& 710 & 9& 810&& \\ 2& 011 & 3& 111 & 4& 211 & 5& 311 & 6& 411 & 7& 511 & 8& 611 & 9& 711 & & && \\ 3& 012 & 4& 112 & 5& 212 & 6& 312 & 7& 412 & 8& 512 & 9& 612 & & & & && \\ 4& 013 & 5& 113 & 6& 213 & 7& 313 & 8& 413 & 9& 513 & & & & & & && \\ 5& 014 & 6& 114 & 7& 214 & 8& 314 & 9& 414 & & & & & & & & && \\ 6& 015 & 7& 115 & 8& 215 & 9& 315 & & & & & & & & & & && \\ 7& 016 & 8& 116 & 9& 216 & & & & & & & & & & & & && \\ 8& 017 & 9& 117 & & & & & & & & & & & & & & && \\ 9& 018 & & & & & & & & & & & & & & & & && \\ \hline 2& 020 & 3& 120 & 4& 220 & 5& 320 & 6& 420 & 7& 520 & 8& 620 & 9& 720 & & && \\ 3& 021 & 4& 121 & 5& 221 & 6& 321 & 7& 421 & 8& 521 & 9& 621 & & & & && \\ 4& 022 & 5& 122 & 6& 222 & 7& 322 & 8& 422 & 9& 522 & & & & & & && \\ 5& 023 & 6& 123 & 7& 223 & 8& 323 & 9& 423 & & & & & & & & && \\ 6& 024 & 7& 124 & 8& 224 & 9& 324 & & & & & & & & & & && \\ 7& 025 & 8& 125 & 9& 225 & & & & & & & & & & & & && \\ 8& 026 & 9& 126 & & & & & & & & & & & & & & && \\ 9& 027 & & & & & & & & & & & & & & & & && \\ \hline 3& 030 & 4& 130 & 5& 230 & 6& 330 & 7& 430 & 8& 530 & 9& 630 & & & & && \\ 4& 031 & 5& 131 & 6& 231 & 7& 331 & 8& 431 & 9& 531 & & & & & & && \\ 5& 032 & 6& 132 & 7& 232 & 8& 332 & 9& 432 & & & & & & & & && \\ 6& 033 & 7& 133 & 8& 233 & 9& 333 & & & & & & & & & & && \\ 7& 034 & 8& 134 & 9& 234 & & & & & & & & & & & & && \\ 8& 035 & 9& 135 & & & & & & & & & & & & & & && \\ 9& 036 & & & & & & & & & & & & & & & & && \\ \hline 4& 040 & 5& 140 & 6& 240 & 7& 340 & 8& 440 & 9& 540 & & & & & & && \\ 5& 041 & 6& 141 & 7& 241 & 8& 341 & 9& 441 & & & & & & & & && \\ 6& 042 & 7& 142 & 8& 242 & 9& 342 & & & & & & & & & & && \\ 7& 043 & 8& 143 & 9& 243 & & & & & & & & & & & & && \\ 8& 044 & 9& 144 & & & & & & & & & & & & & & && \\ 9& 045 & & & & & & & & & & & & & & & & && \\ \hline 5& 050 & 6& 150 & 7& 250 & 8& 350 & 9& 450 & & & & & & & & && \\ 6& 051 & 7& 151 & 8& 251 & 9& 351 & & & & & & & & & & && \\ 7& 052 & 8& 152 & 9& 252 & & & & & & & & & & & & && \\ 8& 053 & 9& 153 & & & & & & & & & & & & & & && \\ 9& 054 & & & & & & & & & & & & & & & & && \\ \hline 6& 060 & 7& 160 & 8& 260 & 9& 360 & & & & & & & & & & && \\ 7& 061 & 8& 161 & 9& 261 & & & & & & & & & & & & && \\ 8& 062 & 9& 162 & & & & & & & & & & & & & & && \\ 9& 063 & & & & & & & & & & & & & & & & && \\ \hline 7& 070 & 8& 170 & 9& 270 & & & & & & & & & & & & && \\ 8& 071 & 9& 171 & & & & & & & & & & & & & & && \\ 9& 072 & & & & & & & & & & & & & & & & && \\ \hline 8& 080 & 9& 180 & & & & & & & & & & & & & & && \\ 9& 081 & & & & & & & & & & & & & & & & && \\ \hline 9& 090 & & & & & & & & & & & & & & & & && \\ \hline sum &= 55 & sum &= 45 & sum &= 36 & sum &= 28 & sum &= 21 & sum &= 15 & sum &= 10 & sum &= 6 & sum &= 3 & sum &= 1 \\ \hline \end{array} $}}$$

sum = 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 220

sum - 1 (0000) = 219

how many 4-digit numbers are there such that the thousands digit is equal to the sum of the other three digits?

  $$\small{\text{4-digit-number:~$abcd = a10^3+b10^2+c10+d\qquad a = b+c+d$}} \\ \small{\text{$\Rightarrow (b+c+d)10^3+bcd_{10} < 10000$}} \\ \small{\text{$\Rightarrow (b+c+d)10^3< 10000-bcd_{10}$}} \\ \small{\text{$\Rightarrow (b+c+d)< \dfrac{10000-bcd_{10}}{ 10^3 }$}} \\ \small{\text{$\Rightarrow (b+c+d)< 10 - \underbrace{ \dfrac{ bcd_{10} }{ 1000 } }_{ 0\dots 0.999} $}} \\\\ \small{\text{$\Rightarrow \boxed{\mathbf{(b+c+d)< 10} } $}} \\\\$$

$$\small{\text{$b+c+d < 10 \qquad \begin{array}{|r|c|l|} \hline (b=0) & c & d \\ \hline 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 3 \\ 0 & 0 & 4 \\ 0 & 0 & 5 \\ 0 & 0 & 6 \\ 0 & 0 & 7 \\ 0 & 0 & 8 \\ 0 & 0 & 9 \\ \hline 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 1 & 4 \\ 0 & 1 & 5 \\ 0 & 1 & 6 \\ 0 & 1 & 7 \\ 0 & 1 & 8 \\ \hline 0 & 2 & 0 \\ 0 & 2 & 1 \\ 0 & 2 & 2 \\ 0 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 2 & 5 \\ 0 & 2 & 6 \\ 0 & 2 & 7 \\ \hline 0 & 3 & 0 \\ 0 & 3 & 1 \\ 0 & 3 & 2 \\ 0 & 3 & 3 \\ 0 & 3 & 4 \\ 0 & 3 & 5 \\ 0 & 3 & 6 \\ \hline 0 & 4 & 0 \\ 0 & 4 & 1 \\ 0 & 4 & 2 \\ 0 & 4 & 3 \\ 0 & 4 & 4 \\ 0 & 4 & 5 \\ \hline 0 & 5 & 0 \\ 0 & 5 & 1 \\ 0 & 5 & 2 \\ 0 & 5 & 3 \\ 0 & 5 & 4 \\ \hline 0 & 6 & 0 \\ 0 & 6 & 1 \\ 0 & 6 & 2 \\ 0 & 6 & 3 \\ \hline 0 & 7 & 0 \\ 0 & 7 & 1 \\ 0 & 7 & 2 \\ \hline 0 & 8 & 0 \\ 0 & 8 & 1 \\ \hline 0 & 9 & 0 \\ \hline \end{array} \qquad sum_{b=0} = 1+2+3+4+\dots +10 $}}$$

$$\small{\text{$ \begin{array}{lclcr} b &\qquad & sum \\ \hline 0 & & 1+2+3+4+5+6+7+8+9 +10 & = & 55\\ 1 & & 1+2+3+4+5+6+7+8 + 9 & = & 45\\ 2 & & 1+2+3+4+5+6+7+ 8 & = & 36\\ 3 & & 1+2+3+4+5+6+7 & = & 28\\ 4 & & 1+2+3+4+5+ 6 & = & 21\\ 5 & & 1+2+3+4+ 5 & = & 15\\ 6 & & 1+2+3+4 & = & 10\\ 7 & & 1+2+3 & = & 6\\ 8 & & 1+2 & = & 3\\ 9 & & 1 & = & 1\\ \hline & & sum &=& 220 \end{array} $}}$$

 There are 220 - (abcd = 0000) = 219  4-digit numbers

Hallo Melody, i think add for the first digit 5 the following digits 221 with 3 numbers of ways and you get

216+3 also 219  4-digit numbers

(a+b)+(b+c)+(a+c)=2(a+b+c)=10+20+24=54

a+b+c=54/2=27

161051, 102487, 65219, 41503, continue the sequence plz ?

This is a geometric sequence with ratio 

$$\small{\text{$r = \dfrac{102487}{161051} = \dfrac{65219}{102487} = \dfrac{41503}{65219} = \dfrac{63}{99} $}}$$

$$t_5=t_4\cdot r=41503 \cdot r=26411$$

$$\small{\text{ Given $m = 2n + 1$, what integer between 0 and $m$ is the inverse of $ 2 \pmod{ m}$ }}\\ \small{\text{ Answer in terms of $n$. }}$$

$$\small{\text{ $2x \equiv1 \pmod{m}$ }}\\ \small{\text{ $2x \equiv1 \pmod{2n+1}$ }}\\ \small{\text{ $2x-1 = 2n+1$ }}\\ \small{\text{ $2x = 2n+2$ }}\\ \small{\text{ $\mathbf{x = n+1}$ }}\\$$

$$\small{\text{ The inverse of $a$ modulo 39 is $b$. }}\\ \small{\text{ What is the inverse of $4a$ modulo 39 in terms of $b$? }}\\ \small{\text{ Give your answer as an expression in terms of $b$. }}$$

$$\small{\text{ $a\cdot b \equiv 1 \pmod{39}$ and $4a\cdot x \equiv 1 \pmod{39}$ so $ab=4ax$ and $x=\frac{1}{4}b$ }}\\ \small{\text{ The inverse of $4a \pmod{39}$ is $\frac{1}{4}b$ }}$$

$$\small{\text{ Let $x$ and $y$ be integers satisfying $41x + 53y = 12$. Find the residue of $x$ modulo 53. }}\\ \small{\text{ (Express your answer as an integer from 0 to 52.) }}$$

$$\small{\text{ If $x=264$ and $y=-204$ then $41x + 53y = 12$. }}\\ \small{\text{ The residue of $x\pmod{ 53 } = 52$. }}$$