heureka

In the diagram below
\(AM = BM = CM\), and
\(\angle BMC+\angle A = 210^\circ\).
Find \(\angle B\) in degree

\(\begin{array}{|rcll|} \hline \angle BMC+\angle A &=& 210^\circ \\ \mathbf{\angle BMC} &=& \mathbf{210^\circ-\angle A} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In }\triangle AMC \\ \hline (180^\circ-\angle BMC)+2*(\angle A) &=& 180^\circ \\ 2*(\angle A)&=& \angle BMC \quad | \quad \mathbf{\angle BMC=210^\circ-\angle A}\\ 2*(\angle A) &=& 210^\circ-\angle A \\ 3*(\angle A) &=& 210^\circ \\ \mathbf{\angle A} &=& \mathbf{70^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In }\triangle BMC \\ \hline \angle BMC+2*(\angle B) &=& 180^\circ \quad | \quad \angle BMC = 2*(\angle A) \\ 2*(\angle A)+2*(\angle B) &=& 180^\circ \\ \angle A+\angle B &=& 90^\circ \\ \angle B &=& 90^\circ -\angle A \\ \angle B &=& 90^\circ -70^\circ \\ \mathbf{\angle B} &=& \mathbf{20^\circ} \\ \hline \end{array}\)

Question: Compute

\(\dfrac{5}{4 \cdot 9} + \dfrac{7}{9 \cdot 16} + \dfrac{9}{16 \cdot 25} + \dfrac{11}{25 \cdot 36}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{5}{4 \cdot 9} + \dfrac{7}{9 \cdot 16} + \dfrac{9}{16 \cdot 25} + \dfrac{11}{25 \cdot 36}} \\\\ &=& \dfrac{5}{9 \cdot 4} + \dfrac{7}{16 \cdot 9} + \dfrac{9}{25 \cdot 16} + \dfrac{11}{36 \cdot 25} \\\\ &=& \dfrac{9-4}{9 \cdot 4} + \dfrac{16-9}{16 \cdot 9} + \dfrac{25-16}{25 \cdot 16} + \dfrac{36-25}{36 \cdot 25} \\\\ &=&\small{ \dfrac{9}{9 \cdot 4} -\dfrac{4}{9 \cdot 4} + \dfrac{16}{16 \cdot 9} -\dfrac{9}{16 \cdot 9} + \dfrac{25}{25 \cdot 16} -\dfrac{16}{25 \cdot 16} + \dfrac{36}{36 \cdot 25} -\dfrac{25}{36 \cdot 25}} \\\\ &=& \dfrac{1}{4} -\dfrac{1}{9} + \dfrac{1}{9} -\dfrac{1}{16} + \dfrac{1}{16} -\dfrac{1}{25} + \dfrac{1}{25} -\dfrac{1}{36} \\\\ &=& \dfrac{1}{4} -\dfrac{1}{36} \\\\ &=& \dfrac{36-4}{4*36} \\\\ &=& \dfrac{32}{4*36} \\\\ &=& \dfrac{8}{36} \\\\ &=& \mathbf{\dfrac{2}{9}} \\ \hline \end{array}\)

\(\dfrac{5}{4 \cdot 9} + \dfrac{7}{9 \cdot 16} + \dfrac{9}{16 \cdot 25} + \dfrac{11}{25 \cdot 36} = \mathbf{\dfrac{2}{9}}\)

If

\(f(x) = \dfrac{a}{x + 2}\),

solve for the value of \(a\) such that \(f(0) = f^{-1}(-3a)\).

My attempt:
\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{a}{x + 2} \\\\ f(0) &=& \dfrac{a}{0 + 2} \\\\ \mathbf{f(0)} &=& \mathbf{\dfrac{a}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{a}{x+2} \\\\ x &=& \dfrac{a}{f^{-1}(x)+2} \\\\ x\Big(f^{-1}(x)+2\Big) &=& a \\\\ xf^{-1}(x)+2x &=& a \\\\ xf^{-1}(x) &=& a-2x \\\\ f^{-1}(x) &=& \dfrac{a-2x}{x} \quad | \quad x =-3a\\\\ f^{-1}(-3a) &=& \dfrac{a-2(-3a)}{-3a} \\\\ f^{-1}(-3a) &=& \dfrac{a+6a}{-3a} \\\\ f^{-1}(-3a) &=& \dfrac{7a}{-3a} \\\\ \mathbf{f^{-1}(-3a)} &=& \mathbf{-\dfrac{7}{3}} \\\\ \hline f(0) &=& f^{-1}(-3a) \\\\ \dfrac{a}{2} &=& -\dfrac{7}{3} \\\\ \mathbf{a} &=& \mathbf{-\dfrac{14}{3}} \\ \hline \end{array}\)

If

\(h(y) = \dfrac{1 + y}{2 - y}\),

then what is the value of \(h^{-1}(15)\)?

My attempt:

\(\begin{array}{|rcll|} \hline h^{-1}\Big(h(y)\Big) &=& y \\\\ &&\boxed{h(y) = 15 \\ \frac{1+y}{2-y}=15\\ 1+y = 15(2-y) \\ 1+y=30-15y\\ 16y = 29 \\ \mathbf{y=\frac{29}{16}} } \\\\ \mathbf{h^{-1}(15)} &=& \mathbf{\dfrac{29}{16}} \\ \hline \end{array}\)

If

\(a + b = 7\) and \(a^3 + b^3 = 42 + ab\),

what is the value of the sum \(\dfrac{1}{a} + \dfrac{1}{b}\)?

\(\begin{array}{|rcll|} \hline \left(a + b\right)^3 =&=& a^3+3a^2b+3ab^2+b^3 \\ \left(a + b\right)^3 =&=& a^3+b^3+3ab(a+b) \\ 7^3 =&=& 42+ab+3ab*7 \\ 343 =&=& 42+22ab \\ 22ab=&=& 343-42\\ ab &=& \dfrac{301}{22} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{a+b}{ab} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{7}{\frac{301}{22}} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{7*22}{301} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{22}{43} \\ \hline \end{array}\)

How many zeros are at the end of \((100!)(200!)(300!)(400!)\) when multiplied out?

\(\begin{array}{|rcll|} \hline \text{zeros} &=& \small{ \lfloor\frac{100}{5}\rfloor +\lfloor\frac{100}{25}\rfloor +\lfloor\frac{200}{5}\rfloor +\lfloor\frac{200}{25}\rfloor +\lfloor\frac{200}{125}\rfloor +\lfloor\frac{300}{5}\rfloor +\lfloor\frac{300}{25}\rfloor +\lfloor\frac{300}{125}\rfloor +\lfloor\frac{400}{5}\rfloor +\lfloor\frac{400}{25}\rfloor +\lfloor\frac{400}{125}\rfloor } \\\\ &=& 20+4+40+8+1+60+12+2+80+16+3 \\ \mathbf{\text{zeros}} &=& \mathbf{246} \\ \hline \end{array}\)

Calculate \(40^{13} \pmod{85}\).
Express your answer as a non-negative integer that is less than \(85\).

\(\begin{array}{|rcll|} \hline 40^{3} \pmod{85} &\equiv& 80 \pmod{85} \\ &\equiv& 80-85 \pmod{85} \\ &\equiv& -5 \pmod{85} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 40^{13} \pmod{85} &\equiv& 40^{3*4+1} \pmod{85} \\ &\equiv& 40^{3*4}*40 \pmod{85} \\ &\equiv& \left(40^3\right)^4*40 \pmod{85} \\ &\equiv& \left(-5\right)^4*40 \pmod{85} \\ &\equiv& 625*40 \pmod{85} \\ &\equiv& 25000 \pmod{85} \\ \mathbf{40^{13} \pmod{85}} &\equiv& \mathbf{ {\color{red}10} \pmod{85}} \\ \hline \end{array}\)

Given positive integers \(x\) and \(y\) such that \(x\neq y\) and \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{20}\),
what is the smallest possible value for \(x + y\)?

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{20} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{20} \\\\ \mathbf{xy} &=& \mathbf{20*(x+y)} \\ \hline \end{array}\)

\(\large{AM\ge GM} \)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 20*(x+y) \\ (x+y)^2 &\ge& 4*20*(x+y) \\ x+y &\ge& 4*20 \\ \mathbf{x+y } &\ge& \mathbf{80} \\ \hline \end{array}\)

What is the largest integer \(n\) such that \(3^n\) is a factor of

\(1 * 3 * 5 * \ldots * 97 * 99 * 101 * 103 *\ldots * 197 * 199\)?

\(\begin{array}{|rcll|} \hline && 1*3*5*7*9*11 \ldots *195*197*199 \\\\ &=& \dfrac{ 1*2*3*4*5* \ldots *196*197*198*199} {2*4*6*8*10* \ldots *194*196*198 } \\\\ &=& \frac{ 1*2*3*4*5* \ldots *196*197*198*199} {(2*1)*(2*2)*(2*3)*(2*4)*(2*5)* \ldots *(2*97)*(2*98)*(2*99) } \\\\ &=& \dfrac{ 1*2*3*4*5* \ldots *196*197*198*199} {2^{99}*1*2*3*4*5* \ldots *97*98*99 } \\\\ &=& \dfrac{ 199!} {2^{99}*99! } \\\\ && \boxed{ 199! = \ldots * \left.3^{ \lfloor\frac{199}{3}\rfloor + \lfloor\frac{199}{9}\rfloor + \lfloor\frac{199}{27}\rfloor + \lfloor\frac{199}{81}\rfloor } \right. * \ldots \\ 199! = \ldots * \left.3^{66 +22 +7 +2 }\right. * \ldots \\ 199! = \ldots * \left.3^{97}\right. * \ldots \\\\ 99! = \ldots * \left.3^{ \lfloor\frac{99}{3}\rfloor + \lfloor\frac{99}{9}\rfloor + \lfloor\frac{99}{27}\rfloor + \lfloor\frac{99}{81}\rfloor } \right. * \ldots \\ 99! = \ldots * \left.3^{33 +11 +3+1 }\right. * \ldots \\ 99! = \ldots * \left.3^{48}\right. * \ldots \\ } \\\\ &=& \dfrac{ \ldots * \left.3^{97}\right. * \ldots} {2^{99}*\ldots * \left.3^{48}\right. * \ldots } \\\\ &=& \dfrac{ \ldots * \left.3^{97-48}\right. * \ldots} {2^{99}*\ldots} \\\\ &=& \dfrac{ \ldots * \left.3^{\color{red}49}\right. * \ldots} {2^{99}*\ldots} \\ \hline \end{array}\)

\(\mathbf{n=49}\)

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