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 

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

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

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

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

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

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

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

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