HCF/GCD of Two numbers Using Python

❤❤

Sir last print ki bad ( ) me f ka ya Matlab hai

thank you and sorry both
I will say the reason for sorry if you reply me

Aap achhe se Nahi samjha pate. Ho. Sirf bol dete ho Ki . I hope aap samjh gye . But students Nahi samjh pata h

Anyone please help me out
I think that for loop must be run in reverse order in order to get highest value otherwise code would just return I=1 because ultimately that's the first value which satisfies if condition
So I think code she be like this
def hcf(a, b):

smaller = min(a, b)
hcf = 1
for i in range(smaller,0,-1):
if a%i == 0 and b%i == 0:
hcf = i
break
return hcf

a=int(input("1st no.\n"))
b=int(input("2nd no.\n"))
print(hcf(a,b))

Last line
print(f"The gcd of these two number is {hcf}")
Why do u use' f ' key

def hcf(a, b):
highest_common_factor = 1
for numbers in range(2, max(a, b)):
if a % numbers == 0 and b % numbers == 0:
highest_common_factor = numbers
return highest_common_factor

Thank you sir 😎😍💝😍

What a lovely explanation

Thank you so much bhai

I have made my own code to find the HCF/GCD using sets in python.

thanks

import math

math.gcd(40,12)

output--4

this is right for gcd or not and why we are not using this , what is difference ?

# Here is a simpler way:
a = int(input("Enter first:"))
b = int(input("Enter second:"))

while a % b != 0:
remainder = a % b
a = b
b = remainder

print("HCF is:", b)

I can get LCM numbers from input using python

harry bhai thanku

thank you very helpful

this code will shoe error if we takes input (0,b) here b is any number :
def gcd(self, A, B):
if A > B:
smaller = B
bigger=A
else:
smaller = A
bigger=B
if smaller==0:
return bigger
for i in range(1, smaller+1):


if((A % i == 0) and (B % i == 0)):
k=i

return k

super explanation bro ,maza aagaya

My program:
a = int(input("Enter n: " ))
l1 = [ ]
for i in range(1,a+1):
b = int(input("Enter the number: "))
l1.append(b)
c = int(min(l1))
# HCF
for i in range(1, c+1):
for numbers in l1:
if numbers%i == 0:
HCF = i

print("The HCF is " , HCF)

def GCD(a,b):
minNum = min(a,b)
maxNum = max(a,b)
rem = 1
while True:
rem = maxNum % minNum
if(rem == 0):
break
maxNum = minNum
minNum = rem

return minNum


a = int(input("Enter first number : "))
b = int(input("Enter second number : "))

result = GCD(a,b)
print(result)



# we can do like that ...!!

Bro I have another approach, instead of looping from 1 till min number and updating hcf every time, we can loop from the min of two numbers till 1 and if we find a factor for both, we assign it as hcf and break the loop. I guess this approach might reduce the num of iterations.

# this is program to get gcd of two numbers
# author ♠♠kartik♠♠
lst = [1, ]
def gcd(num1 , num2):

if num2 > num1:
upperbound = num1
else:
upperbound = num2

for i in range(2 , upperbound+1):
if num1%i == 0 and num2%i == 0:
lst.append(i)
else:
pass
if _name_ == '__main__':

num1 = int(input("Enter the number1\n"))
num2 = int(input("Enter the number1\n"))
gcd(num1,num2)
print(f"Commom factor of '{num1}' and '{num2}' is {lst}")
greatest_gcd = lst[-1]
print(f"The GCD of the number '{num1}' and '{num2}' is '{greatest_gcd}'")
###########################☻R##################
import math
gcd_number = math.gcd(num1 , num2)
print(f"GCD of the number is {gcd_number}")

Please also teach block based coding

Great one vro!=

Very very useful 👍👍👍🎉🎉 Thanks

Sir your method is good but I have better method with time complexity
num1=int(input("Enter firat number\n"))
num2=int(input("Wnter second number\n"))

if num1>num2:
mn=num2

else:
mn=num1

for i in range(1,int((mn/2)+1)):
if num1%i==0 and num2%i==0:
hcf=i

if num2%num1==0:
hcf=num1

print(f"Your HCF is {hcf}")

Thanks sir

# Find HCF Function
def HCF(n1,n2):
min = n1
while (True):
if (n1 % min == 0 and n2 % min == 0):
return min
min -= 1

# Main function
if __name__=="__main__":
n1 = int(input("Enter The First Number : "))
n2 = int(input("Enter The Second Number : "))
n1, n2 = min(n1, n2), max(n1, n2)
hcf = HCF(n1, n2)
print(f"{n1} & {n2} HCF is {hcf}")

# Abhishek Sharma

VERY INTERSTING SIR

n1 = int(input("Enter the first number: "))
n2 = int(input("Enter the second number: "))
num_list = [n1,n2]
maximum_number = max(num_list)
# print(maximum_number)
num = 0
list_n1_from_loop = []
list_n2_from_loop = []
for i in range(maximum_number):
num = num + 1
if n1 % num == 0:
list_n1_from_loop.insert(0,num)
max_list_n1 = max(list_n1_from_loop)
if n2 % num == 0:
list_n2_from_loop.insert(0,num)
max_list_n2 = max(list_n2_from_loop)

convertset_n1 = set(list_n1_from_loop)
convertset_n2 = set(list_n2_from_loop)
intersection = convertset_n1.intersection(convertset_n2)
# print(sorted(intersection))
convert_intersection_list = list(intersection)
max_intersection = max(convert_intersection_list)
print(f"The maximum and similar dividable number of {n1} and {n2} with remainder 0 is: {max_intersection}\nTHANK YOU!!!!!!!!!")

2 se zyada numbers ka kya karoge?

Instead of updating the maximum value in for loop...we can simply run a loop in reverse direction from smaller number to 1.... correct?

sir i tried it myself and i came up with a different logic . Please check if I m correct?
a = int(input("Enter the first number : "))
b = int(input("Enter the second number : "))
hcf=1
i=min(a,b)
while i>1:
if a%i==0 and b%i ==0:
hcf=i
break
i=i-1

print(f"The HCF of {a} & {b} is : {hcf} ")
I have learnt Python from your channel only!!
you're the BEST !!

Available with 3.9 and dict union ... beautiful feature, faster than previous

guys u dont need to to take out the max number. you can just range from 1 to any of the two number and it will still work. example
for i in range(1, num1):
if num1%i==0 and num2%i==0:
hcf=i;

Bro we r nt getting it properly because we r nt able to see this lecture from outside home because of lecture is on dark display so plzz attention in next lecture

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Plz like everyone

Harry bhai GCD ko Euclid division wale method se krke samjha do bhai.. Bahut problem aa rahi hai smjhne me

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