If \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}....\frac{1}{29} =\frac{N}{29!}\) Find the remainder when N is divided by 19.
We have the sum: \[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \ldots + \frac{1}{29} = \frac{N}{29!}.\] First, let's rewrite the given expression using a common denominator. The common denominator of the fractions from 1 to 29 is \(29!\), so: \[\frac{29!}{29!} + \frac{14 \cdot 29!}{2 \cdot 29!} + \frac{29!}{3 \cdot 29!} + \frac{29!}{4 \cdot 29!} + \frac{29!}{5 \cdot 29!} + \frac{29!}{6 \cdot 29!} + \frac{29!}{7 \cdot 29!} + \frac{29!}{8 \cdot 29!} + \frac{29!}{9 \cdot 29!} + \ldots + \frac{29!}{29 \cdot 29!} = \frac{N}{29!}\]. Simplifying each fraction: \[1 + 14 + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \ldots + \frac{1}{29} = \frac{N}{29!}.\] Now, let's find the sum of the fractions: \[\frac{N}{29!} = 15 + \left(\frac{1}{3} + \frac{1}{6}\right) + \left(\frac{1}{4} + \frac{1}{8}\right) + \left(\frac{1}{5} + \frac{1}{10}\right) + \ldots + \left(\frac{1}{29} + \frac{1}{58}\right).\] Notice that we have pairs of fractions within parentheses, where the second fraction is half of the first fraction. This simplifies to: \[\frac{N}{29!} = 15 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots + \frac{1}{2}.\] There are a total of 28 fractions in this form. So: \[\frac{N}{29!} = 15 + 28 \cdot \frac{1}{2} = 15 + 14 = 29.\] Now, to find the remainder when \(N\) is divided by 19, we will calculate \(29 \mod 19\): \[29 \mod 19 = 10.\] So, the remainder when \(N\) is divided by 19 is \(10\). Therefore, the correct option is(B): \(10\).
