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Bunuel
If \(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} = \frac{m}{3*5*7*9*11}\), what is the remainder when m is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

We can take the LCM of the fractions on the RHS. The numerator of each term (\(\frac{1}{3}, \frac{1}{7} \) and \(\frac{1}{9}\)) will have 5 in the numerator. The only term that will not have 5 in its numerator is 1/5.

The denominators of the RHS and LHS will cancel out.

The numerator of 1/5 will be 7 * 3 * 9 * 11

The remainder of m when divided by 5 is the same as the remainder when (7*3*9*11) is divided by 5.

We can use the multiplicative property -

Remainder of (\(\frac{7*3*9*11}{5}\)) = Rem(\(\frac{7}{5}\)) * Rem(\(\frac{3}{5}\)) * Rem(\(\frac{9}{5}\)) * Rem(\(\frac{11}{5}\))

= 2 * 3 * 4 * 1

= Rem(\(\frac{24}{5}\)) = 4

Option E
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Multiply both sides by 3x5x7x9x11.

It will result m = 5k + 3x7x9x11

Just find the remainder for 3x7x9x11 when divided by 5 using remainder theorem, it will be 4.

Alternatively 3x7x9x11 will have it's units digit as 9.
Thus 9/5 will leave remainder as 4.

Hence E is the correct answer.

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\(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} = \frac{m}{3*5*7*9*11}\)

Multiplying with 3*5*7*9*11 on both the sides we get

\(\frac{ 3*5*7*9*11 }{3} + \frac{ 3*5*7*9*11 }{5} + \frac{ 3*5*7*9*11 }{7} + \frac{ 3*5*7*9*11 }{9} + \frac{ 3*5*7*9*11 }{11} = \frac{ 3*5*7*9*11 * m}{3*5*7*9*11}\)

=> Integer Multiple of 5(i.e 5*7*9*11) + 3*7*9*11 + Integer Multiple of 5(i.e 3*5*9*11) + Integer Multiple of 5(i.e 3*5*7*11) + Integer Multiple of 5(i.e 3*5*7*9) = m
Remainder of m by 5 = 0 + Remainder of 3*7*9*11 by 5 + 0 + 0 + 0 = Remainder of 3*7*9*11 by 5

Remainder of 3*7*9*11 by 5 = Remainder of units' digit of 3*7*9*11 by 5 = Remainder of 1 by 5 = 1

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Remainders

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