heureka

What is the remainder when 333^{333} is divided by 11?

\(\begin{array}{|rcll|} \hline && 333^{333} \pmod{11} \quad & | \quad 333 \equiv 3 \pmod{11} \\ &\equiv & 3^{333} \pmod{11} \quad & | \quad \text{Fermat's little theorem } 3^{10} \equiv 1 \pmod{11} \\ &\equiv & 3^{10\cdot 33+3} \pmod{11} \\ &\equiv & \left(3^{10}\right)^{33} 3^3 \pmod{11} \\ &\equiv & (1)^{33} 3^3 \pmod{11} \\ &\equiv & 1\cdot 3^3 \pmod{11} \\ &\equiv & 3^3 \pmod{11} \\ &\equiv & 27 \pmod{11} \\ &\mathbf{\equiv} & \mathbf{5 \pmod{11}} \\ \hline \end{array}\)

Thank you, CPhill

excellent

Heureka

What is the radius of the circle inscribed in triangle ABC if AB=22, AC=12, BC=14?

Express your answer in simplest radical form.

\(\text{Let $c = AB = 22 $} \\ \text{Let $b = AC = 12 $} \\ \text{Let $a = BC = 14 $} \\ \text{Let $r =$ radius of the circle inscribed. }\)

Formula:

\(\begin{array}{|rcll|} \hline \displaystyle r = \sqrt{ \dfrac{(s-a)(s-b)(s-c)}{s} } \qquad \text{ with } \qquad s=\dfrac{a+b+c}{2} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline s &=& \dfrac{14+12+22}{2} \\\\ &=& \dfrac{48}{2} \\\\ &=& 24 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s-a &=& 24- 14 \\ &=& 10 \\\\ s-b &=& 24 - 12 \\ &=& 12 \\\\ s-c &=& 24 - 22 \\ &=& 2 \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline r &=& \sqrt{ \dfrac{(s-a)(s-b)(s-c)}{s} } \\\\ &=& \sqrt{ \dfrac{10\cdot 12 \cdot 2}{24} } \\\\ &=& \sqrt{ \dfrac{10\cdot 24}{24} } \\\\ &=& \sqrt{ 10 } \\ \hline \end{array}\)

The radius of the cricle inscribed in triangle ABC is \(\sqrt{10}\)

Hi Melody

"Find the coefficient of x^3 y^3 z^2 in the expansion of (3x+5y+7z)^8".

Set \(a = 3x\)

Set \(b = 5y\)

Set \(c = 7z\)

Set \(n = 8\)

Set \(i = 3\)

Set \(j = 3\)

Set \(k = 2\)

Trinomial Coefficient of  \(x^3 y^3 z^2\):
\(\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2}(3x)^3\cdot(5y)^3\cdot (7z)^2 &=& \dfrac{8!}{3!3!2!} (3x)^3\cdot(5y)^3\cdot (7z)^2 \\\\ &=& \dfrac{8!}{3!3!2!} \cdot 3^35^37^2 x^3y^3z^2 \quad & | \quad \dfrac{8!}{3!3!2!} = 560 \\\\ &=& 560\cdot 3^35^37^2 x^3y^3z^2 \\\\ &=& 560\cdot 27 \cdot 125 \cdot 49 x^3y^3z^2 \\\\ &\mathbf{=}& \mathbf{ 92610000 x^3y^3z^2 } \\ \hline \end{array}\)

\(\text{The coefficient of $x^3 y^3 z^2$ in the expansion of $(3x+5y+7z)^8$ is $\mathbf{92~ 610~ 000}$ } \)

How many positive four-digit integers n have the property that the three-digit number obtained by removing

the leftmost digit is one-ninth of n?

\(\begin{array}{|lrcll|} \hline (1) & n_4 &=& 10^3a + n_3 \\ \\ \hline \\ (2) & n_3 &=& \dfrac{n_4}{9} \\\\ & 9n_3 &=& n_4 \\\\ & 9n_3 &=& 10^3a + n_3 \\\\ & 8n_3 &=& 10^3a \quad & | \quad : 8 \\\\ & \mathbf{n_3} & \mathbf{=} & \mathbf{125a} \\ \hline \end{array} \)

\(\begin{array}{|r|c|l|l|} \hline & n_3=125a & \\ a & n_3\lt 1000 & n_4 &\\ \hline 1 & 125 & 1125 & \frac{1125}{9} = 125 \\ 2 & 250 & 2250 & \frac{2250}{9} = 250 \\ 3 & 375 & 3375 & \frac{3375}{9} = 375 \\ 4 & 500 & 4500 & \frac{4500}{9} = 500 \\ 5 & 625 & 5625 & \frac{5625}{9} = 625 \\ 6 & 750 & 6750 & \frac{6750}{9} = 750 \\ 7 & 875 & 7875 & \frac{7875}{9} = 875 \\ \hline \end{array}\)

Find the smallest positive  \(N\) such that

\(N\equiv6(mod 12) \\ N\equiv6(mod 18) \\ N\equiv6(mod 24) \\ N\equiv6(mod 30 ) \\ N\equiv6(mod 60) \)

\(\begin{array}{|rcll|} \hline & N &\equiv& 6 \pmod {12} \\ & N &\equiv& 6 \pmod {18} \\ & N &\equiv& 6 \pmod {24} \\ & N &\equiv& 6 \pmod {30} \\ & N &\equiv& 6 \pmod {60} \\\\ \Rightarrow & N &\equiv& 6 \pmod{\text{lcm}(12,18,24,30,60)} \\ & N &\equiv& 6 \pmod{360} \\ & \mathbf{N} & \mathbf{=} & \mathbf{6+360m,\ \quad m \in Z} \\ \hline \end{array} \)

\(\text{The smallest positive $ N = 6,\ $ if $m = 0$ }\)

When N is divided by 10, the remainder is a. When N is divided by 13, the remainder is b.
What is N modulo 130, in terms of a and b?
(Your answer should be in the form ra+sb, where r and s are replaced by nonnegative integers less than 130.)

\(\begin{array}{|rclcl|} \hline N &\equiv& a \pmod{10} \quad &\text{or}& \quad N= a+10n,\ n \in Z \\ N &\equiv& b \pmod{13} \quad &\text{or}& \quad N= a+13m,\ m \in Z \\ \\ \hline \\ N = a+10n &=& b+13m \\ 10n &=& 13m+b-a \\\\ \mathbf{n} & \mathbf{=}& \mathbf{\dfrac{13m+b-a}{10}} \\\\ n &=& \dfrac{10m+3m+b-a}{10} \\\\ n &=& \dfrac{10m}{10} +\dfrac{3m+b-a}{10} \\\\ n &=& m +\underbrace{\dfrac{3m+b-a}{10}}_{=c} \\\\ c &=& \dfrac{3m+b-a}{10} \\\\ 10c &=& 3m+b-a \\\\ 3m &=& 10c+a-b \\\\ m &=& \dfrac{10c+a-b}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \\\\ m &=& \dfrac{9c+c+a-b}{3} \\\\ m &=& \dfrac{9c}{3} + \dfrac{c+a-b}{3} \\\\ m &=& 3c + \underbrace{\dfrac{c+a-b}{3}}_{=d} \\\\ d &=& \dfrac{c+a-b}{3} \\\\ 3d &=& c+a-b \\\\ \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\ \hline \end{array} \)

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

\(\begin{array}{|rcll|} \hline \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \quad & | \quad \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\\\ m &=& \dfrac{10(3d+b-a)+a-b}{3} \\\\ m &=& \dfrac{30d+10b-10a+a-b}{3} \\\\ m &=& \dfrac{30d+9b-9a}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline N &=& b+13m \quad & | \quad \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ N &=& b+13(10d+3b-3a) \\ N &=& 130d -39a + 40b \\ \hline \end{array}\)

\(\mathbf{N \pmod{130} =\ ?}\)

\(\begin{array}{|rcll|} \hline N &=& 130d -39a + 40b \quad & | \quad \pmod{130} \\ N &\equiv& 0 -39a + 40b \pmod{130} \\ N &\equiv& -39a + 130a + 40b \pmod{130} \\ \mathbf{N} & \mathbf{\equiv} & \mathbf{91a + 40b \pmod{130}} \\ \hline \end{array}\)

\(N \pmod{130} = 91a + 40b\)

In the sequence {2001, 2002, 2003,...} each term after the third is found

by subtracting the previous term from the sum of the two terms that precede that term.

For example, the fourth term is 2001+2002-2003=2000.

What is the value of the 2004th term? 

2. Solution:

\(\huge{ \begin{equation} a_n = \begin{cases} 2000+n \text{,} & \text{if } \ n \ \text{ is odd} \\ 2004-n \text{,} & \text{if } \ n \ \text{ is even} \\ \end{cases} \end{equation} }\)

\(\large{ \begin{array}{|rcll|} \hline a_{2004} &=& 2004 - 2004 \\ &=& 0 \\ \hline \end{array} }\)

In the sequence {2001, 2002, 2003,...} each term after the third is found

by subtracting the previous term from the sum of the two terms that precede that term.

For example, the fourth term is 2001+2002-2003=2000.

What is the value of the 2004th term? 

\(\begin{array}{|lcll|} \hline a_1 &=& 2001 \\ a_2 &=& 2002 \\ a_3 &=& 2003 \\ a_4 &=& -1\cdot a_3 + 1\cdot a_2 + 1\cdot a_1 \\ a_5 = -1\cdot a_4 + 1\cdot a_3 + 1\cdot a_2 &=& +2\cdot a_3 + 0\cdot a_2 - 1\cdot a_1 \\ a_6 = -1\cdot a_5 + 1\cdot a_4 + 1\cdot a_3 &=& -2\cdot a_3 + 1\cdot a_2 + 2\cdot a_1 \\ a_7 = -1\cdot a_6 + 1\cdot a_5 + 1\cdot a_4 &=& +3\cdot a_3 + 0\cdot a_2 - 2\cdot a_1 \\ a_8 = -1\cdot a_7 + 1\cdot a_6 + 1\cdot a_5 &=& -3\cdot a_3 + 1\cdot a_2 + 3\cdot a_1 \\ a_9 = -1\cdot a_8 + 1\cdot a_7 + 1\cdot a_6 &=& +4\cdot a_3 + 0\cdot a_2 - 3\cdot a_1 \\ a_{10} = -1\cdot a_9 + 1\cdot a_8 + 1\cdot a_7 &=& -4\cdot a_3 + 1\cdot a_2 + 4\cdot a_1 \\ a_{11} = -1\cdot a_{10} + 1\cdot a_{9} + 1\cdot a_{8} &=& +5\cdot a_3 + 0\cdot a_2 - 4\cdot a_1 \\ a_{12} = -1\cdot a_{11} + 1\cdot a_{10} + 1\cdot a_{9} &=& -5\cdot a_3 + 1\cdot a_2 + 5\cdot a_1 \\ a_{13} = -1\cdot a_{12} + 1\cdot a_{11} + 1\cdot a_{10} &=& +6\cdot a_3 + 0\cdot a_2 - 5\cdot a_1 \\ a_{14} = -1\cdot a_{13} + 1\cdot a_{12} + 1\cdot a_{11} &=& -6\cdot a_3 + 1\cdot a_2 + 6\cdot a_1 \\ \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} & \mathbf{=} & \mathbf{ -(-1)^n\lfloor \dfrac{n-1}{2}\rfloor \cdot a_3 + \lfloor\dfrac{1+(-1)^n}{2}\rfloor \cdot a_2+(-1)^n\lfloor \dfrac{n-2}{2} \rfloor \cdot a_1 } \\\\ a_{2014} &=& \small{ -(-1)^{2014}\lfloor \dfrac{2014-1}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+(-1)^{2014}}{2}\rfloor \cdot 2002 +(-1)^{2014}\lfloor \dfrac{2014-2}{2} \rfloor \cdot 2001 } \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+1}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{1+1}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -\lfloor \dfrac{2013}{2}\rfloor \cdot 2003 + \lfloor\dfrac{2}{2}\rfloor \cdot 2002 +1\cdot \lfloor \dfrac{2012}{2} \rfloor \cdot 2001 \\\\ &=& -1006 \cdot 2003 + 1 \cdot 2002 +1\cdot 1006 \cdot 2001 \\\\ &=& -2015018 + 2002 + 2013006 \\\\ &=& -2013016 + 2013006 \\\\ &=& 0 \\ \hline \end{array} \)

The value of the 2004th term is 0