Solve {l}{(1+1/2+1/3+1/4+1/5)*(frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6})-(1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{8}+frac{1}{6})}{*(frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5})}

$$\left. \begin{array} { l } { ( 1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 } ) \times ( \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } ) - ( 1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 8 } + \frac { 1 } { 6 } ) } \\ { \times ( \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 } ) } \end{array} \right.$$

Evaluate

$$\frac{631}{2400}\approx 0.262916667$$

Solution Steps

Convert 1 to fraction \frac{2}{2}.

$$\left(\frac{2}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{2}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\left(\frac{2+1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 2 and 1 to get 3.

$$\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 2 and 3 is 6. Convert \frac{3}{2} and \frac{1}{3} to fractions with denominator 6.

$$\left(\frac{9}{6}+\frac{2}{6}+\frac{1}{4}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{9}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\left(\frac{9+2}{6}+\frac{1}{4}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 9 and 2 to get 11.

$$\left(\frac{11}{6}+\frac{1}{4}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 6 and 4 is 12. Convert \frac{11}{6} and \frac{1}{4} to fractions with denominator 12.

$$\left(\frac{22}{12}+\frac{3}{12}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{22}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$\left(\frac{22+3}{12}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 22 and 3 to get 25.

$$\left(\frac{25}{12}+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 12 and 5 is 60. Convert \frac{25}{12} and \frac{1}{5} to fractions with denominator 60.

$$\left(\frac{125}{60}+\frac{12}{60}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{125}{60} and \frac{12}{60} have the same denominator, add them by adding their numerators.

$$\frac{125+12}{60}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 125 and 12 to get 137.

$$\frac{137}{60}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 2 and 3 is 6. Convert \frac{1}{2} and \frac{1}{3} to fractions with denominator 6.

$$\frac{137}{60}\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{3}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{137}{60}\left(\frac{3+2}{6}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 3 and 2 to get 5.

$$\frac{137}{60}\left(\frac{5}{6}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 6 and 4 is 12. Convert \frac{5}{6} and \frac{1}{4} to fractions with denominator 12.

$$\frac{137}{60}\left(\frac{10}{12}+\frac{3}{12}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{10}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$\frac{137}{60}\left(\frac{10+3}{12}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 10 and 3 to get 13.

$$\frac{137}{60}\left(\frac{13}{12}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 12 and 5 is 60. Convert \frac{13}{12} and \frac{1}{5} to fractions with denominator 60.

$$\frac{137}{60}\left(\frac{65}{60}+\frac{12}{60}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{65}{60} and \frac{12}{60} have the same denominator, add them by adding their numerators.

$$\frac{137}{60}\left(\frac{65+12}{60}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 65 and 12 to get 77.

$$\frac{137}{60}\left(\frac{77}{60}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 60 and 6 is 60. Convert \frac{77}{60} and \frac{1}{6} to fractions with denominator 60.

$$\frac{137}{60}\left(\frac{77}{60}+\frac{10}{60}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{77}{60} and \frac{10}{60} have the same denominator, add them by adding their numerators.

$$\frac{137}{60}\times \frac{77+10}{60}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 77 and 10 to get 87.

$$\frac{137}{60}\times \frac{87}{60}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Reduce the fraction \frac{87}{60} to lowest terms by extracting and canceling out 3.

$$\frac{137}{60}\times \frac{29}{20}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Multiply \frac{137}{60} times \frac{29}{20} by multiplying numerator times numerator and denominator times denominator.

$$\frac{137\times 29}{60\times 20}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Do the multiplications in the fraction \frac{137\times 29}{60\times 20}.

$$\frac{3973}{1200}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Convert 1 to fraction \frac{2}{2}.

$$\frac{3973}{1200}-\left(\frac{2}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{2}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\left(\frac{2+1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 2 and 1 to get 3.

$$\frac{3973}{1200}-\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 2 and 3 is 6. Convert \frac{3}{2} and \frac{1}{3} to fractions with denominator 6.

$$\frac{3973}{1200}-\left(\frac{9}{6}+\frac{2}{6}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{9}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\left(\frac{9+2}{6}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 9 and 2 to get 11.

$$\frac{3973}{1200}-\left(\frac{11}{6}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 6 and 4 is 12. Convert \frac{11}{6} and \frac{1}{4} to fractions with denominator 12.

$$\frac{3973}{1200}-\left(\frac{22}{12}+\frac{3}{12}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{22}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\left(\frac{22+3}{12}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 22 and 3 to get 25.

$$\frac{3973}{1200}-\left(\frac{25}{12}+\frac{1}{8}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 12 and 8 is 24. Convert \frac{25}{12} and \frac{1}{8} to fractions with denominator 24.

$$\frac{3973}{1200}-\left(\frac{50}{24}+\frac{3}{24}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{50}{24} and \frac{3}{24} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\left(\frac{50+3}{24}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 50 and 3 to get 53.

$$\frac{3973}{1200}-\left(\frac{53}{24}+\frac{1}{6}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 24 and 6 is 24. Convert \frac{53}{24} and \frac{1}{6} to fractions with denominator 24.

$$\frac{3973}{1200}-\left(\frac{53}{24}+\frac{4}{24}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{53}{24} and \frac{4}{24} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\frac{53+4}{24}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 53 and 4 to get 57.

$$\frac{3973}{1200}-\frac{57}{24}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Reduce the fraction \frac{57}{24} to lowest terms by extracting and canceling out 3.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 2 and 3 is 6. Convert \frac{1}{2} and \frac{1}{3} to fractions with denominator 6.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{4}+\frac{1}{5}\right)$$

Since \frac{3}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{3+2}{6}+\frac{1}{4}+\frac{1}{5}\right)$$

Add 3 and 2 to get 5.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{5}{6}+\frac{1}{4}+\frac{1}{5}\right)$$

Least common multiple of 6 and 4 is 12. Convert \frac{5}{6} and \frac{1}{4} to fractions with denominator 12.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{10}{12}+\frac{3}{12}+\frac{1}{5}\right)$$

Since \frac{10}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{10+3}{12}+\frac{1}{5}\right)$$

Add 10 and 3 to get 13.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{13}{12}+\frac{1}{5}\right)$$

Least common multiple of 12 and 5 is 60. Convert \frac{13}{12} and \frac{1}{5} to fractions with denominator 60.

$$\frac{3973}{1200}-\frac{19}{8}\left(\frac{65}{60}+\frac{12}{60}\right)$$

Since \frac{65}{60} and \frac{12}{60} have the same denominator, add them by adding their numerators.

$$\frac{3973}{1200}-\frac{19}{8}\times \frac{65+12}{60}$$

Add 65 and 12 to get 77.

$$\frac{3973}{1200}-\frac{19}{8}\times \frac{77}{60}$$

Multiply \frac{19}{8} times \frac{77}{60} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3973}{1200}-\frac{19\times 77}{8\times 60}$$

Do the multiplications in the fraction \frac{19\times 77}{8\times 60}.

$$\frac{3973}{1200}-\frac{1463}{480}$$

Least common multiple of 1200 and 480 is 2400. Convert \frac{3973}{1200} and \frac{1463}{480} to fractions with denominator 2400.

$$\frac{7946}{2400}-\frac{7315}{2400}$$

Since \frac{7946}{2400} and \frac{7315}{2400} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7946-7315}{2400}$$

Subtract 7315 from 7946 to get 631.

$$\frac{631}{2400}$$

Factor

$$\frac{631}{2 ^ {5} \cdot 3 \cdot 5 ^ {2}} = 0.2629166666666667$$