Inverse sums

Algebra Level 3

Consider the following sum of finitely many fractions,

S = 1 a \frac{1}{a} + 1 b \frac{1}{b} + 1 c \frac{1}{c} + 1 d \frac{1}{d} + 1 e \frac{1}{e} + 1 f \frac{1}{f} + . . . . . . . . . 1 k \frac{1}{k}

Here, a , b , c , d , e , f ,......, k are some positive integers with no prime divisor greater than 7.

If M N \frac{M}{N} is the smallest real number which is always greater than S, then find M + N.


The answer is 43.

Avyukta Manjunatha Vummintala by Avyukta Manjunatha Vummi… , 19

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1 solution

When you think long enough, you realize that M N \frac{M}{N} is nothing but ( multiply directly to convince yourself.)

( 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 2 2 \frac{1}{2^2} + 1 2 3 \frac{1}{2^3} + .........)( 1 1 \frac{1}{1} + 1 3 \frac{1}{3} + 1 3 2 \frac{1}{3^2} + 1 3 3 \frac{1}{3^3} + .........)( 1 1 \frac{1}{1} + 1 5 \frac{1}{5} + 1 5 2 \frac{1}{5^2} + 1 5 3 \frac{1}{5^3} + .........)( 1 1 \frac{1}{1} + 1 7 \frac{1}{7} + 1 7 2 \frac{1}{7^2} + 1 7 3 \frac{1}{7^3} + .........) = 2 1 \frac{2}{1} X 3 2 \frac{3}{2} X 5 4 \frac{5}{4} X 7 6 \frac{7}{6}

= 35 8 \frac{35}{8} .

35 + 8 = 43

By the way, the above problem is inspired by a problem from AOPS.

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