# Task
Given two integers `a` and `b`, return the sum of the numerator and the denominator of the reduced fraction `a/b`.

# Example

For `a = 2, b = 4`, the result should be `3`

Since `2/4 = 1/2 => 1 + 2 = 3`.

For `a = 10, b = 100`, the result should be `11`

Since `10/100 = 1/10 => 1 + 10 = 11`.

For `a = 5, b = 5`, the result should be `2`

Since `5/5 = 1/1 => 1 + 1 = 2`.

# Input/Output


- `[input]` integer `a`

The numerator, `1 ≤ a ≤ 2000`.


- `[input]` integer `b`

The denominator, `1 ≤ b ≤ 2000`.


- `[output]` an integer

The sum of the numerator and the denominator of the reduces fraction.

Difficulty: introductory

Starter code:

def fraction(a, b):
	

Solutions:
from fractions import gcd

def fraction(a,b):
    return (a+b)//gcd(a,b)

from math import gcd
def fraction(a, b):
    return a/gcd(a,b) + b/(gcd(a,b))


def fraction(a, b):
    from math import gcd
    return (a + b) / gcd(a, b)

from fractions import gcd

def fraction(a, b):
    d = gcd(a, b)
    return a//d + b//d

from fractions import Fraction
def fraction(a, b):
  #coding and coding..
  f = Fraction(a,b)
  return f.numerator + f.denominator

from fractions import Fraction
def fraction(a, b):
    return Fraction(a,b).denominator+Fraction(a,b).numerator

import fractions
def fraction(a, b):
    return sum([int(s) for s in str(fractions.Fraction(a,b)).split('/')])

def fraction(a, b):
  #coding and coding..
  c=a+b
  if b>a:
      a,b=b,a
  while b!=0:
      a,b=b,a%b
  return int(c/a)

def fraction(a, b):
    # check if one arguments can serve as the denominator
    if max([a, b]) % min([a, b]) == 0:
        return (a + b) // min([a, b])
    # otherwise start from the half of the smallest argument and search it
    else:
        x = min([a, b]) // 2
        while True:
            if a % x == 0 and b % x == 0:
                break
            x -= 1
        return (a + b) // x

import math

def fraction(a, b):
    g = math.gcd(a, b)
    return a / g + b / g