A-1 To Be Proper This Year

Algebra Level 4

What is the sum of all proper fractions with denominator less than 2016?

The answer is 1014552.5.

2 solutions

Brian Dela Torre
Feb 1, 2016

Let's start at this one, d < 2016

In this case, we have 2015, 2014, 2013, ... , 4, 3, 2 as denominator (1 is not included since we are talking about proper fraction.

Let's find first the sum of all proper fraction with denominator 2015, 2014, 2013 respectively, then

$\frac{1}{2015}$ + $\frac{2}{2015}$ + $\frac{3}{2015}$ + ... + $\frac{2013}{2015}$ + $\frac{2014}{2015}$ = $\frac{\frac{(2014)(2015)}{2}}{2015}$ = 1007

$\frac{1}{2014}$ + $\frac{2}{2014}$ + $\frac{3}{2014}$ + ... + $\frac{2012}{2014}$ + $\frac{2013}{2014}$ = $\frac{\frac{(2013)(2014)}{2}}{2014}$ = 1006.5

$\frac{1}{2013}$ + $\frac{2}{2013}$ + $\frac{3}{2013}$ + ... + $\frac{2011}{2013}$ + $\frac{2012}{2013}$ = $\frac{\frac{(2012)(2013)}{2}}{2013}$ = 1006

Notice that the sum obtained of each common difference of - 0.5 . Using Arithmetic Sum of $S_{n}$ = $\frac{n}{2}$ ( $a_{1}$ + $a_{n}$ ) with last term 0.5 ( sum of proper fraction with denominator 2) and n = 2014 (number of terms)

Then we have $S_{2014}$ = $\frac{2014}{2}$ ( 0.5 + 1007 ) = (1007)( $\frac{2015}{2}$ ) = 1014552.5

Hence the sum of all proper fraction less than 2016 is $\boxed{1014552.5}$

It's possible to solve this problem without assuming that because the first few terms are in AP the entire series must be in AP. Here is such a solution: $\sum _{ n=2 }^{ 2015 }{ \frac { \sum _{ k=1 }^{ n-1 }{ k } }{ n } } =\sum _{ n=2 }^{ 2015 }{ \frac { \frac { (n-1)(n) }{ 2 } }{ n } } =\sum _{ n=2 }^{ 2015 }{ \frac { n-1 }{ 2 } } =\sum _{ n=1 }^{ 2014 }{ \frac { n }{ 2 } } =\frac { 1 }{ 2 } \sum _{ n=1 }^{ 2014 }{ n } =(\frac { 1 }{ 2 } )(\frac { (2014)(2015) }{ 2 } )\\ =1,014,552.5$

Chris Callahan - 5 years, 4 months ago

But you are counting many, many fractions twice! For instance, 1007/2014=1/2.

I believe the answer to this question must be much smaller when you take into account the fact that there are repetitions. It can be done using the sum of the Euler totient function, along these lines.

from fractions import Fraction
from sympy.ntheory import totient

maxD=2016

sumtotient=0
for n in range(1,maxD):
    sumtotient+=totient(n)
print (n,1.*(Fraction(sumtotient-1,2)))

Result is 617431.5.

Bill Bell - 5 years, 4 months ago

I too first obtained this result, but, according to encyclopedia britannica, the only requirement for a fraction to be a proper fraction is that "the numerator is less than the denominator". It need not be in reduced form and thus 1007/2014 is a proper fraction with denominator equal to 2014 and 1/2 is a proper fraction with denominator equal to 2. While they are the same proper fraction, both must be counted within their subset of proper fractions with the same denominator.

Chris Callahan - 5 years, 4 months ago

Thanks Sir Chris for your comments. This proves that my question and solution is correctly stated.

Brian Dela Torre - 5 years, 4 months ago

Thank you. I was thinking of this more or less overnight. Somehow I knew I must be wrong because the question has a '4' rating. When I was doing the calculation I was hoping I would find that someone had a much cleverer way of obtaining the result than I had.

Bill Bell - 5 years, 4 months ago

Aditya Dhawan
Feb 19, 2016

$The\quad number\quad of\quad proper\quad fractions\quad with\quad denominator\quad k=\frac { 1 }{ k } \left( 1+2+3...(k-1) \right) =\frac { 1 }{ k } \sum _{ n=1 }^{ k-1 }{ n } =\frac { k-1 }{ 2 } \\ Thus\quad we\quad need\quad to\quad compute\quad the\quad following:\\ \sum _{ k=2 }^{ 2015 }{ \frac { k-1 }{ 2 } } =\quad \frac { 1 }{ 2 } \left( \sum _{ k=1 }^{ 2015 }{ k } -\sum _{ k=1 }^{ 2015 }{ 1 } \right) =\frac { 1 }{ 2 } \left( 2015\times 1008-2015 \right) =\quad \frac { 2015\times 1007 }{ 2 } =\quad \boxed { 1014552.5 }$

A-1 To Be Proper This Year

The answer is 1014552.5.

2 solutions

0 pending reports