How to deal with the reciprocal?

Number Theory Level 3

Denote d 1 , d 2 , d 3 , … , d n d_1, d_2, d_3, \ldots , d_n as the distinct positive integral divisors of the number 10500 10500 .

What is the range of the sum: S = ∑ k = 1 n 1 d k \displaystyle S = \sum_{k=1}^{n} \frac 1 {d_k} ?

Bonus : Find the exact value of S S .

4 < S ≤ 5 4 < S \leq 5 S > 5 S> 5 0 < S ≤ 3 0 <S \leq 3 3 < S ≤ 4 3 < S \leq 4

View wiki
Mayank Singh by Mayank Singh , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Phoebe Liu
Nov 28, 2015

The dumb way to do it: The prime factorization of 10500 is (2^2)(3)(5^3)(7). (The number of positive integral divisors of 10500 is the product of one more than each exponent in the prime factorization: (3)(2)(4)(2)=48). Made a list of all the possible factors of 10500 that are significant in size (i.e. not 1/10500) (combinations from the PF), took the reciprocal of each one, and added. :)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

That's very tedious and time consuming. Just do this: (1+1/2+1/4+1/8)(1+1/3)(1+1/5+1/25+1/125)(1+1/7). You can Calculate each bracket using GP.

Kushagra Sahni - 5 years, 2 months ago
Vaibhav Bhatt
Dec 31, 2017

I don't know if I'm correct but I did it by considering the sum of divisors of 1 10500 \frac{1}{10500} - ∑ k = 1 n \displaystyle \sum_{k=1}^n 1 d k \frac{1}{d_{k}} = [ 1 + 1 2 + 1 2 2 ] [ 1 + 1 3 ] [ 1 + 1 5 + 1 5 2 + 1 5 3 ] [ 1 + 1 7 ] = 3.328 = [1+\frac{1}{2}+\frac{1}{2^{2}}][1+\frac{1}{3}][1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}][1+\frac{1}{7}]=\boxed{3.328}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...