heureka

In the prime factorization of \(109!\), what is the exponent of \(3\)?

(Reminder: The number \(n!\) is the product of the integers from \(1\) to \(n\).

For example, \(5!=5\cdot 4\cdot3\cdot2\cdot 1= 120\).)

\(\begin{array}{|rcll|} \hline \dfrac{109}{3^1} &=& \lfloor 36.3333333333 \rfloor \\ &=& 36 \\ \hline \dfrac{109}{3^2} &=& \lfloor 12.1111111111 \rfloor \\ &=& 12 \\ \hline \dfrac{109}{3^3} &=& \lfloor 4.03703703704 \rfloor \\ &=& 4 \\ \hline \dfrac{109}{3^4} &=& \lfloor 1.34567901235 \rfloor \\ &=& 1 \\ \hline \dfrac{109}{3^5} &=& \lfloor 0.44855967078 \rfloor \\ &=& 0 \\ \ldots \\ &=& 0 \\ \hline \text{sum} && 36+12+4+1= \mathbf{53} \\ \hline \end{array} \)

\(\mathbf{109!} = 2^{104}×\mathbf{3^{\color{red}53}}×5^{25}×7^{17}×11^9×13^8×17^6×19^5×23^4×\cdots\)

Let \(a\) and \(b\) be positive real numbers such that \(a^b = b^a,\ b = 9a\),
Then \(a\) can be expressed in the form \(sqrt(n,m)\) where \(m\) and \(n\) are positive integers
and \(n\) is as small as possible.

Find \(m + n\)

\(\begin{array}{|rcll|} \hline \mathbf{a^b} &=& \mathbf{b^a} \quad | \quad \mathbf{b = 9a} \\\\ a^{9a} &=& (9a)^a \\\\ a^{9a*\frac{1}{a}} &=& \left(9a\right)^{a*\frac{1}{a}} \\\\ a^{9} &=& 9a \quad | \quad *(a^{-1}) \\\\ a^{9}a^{-1} &=& 9a^1a^{-1} \\\\ a^{9-1} &=& 9a^{1-1} \\\\ a^{8} &=& 9a^0 \quad | \quad a^0 = 1 \\\\ a^{8} &=& 9 \\\\ a^{8*\frac{1}{8}} &=& 9^{\frac{1}{8}} \\\\ a &=& 9^{\frac{1}{8}} \\\\ a &=& (3^2)^{\frac{1}{8}} \\\\ a &=& 3^{2*\frac{1}{8}} \\\\ a &=& 3^{\frac{2}{8}} \\\\ a &=& 3^{\frac{1}{4}} \\\\ \mathbf{a} &=& \mathbf{\sqrt[4]{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline m+n &=& 3+4 \\ \mathbf{m+n} &=& \mathbf{7} \\ \hline \end{array}\)

Post New Answer

\(\log_{10}(x) = 3 + \log_{10}(y)\)

find \(\dfrac{x}{y}\)

Formula:

\(\boxed{\log\left({\dfrac{x}{y}}\right) = \log(x)-\log(y)}\)

\(\begin{array}{|rcll|} \hline \log_{10}(x) &=& 3 + \log_{10}(y) \quad | \quad -\log_{10}(y) \\\\ \log_{10}(x) -\log_{10}(y) &=& 3 \\\\ \log_{10}\left({\dfrac{x}{y}}\right) &=& 3 \\\\ \dfrac{x}{y} &=& 10^3 \\\\ \mathbf{ \dfrac{x}{y} } &=& \mathbf{1000} \\ \hline \end{array}\)

Algebra
Compute the sum \(\mathbf{\dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \cdots}\)

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\ \begin{array}{|lcll|} \hline s_n = \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

Formula:
\(\begin{array}{|lcll|} \hline \text{in general}:\ \frac{1}{n(n+d)} = \frac{1}{d}\left(\frac{1}{n}- \frac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

we rearrange:
\(\begin{array}{|rcll|} \hline \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{2}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{2}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n}\times \dfrac{1}{n+1} - \dfrac{2}{n}\times \dfrac{1}{n+2} \\\\ &=& 2\times \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- 2\times \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n} - \dfrac{2}{n+1} -\dfrac{1}{n} + \dfrac{1}{n+2} \\\\ \mathbf{\dfrac{2}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2} } \\ \hline \end{array}\)

telescoping series
\(\begin{array}{|rcll|} \hline s_n &=& \mathbf{\dfrac{1}{1}} &\mathbf{-}& \mathbf{\dfrac{2}{2}} &\color{red}+& \color{red}\dfrac{1}{3} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{2}} &\color{red}-& \color{red}\dfrac{2}{3} &\color{blue}+& \color{blue}\dfrac{1}{4} \\\\ &\color{red}+& \color{red}\dfrac{1}{3} &\color{blue}-& \color{blue}\dfrac{2}{4} &\color{red}+& \color{red}\dfrac{1}{5} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{4} &\color{red}-& \color{red}\dfrac{2}{5} &\color{green}+& \color{green}\dfrac{1}{6} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{n-2} &\color{green}-& \color{green}\dfrac{2}{n-1} &\color{red}+& \color{red}\dfrac{1}{n} \\\\ &\color{green}+& \color{green}\dfrac{1}{n-1} &\color{red}-& \color{red}\dfrac{2}{n} &\mathbf{+}& \mathbf{\dfrac{1}{n+1}} \\\\ &\color{red}+& \color{red}\dfrac{1}{n} &\mathbf{-}& \mathbf{\dfrac{2}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{n+2}} \\ \hline \end{array}\)

The part of each term cancelling with part of the next two diagonal terms:
Example:
\(\begin{array}{|lcll|} \hline \frac{1}{3}-\frac{2}{3}+\frac{1}{3} = 0 \\ \frac{1}{4}-\frac{2}{4}+\frac{1}{4} = 0 \\ \frac{1}{5}-\frac{2}{5}+\frac{1}{5} = 0 \\ \ldots \\ \frac{1}{n}-\frac{2}{n} + \frac{1}{n} = 0 \\ \hline \end{array}\)

So s_n is, we have all black terms left :
\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{1}{1}-\dfrac{2}{2}+\dfrac{1}{2} + \dfrac{1}{n+1} - \dfrac{2}{n+1} + \dfrac{1}{n+2} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{1}{2} - \dfrac{1}{n+1} + \dfrac{1}{n+2}} \\ \hline \end{array}\)

\(\lim \limits_{n\to \infty} { \dfrac{1}{n+1}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{1}{n+2} } = 0\)

\( \begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{1}{2} - 0 + 0 \\ &=& \dfrac{1}{2} \\ \hline \end{array}\)

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ \dfrac{1}{2} } } \\ \end{array}\)

One painter can paint an Italian restaurant in 10 hours and the second painter takes 8 hours.
How long would it take the two painters together to paint the Italian restaurant (R)?

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{R}{10\ h} + \dfrac{R}{8\ h}} &=& \mathbf{ \dfrac{R}{x} } \\\\ \dfrac{1}{10} + \dfrac{1}{8} &=& \dfrac{1}{x} \quad | \quad \times 80x \\\\ \dfrac{80x}{10} + \dfrac{80x}{8} &=& \dfrac{80x}{x} \\\\ 8x+10x &=& 80 \\\\ 18x &=& 80 \\\\ x &=& \dfrac{80}{18} \\\\ x &=& 4.\overline{4}\ \text{hours} \\\\ \mathbf{x} &=& \mathbf{4\ \text{hours}\ 26 \ \text{minutes}\ 40 \ \text{seconds} } \\ \hline \end{array}\)

Fatimah,Siti and Noraini share 90 marbles in the ratio 4 : 5 : 6.

Calculate the number of marbles that siti receives.

\(\begin{array}{|rcll|} \hline 4x+5x+6x &=& 90 \\ x(4+5+6) &=& 90 \\ 15x &=& 90 \\ x &=& \dfrac{90}{15} \\ \mathbf{x} &=& \mathbf{6} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{Fatimah}: & 4\times \mathbf{6} &=& 24\ \text{marbles} \\ \text{Siti}: & 5\times \mathbf{6} &=& 30\ \text{marbles} \\ \text{Noraini}: & 6\times \mathbf{6} &=& 36\ \text{marbles} \\ \hline \end{array}\)

Probability

Find the number of ordered pairs \((m, n)\) of integers that satisfy

\(mn = 2m + 4n\)

\(\begin{array}{|rcll|} \hline \mathbf{mn} &=& \mathbf{2m + 4n} \\ mn-2m - 4n &=& 0 \\ m(n-2) - 4n &=& 0 \\ m(n-2) - 4n +8 &=& 8 \\ m(n-2) - 4(n-2) &=& 8 \\ \mathbf{(n-2)(m-4)} &=& \mathbf{8} \\ \hline \end{array}\)

4 solutions:
\(\begin{array}{|rcll|} \hline 1) & 8 &=& 1*8 \\ 2) & 8 &=& 2*4 \\ 3) & 8 &=& 4*2 \\ 4) & 8 &=& 8*1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 1) & (n-2)(m-4) &=& 1*8 \\ & n-2 =1 && m-4 = 8 \\ & \mathbf{n =3} && \mathbf{m = 12} \\ \hline \end{array} \begin{array}{|rcll|} \hline 2) & (n-2)(m-4) &=& 2*4 \\ & n-2 =2 && m-4 = 4 \\ & \mathbf{n =4} && \mathbf{m = 8} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 3) & (n-2)(m-4) &=& 4*2 \\ & n-2 =4 && m-4 = 2 \\ & \mathbf{n =6} && \mathbf{m = 6} \\ \hline \end{array} \begin{array}{|rcll|} \hline 4) & (n-2)(m-4) &=& 8*1 \\ & n-2 =8 && m-4 = 1 \\ & \mathbf{n =10} && \mathbf{m = 5} \\ \hline \end{array}\)

Let \(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,\ r_2,\ \ldots,\ r_{100}\) be the roots of \( f(x)\).
Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array}\)

\(\mathbf{\text{vieta:}}\)
For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0\)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:
\(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array}\)