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Arithmetic sequence with 52 terms

sum = 52 ( 2+104)/2 =2756                2756 mod 7 =5

U can remember infinity numbers If u can which is impossible 

Bcz 2 to 9 they are 7 numbers in between even 1st three digits and last 3 digits are same..it is even impossible to remember all numbers they are 7 different numbers in three places of phone and 4th digit is different from everything

"I got the six equations and six unknowns but was intimidated to solve it."

I'm not surprised.  I wouldn't have bothered if I didn't have software available to do it!  Although there is a matrix capability on the web2.0 calculator on the home page here, it just goes up to 4x4 matrices, so wouldn't have helped here.  Probably Wolfram Alpha would do it, but I haven't tried it with this problem.

The smallest possible value of x + y = 45, which you get for x = 9 and y = 36.

The number of possible sequences is 108.

The possible areas of the square are 1, 4.

awesome! I got the six equations and six unknowns but was intimidated to solve it.

y can be 1 or 2    making x 3 or 1   respectively            two pairs of positive integers

Happy July 4th to everyone. :DDD

=^._.^=

See https://web2.0calc.com/questions/help-pls_85427#2

Like so:

Determine the smallest non-negative integer a that satisfies the congruences:

\(a \equiv 2 \pmod{ 3}\\ a \equiv 4 \pmod{ 5}\\ a \equiv 1 \pmod{ 7}\\ a \equiv 8 \pmod{ 9}\)

1. join
\(a \equiv 2 \pmod{ 3} \\ a \equiv 8 \pmod{ 9}\)

\(\begin{array}{|rcll|} \hline a&=&2+3n \\ a&=&8+9m \\ \hline 2+3n&=& 8+9m \\ 3n&=& 6+9m \quad | \quad : 3\\ n&=& 2+3m \\ \hline a&=&2+3n \quad | \quad n=2+3m \\ a&=&2+3(2+3m) \\ a&=&2+6+9m \\ a&=&8+9m \\ \mathbf{a} \equiv \mathbf{ 8 \pmod{ 9} } \\ \hline \end{array} \\ \begin{equation}\left . \begin{aligned} & a \equiv 2 \pmod{ 3} \\ & a \equiv 8 \pmod{ 9} \end{aligned}\right \rbrace \mathbf{a} \equiv \mathbf{ 8 \pmod{ 9} } \end{equation} \)

2. join
\(a \equiv 4 \pmod{ 5} \\ a \equiv 1 \pmod{ 7}\)

\(\begin{array}{|rcl|rcl|rcl|} \hline a&=&4+5r \\ a&=&1+7s \\ \hline 4+5r&=&1+7s \\ 5r &=& 7s-3 \\ r &=& \frac{7s-3}{5} \\ r &=& \frac{5s+2s-3}{5} \\ r &=& s+ \underbrace{ \frac{2s-3}{5} }_{=t} \\ \mathbf{r} &=& \mathbf{s+t} & t &=& \frac{2s-3}{5}\\ & & & 5t &=& 2s-3 \\ & & & 2s &=& 5t+3 \\ & & & s&=& \frac{5t+3}{2} \\ & & & s&=& \frac{4t+t+2+1}{2} \\ & & & s&=& 2t+1 + \underbrace{ \frac{t+1}{2}}_{=u} \\ & & & \mathbf{s}&=& \mathbf{2t+1 + u} \\ & & & & & & u&=&\frac{t+1}{2} \\ & & & & & & 2u&=& t+1 \\ & & & & & & t&=& 2u-1 \\ & & & s &=& 2(2u-1)+1 +u \\ & & & s &=& 5u-1 \\ r &= &5u-1+2u-1 \\ r &= &7u-2 \\ \hline a &=& 4+5r \\ a &=& 4+5(7u-2) \\ a &=& -6+35u \\ \mathbf{a} &\equiv& \mathbf{ -6 \pmod{ 35} } \\ \hline \end{array}\\ \begin{equation}\left . \begin{aligned} & a \equiv 4 \pmod{ 5} \\ & a \equiv 1 \pmod{ 7} \end{aligned}\right \rbrace \mathbf{a} \equiv \mathbf{ -6 \pmod{ 35} } \end{equation}\)

3. join
\(a \equiv 8 \pmod{ 9} \\ a \equiv -6 \pmod{35}\)

\(\begin{array}{|rcl|rcl|rcl|} \hline a&=&8+9v \\ a&=&-6+35w \\ \hline 8+9v&=&-6+35w \\ 9v &=& 35w-14 \\ v&=& \frac{35w-14}{9} \\ v &=& \frac{27w+8w-9-5}{9} \\ v &=& 3w-1+ \underbrace{ \frac{8w-5}{9} }_{=x} \\ \mathbf{v} &=& \mathbf{3w-1+x} & x &=& \frac{8w-5}{9}\\ & & & 9x &=& 8w-5 \\ & & & 8w &=& 9x+5 \\ & & & w&=& \frac{9x+5}{8} \\ & & & w&=& \frac{8x+x+5}{8} \\ & & & w&=& x + \underbrace{ \frac{x+5}{8}}_{=y} \\ & & & \mathbf{w}&=& \mathbf{x+y} \\ & & & & & & y&=&\frac{x+5}{8} \\ & & & & & & 8y&=& x+5 \\ & & & & & & x&=& 8y-5 \\ & & & w &=& 8y-5 +y \\ & & & w &=& 9y-5 \\ v &= &3(9y-5)-1+8y-5 \\ v &= &35y-21 \\ \hline a &=& 8+9v \\ a &=& 8+9(35y-21) \\ a &=& -181+315y \\ \mathbf{a} &\equiv& \mathbf{ -181 \pmod{ 315} } \\ \mathbf{a} &\equiv& \mathbf{ -181+315 \pmod{ 315} } \\ \mathbf{a} &\equiv& \mathbf{ 134 \pmod{ 315} } \\ \hline \end{array}\\ \begin{equation}\left . \begin{aligned} & a \equiv 8 \pmod{ 9} \\ & a \equiv -6 \pmod{35} \end{aligned}\right \rbrace \mathbf{a} \equiv \mathbf{ 134 \pmod{ 315} } \end{equation}\)

The smallest non-negative integer a is 134

What is the smallest integer n, greater than 1, such that
\(\dfrac{1}{n} \pmod{130}\) and \(\dfrac{1}{n} \pmod {9}\)

are both defined?

There is an invertible modulo 130, if \(\gcd(130,n)=1~\)(130 and n are relatively prime)
There is an invertible modulo 9, if \(\gcd(9,n)=1~\) (9 and n are relatively prime)

\(n > 1\)

\(\begin{array}{|c|c|c|c|} \hline n & \gcd(130,n) & \gcd(9,n) \\ \hline 2 & \gcd(130,2)=2 & \gcd(9,2)=\color{red}1 \\ 3 & \gcd(130,3)=\color{red}1 & \gcd(9,3)=3 \\ 4 & \gcd(130,4)=2 & \gcd(9,4)=\color{red}1 \\ 5 & \gcd(130,5)=5 & \gcd(9,5)=\color{red}1 \\ 6 & \gcd(130,6)=2 & \gcd(9,6)=3 \\ \mathbf{7} & \gcd(130,7)=\color{red}1 & \gcd(9,7)=\color{red}1 & \checkmark \\ \hline \end{array}\)

The smallest integer is 7

\(\dfrac{1}{7} \pmod{130} \equiv 93 \text{ and } \dfrac{1}{7} \pmod {9} \equiv 4\)

What is the remainder when \(17^7\) is divided by \(35\)?

\(\small{ \begin{array}{|rcll|} \hline \mathbf{17^{7} \pmod{35}} &\equiv&\left( 17\right)^{2*3+1} \pmod{35} \\ &\equiv& \left( 17^2 \right)^3*17 \pmod{35} \quad &| \quad 17^2=289 \equiv 9 \pmod{35} \\ &\equiv& \left( 9 \right)^3*17 \pmod{35} \\ &\equiv& 729*17 \pmod{35} \quad &| \quad 729 \equiv 29 \pmod{35} \\ &\equiv& 29*17 \pmod{35} \\ &\equiv& 493 \pmod{35} \quad &| \quad 493 \equiv 3 \pmod{35} \\ &\equiv& \mathbf{ 3 \pmod{35} } \\ \hline \end{array} }\)

The reminder is 3

What is the remainder when \(2013^{13}\) is divided by \(13\)?

\(\begin{array}{|rcll|} \hline && \mathbf{2013^{13} \pmod{13}} \quad &| \quad 2013 \equiv 11 \pmod{13} \\ &\equiv&11^{13}\pmod{13} \quad &| \quad 11^4 \equiv 3 \pmod{13} \\ &\equiv&\left( 11\right)^{4*3+1} \pmod{13} \\ &\equiv& \left( 11^4 \right)^3*11 \pmod{13} \\ &\equiv& \left( 3 \right)^3*11 \pmod{13} \\ &\equiv& 27*11 \pmod{13}\quad &| \quad 27 \equiv 1 \pmod{13} \\ &\equiv& 1*11 \pmod{13} \\ &\equiv& \mathbf{ 11 \pmod{13} } \\ \hline \end{array}\)

The reminder is 11

Area of triangle ABC 

This question is WRONG!!!

[ABC] cannot be more than 200 square units!!!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Here's the corrected question:

Triangle ABC is isosceles with AB = BC.  If AB = 20 and [ABC] = 180, then find the perimeter of triangle ABC.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[ABC] = 180           AB = BC = 20

FE = [ABC] / BC = 9

x/9 = 9/(20-x)

x = 10 - √19

AC = 2[sqrt(x² + 9²)] ≈ 21.244

Perimeter ≈ 61.244

What is the remainder when 17^7 is divided by 35?

Hello Guest!

17^7:35 = 11723962.0857

0.0857 x 35 = 3

The remainder is 3

  !

You can use inversion on the problem.

Find the area of the triangle whose two angles are 30 degrees and 45 degrees, and the side included between them has sqrt(3) + 1 length units.

Hello Guest!

\(tan(30°)=\frac{h}{\sqrt{3}+1-h}\\ \frac{ \sqrt{3} }{3} (\sqrt{3}+1-h)=h\\ 1+\frac{ \sqrt{3} }{3}-\frac{ h\sqrt{3} }{3}=h\\ 1+\frac{ \sqrt{3} }{3}=h+\frac{ h\sqrt{3} }{3}\\ 1+\frac{ \sqrt{3} }{3}=h(1+\frac{ \sqrt{3} }{3}) \)

\(h=1\)

\(A=\frac{h}{2}(\sqrt{3}+1)\\ A=\frac{1}{2}(\sqrt{3}+1)\\ A= \color{blue}\frac{\sqrt{3}}{2}+\frac{1}{2}=1.3660\)

  !

The number of possible license plates is 11656.

Hey there, Guest!

1x + 2x + 3x + 4x + 5x + 6x = 1

Therefore 21x equals 1.

x = \(\frac{1}{21}\)

Hope this helped! :)

( ﾟдﾟ)つ Bye

cases: 73 is in the front, middle or end. 

 case where it is the front: 

first - second digit, 1 choice each

3rd digit: 1~9 = 9

4th : 9

case where it is the middle:

first digit has 9 choices, and the 4th also has 9 choices. 2nd and 3rd have one choice. 

case where it is the end. 

same thing, first digit 9 choices, 2nd digit 9 choices, 3 and 4, 1 choice. 

So! (9 times 9)+(9 times 9)+(9 times 9) = ill leave that to you

I am not sure, but I try a few numbers in and the smallest possible value I can get is 

$\boxed{9}$, with $x=-1, y=-1, z=2$