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Algebra Difficulty 2.7 Junior Find the answer

The expression (1+12)(1+13)(1+14)(1+15)(1+16)(1+17)(1+18)(1+19)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{6}\right)\left(1+\frac{1}{7}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{9}\right) is equal to what?

A number or a short expression. Spacing and $ signs are ignored.

Solution

The expression is equal to (32)(43)(54)(65)(76)(87)(98)(109)\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\left(\frac{5}{4}\right)\left(\frac{6}{5}\right)\left(\frac{7}{6}\right)\left(\frac{8}{7}\right)\left(\frac{9}{8}\right)\left(\frac{10}{9}\right) which equals 3⋅4⋅5⋅6⋅7⋅8⋅9⋅102⋅3⋅4⋅5⋅6⋅7⋅8⋅9\frac{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}. Removing common factors from the numerator and denominator, we obtain 102\frac{10}{2} or 5.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.