Question

Let N be the sum of all positive irreducible fractions less than 1 whose denominator is 2021. Find the largest prime number p that divides N.

Answer 1

Prime factors of 2021 are 43 and 47 only. So the irreducible fractions never include the fractions having multiples of 43 or 47 as numerator.

We need to reduce the sum of such fractions, from the sum of all positive fractions below 1, in order to find the sum of the irreducible fractions. Note that the denominator is 2021 in each case.

Then the formula is,

→   S(irreducible fractions) = S(all fractions) - S(multiples of 43 or 47)   ...(1)

Here,

•   S(all fractions) = 1/2021 + 2/2021 + 3/2021 +...+ 2020/2021

•   S(all fractions) = (1 + 2 + 3 +...+ 2020)/2021

•   S(all fractions) = (2020 × 2021 / 2)/2021 = 2020/2 = 1010

and,

•   S(multiples of 43 or 47) = S(multiples of 43) + S(multiples of 47) - S(multiples of 43 and 47)   ...(2)

This formula is found according to the set formula n(A∪B) = n(A) + n(B) - n(A∩B)

Since 2021 = 43 × 47, there are 46 multiples of 43 and 42 multiples of 47 below 2021. So,

• S(multiples of 43) = 43/2021 + 43×2/2021 + 43×3/2021 +...+ 43×46/2021

• S(multiples of 43) = 43(1 + 2 + 3 +...+ 46)/2021

• S(multiples of 43) = 43(46 × 47 / 2)/2021 = 23

and,

• S(multiples of 47) = 47/2021 + 47×2/2021 + 47×3/2021 +...+ 47×42/2021

• S(multiples of 47) = 47(1 + 2 + 3 +...+ 42)/2021

• S(multiples of 47) = 47(42 × 43 / 2)/2021 = 21

and,

• S(multiples of 43 and 47) = S(multiples of 2021) = 0

because there are no multiples of 2021 below 2021.

Then (2) becomes,

•   S(multiples of 43 or 47) = 23 + 21 - 0 = 44

Now (1) becomes,

→   S(irreducible fractions) = 1010 - 44 = 966.

That is, N = 966.

Prime factors of 966 are 2, 3, 7 and 23. The largest prime number that divides 966 is 23. Hence p = 23.