Basic Maths: Test 37

$\begin{align} & \left( \frac{2-1}{2} \right)\,\left( \frac{3-1}{3} \right)\,\left( \frac{4-1}{4} \right)\,.......\left( \frac{120-1}{120} \right) \\ & =\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times ......\times \frac{119}{120} \\ & =\frac{1}{120} \\ \end{align}$

$\begin{align} & 13\frac{1}{2}%\,\,of\,\,100=\frac{27}{2}=13.5 \\ & \\ & 13.\overline{55}=13\frac{55}{99}\simeq 13.555555556(approx) \\ & {{\left( \frac{18.5}{5} \right)}^{2}}={{\left( 3.7 \right)}^{2}}=13.69 \\ & =\sqrt{187}=13.675 \\ \end{align}$

\[\begin{align} & {{\left( 1000 \right)}^{10000}}={{\left( {{10}^{3}} \right)}^{10000}}={{\left( 10 \right)}^{30000}} \\ & {{\left( 100000 \right)}^{1000}}={{\left( {{10}^{5}} \right)}^{1000}}={{10}^{5000}} \\ & {{2}^{100000}}={{\left( {{2}^{10}} \right)}^{10000}}={{\left( 1024 \right)}^{10000}} \\ & {{\left( 1024 \right)}^{10000}}>{{\left( 1000 \right)}^{10000}} \\ & {{3}^{200}}={{\left( {{3}^{2}} \right)}^{100}}={{\left( 9 \right)}^{100}} \\ & {{\left( 9 \right)}^{100}}\le {{\left( 10 \right)}^{100}} \\ \end{align}\]

$\begin{align} & If\,\,the\,\,\,sum\,\,\,\,of\,\,\,three\,\,\,non-zero\,\,\,\operatorname{int}egers\,\,\,be\,\,\,S,\,\,then \\ & First\,\,\,number=\frac{S}{5} \\ & \operatorname{Sec}ond\,\,\,number=\frac{S}{4} \\ & Third\,\,\,number\,=\frac{S}{k} \\ & Therefore\,\,\,\frac{1}{5}+\frac{1}{4}+\frac{1}{k}=1 \\ & \Rightarrow \frac{1}{k}=1-\frac{1}{5}-\frac{1}{4}=\frac{20-4-5}{20} \\ & \Rightarrow \frac{1}{k}=\frac{11}{20} \\ & \Rightarrow k=\frac{20}{11} \\ \end{align}$