Python Armstrong Number Program - In Hindi

[@nirojsubedi]

you are absolutely amazing yarr.. i know you have done lot of practice to teach and hope you will be one of the great programmer in coming days ....i respect you cause you still make video to share information even you have less viewers....thanks

[@codeitup]

Thanks a ton🙂

[@nirojsubedi]

@@codeitup lots of love from Nepal.......

[@codeitup]

@@nirojsubedi Thanks you so much!

[@leon.s.kennedy5198]

He is nice teacher.

[@arpitapandey8645]

Thank you sir❤

[@codeitup]

Most welcome!

[@vineetsingh7547]

Thanks 😊 Sir 👍🙏

[@codeitup]

Most welcome!

[@tanveerbibi6474]

Very nice video hai❤

[@codeitup]

Thank you so much !

[@oxygen3654]

Pranam Gurudev, Aapki sabhi khwaishein poori ho. Aap mahan hai!!

[@codeitup]

🙂🙏

[@rohitgarg4446]

I did like this !! n = int(input("Enter your numbers of even No. = ")) i = n count = 0 while (i>0): i = i // 10 count = count + 1 i = n sum = 0 while (i > 0): sum = sum + (i % 10)**count i = i // 10 if(sum==n): print("Armstrong Number") else: print("Not Armstrong Number")

[@mansirao5768]

easier ig

[@mansirao5768]

what's the use of count ? when your not printing

[@Tripti_Foundation]

actually your explanation with dry run is so much adorable.

[@codeitup]

Glad you think so!

[@eduvista9644]

sir aapka padhane ka tareeka jabardast hai, maza aagya

[@codeitup]

🙂

[@rounakverma8115]

Thanks sir

[@codeitup]

Welcome Rounak!

[@rosy9003]

Ur teaching way is super awesome... U made this program easiest for me.... Thanks sir 🙏🏻

[@codeitup]

Glad to hear that!

[@debarghyaadhikari5558]

NICE SIR

[@codeitup]

Thanks dear😊

[@motasembillatanjid8483]

Thanks you sir!! You are great

[@codeitup]

Thanks & most welcome dear😊

[@SulemanTheTraveller]

Sir ap ka style hi amazing hy. thumbs up. by the way armstrong number find krnay k liye mujay last video wala method easy lga hy

[@codeitup]

Thanks Malik!

[@iitianvibes_1]

mindblowing series

[@codeitup]

Thanks Ashish!

[@opgaming-jh4hu]

sir your teching style is best also explanation thank u sir u are the best teacher

[@codeitup]

Thanks & most welcome Hitesh!

[@paristadebbarma6359]

I'm just starting loops and your video helps me a lot. thanks

[@codeitup]

Glad to help you😊

[@shriyawayangankar393]

n=int(input("Enter the number: ")) def arm(n): i=n count=0 while (i>0): digit=i%10 count+ =digit**3 i//=10 if n==count: print (n, "Armstrong") else: print(n, "Not armstrong") arm(n)

[@AmanBaba-v3e]

Bhout easy h

[@Anitarawat03]

Century comment🎉

[@chandranathrumabhowmick8752]

Please do a total video on threads your explanation is high that influenced me so much

[@mansirao5768]

i'm not getting this program , it is lengthier so i would prefer the previous way. although thanks for the explanation sir..

[@kanduladivya5804]

very good explanation sir,thanku verymuch.

[@codeitup]

You're most welcome !

[@Juhi1910]

Thank you sir

[@codeitup]

Most welcome!

[@aneelaamin2235]

very good sir, this video is very helpful for me, thanks,

[@codeitup]

Glad to help you Aneela!

[@praneetkumarjha215]

Perfect content perfect teacher perfect teaching style ... On the way to python

[@codeitup]

Thanks Praneet😊

[@Tripti_Foundation]

n = int(input('Enter a number:=')) order = len(str(n)) org = n sum = 0 while n>0: sum = sum + (n%10)**order n=n//10 if org == sum: print("Number is Armstrong") else: print("Number is not Armstrong") This is the alternative to fird Armstrong number. if set the range between 1 to 500 or more use for loop.

[@fenrir834]

hands down this is the best channel to learn python for Hindi-speaking beginners.

[@codeitup]

So nice of you, God bless you!

[@harshasri3711]

Excellent teaching but please teach in English also.....

[@manav.2907]

#Sir we can use this also n=int(input("enter the number:")) i=n sum=0 x=len(str(n)) while(n>0): sum=sum+(n%10)**x n=n//10 if i==sum: print("yes the number is an armstrong") else: print("no the number is not an armstrong")

[@AksaMWani]

Good one!!

[@mayurlavangare9432]

nice one

[@sanchitapadole3512]

Sir exercise bhi diya kro..

[@SurajKumar-jw7jo]

Sir isse previous question me aap ne is online board (digitally) samjaya tha . Aur oski programming thodi alag thi aur isme thodha alag he . So dono question me difference kya he .

[@codeitup]

This is for any number of digits. The previous was only for three digits number.

[@dipendrasingh1586]

Sir,video banate rahna👍🏻👍🏻👍🏻

[@fenrir834]

just a suggestion to do this differently. first make a loop to count the number of digits, let the no.' of digits be n then write a seperate loop to take the sum of the digits^n.

[@youthoob3573]

its okay , everyone has their own method, calm down

[@fenrir834]

@@youthoob3573 Sure mate, i have been loosing my shit since two years as you can clearly tell from my comment two years ago, I am calm now since you asked me to

[@4k45hxd9]

Sir iske phle wala mathod bhi sahi h na ?

[@codeitup]

Wo sirf 3 digit numbers ke lie tha, ye any number of digit ke lie hai.

[@4k45hxd9]

@@codeitup okk sir thanks

[@AHMAD-yd2ib]

I HAVE A HUNDRED TIMES SIMPLER VERSION OF CODE : 🔥 (Using just 2 while loops and a pair of if-else statements) n = int(input('enter a number: ')) z = n k = n rev = 0 count = 0 while k>0: count += 1 k = k//10 while n>0: rev = rev + (n%10)**count n = n//10 if rev == z: print('The number is Armstrong') else: print('The number is not Armstrong')

[@Ansh_Rajpoot_]

sir , can we use this code >is it correct>>>>>>>>>>>>> a = int(input('ener a num')) i = a sum = 0 len = len(str(a)) while i>0: sum = sum + (i%10)** len i = i//10 if a==sum: print(' The num is armtrong') else : print('not armstrong') its working fine like yours.

[@codeitup]

If it is working fine then perfect Ansh. A single program may have different logic to solve.

[@AmanBaba-v3e]

Sir ye BCA valo ke according h na

[@riyajoshi8229]

Sir I use the same program with same logics in pycharm, but i am unable to get the desired answers though indentation and other things are good in place.

[@ashoknandurka7204]

What's the difference between previous one??

[@skybird7274]

i=int(input("enter the number")) o=i count=0 while (i>0): i=i//10 count+=1 i=o sum1=0 while (i>0): sum1=sum1+(i%10)**count i=i//10 if sum1==o: print("YES") else: print("NO")

[@harshkumar-rh8kk]

bhai sahab agar ese coding sikhoge toh jo ata hoga vo bhi bhul jaynge ye itna complicated nahi hai jitna apne bana diya

[@aryasharma5571]

sir isse phle bhi ek video hai armstrong ki....usme count variable ke bina karvaya hai apne. which one is more correct?🙂😃

[@codeitup]

Last video is for 3 digit only. This video is for any number of digit. SO, this program is better.

[@aryasharma5571]

@@codeitup thanx sir❤️❤️..

[@lovelyaggarwal9828]

Sir m programming 1st tym kr rhi hu..in formula ko learn krna padta h kya..?

[@codeitup]

Yes Lovely, kuch formula aapko yaad karne hoge.

[@rameshchandsharma6015]

Sir video banate rahie

[@codeitup]

Sure Ramesh !

[@meow-rd6op]

💥💥

[@codeitup]

🙂

[@MrKP-nm3qy]

Sir amstrong wali ek or vdo h jisme ek particular cube k condition p h kya wo bhi shi h kya

[@codeitup]

Wo sirf 3 digit number ke lie hai...

[@sunnyyadav4664]

sir regular expression pe series bnado please

[@codeitup]

Sure !

[@abhinavkislay9442]

Sir can u please make playlist on oops also

[@codeitup]

Sure but it will take some time !

Sir I have an extra method to code this, pls let me know whether I am wrong or right? and mine one is using string function's feature. n = int(input("Enter a number: ")) sum = 0 i = n a = len(str(n)) while i > 0: digit = i % 10 sum += digit ** a i //= 10 if n == sum: print(n, "is an Armstrong number") else: print(n, "is not an Armstrong number")

[@Ansh_Rajpoot_]

Sir can we use len () function instead of looop

[@codeitup]

Yes you can ! But in exams you can not use pre defined functions.

[@AMRITPALSINGH-so2lj]

n=int(input("ENTER NUMBER:")) i=n count=0 while(i>0): i=i//10 count=count+1 i=n sum=0 while(i>0): sum=sum+(i%10)**count i=i//10 if(sum==n): print("ARMSTRONG") else: print("NOT ARMSTRONG") sir aisa kar sakte hai?

[@codeitup]

Yes, kar sakte hai !

[@meow-rd6op]

Sir previous armstrong ki video mai toh kuch aur coding thi?

[@codeitup]

Previous coding sirf 3 digit number ko hi armstrong ke lie check kar sakta tha. Ye any number of digit ke lie hai.

[@meow-rd6op]

@@codeitup ohkh sir.thnx

[@Iamkaran553]

Sir mai 12th class mai hu , sir mai abhi python study kr rha hu 117 video lecture wali playlist dekh rha hu,to kya sir mujhe class 12 python wali bhi playlist dekhni pdegi, in dono playlist mai same content ya different hai

[@asitchavda9191]

Sir aise bhi kar sakte he na?? i=int(input("enter the number=")) a=n=i count=0 while(i>0): i=i//10 count=count+1 sum=0 while(a>0): sum=sum+(a%10)**count a=a//10 if sum==n: print("number is Armstrong") else: print("number is not Armstrong")

[@ShadabAnsari-dt4ni]

i=int(input('enter value')) org=i count=0 while i>0: i=i//10 count=count+1 org=i sum=0 while i>0: digit=i%10 x=1 prod=1 while x>=count: prod=prod*digit x=x+1 sum=sum+prod i=i//10 if org==sum: print('Arm') else: print('Not Arm')

[@aashishgupta1580]

n5 = input("Enter the number :") a = int(n5) b = a n5 = list(n5) i = len(n5) sum3 = 0 while a>0: sum3 = sum3 + (a % 10)**i a = a//10 if b == sum3: print("Number is Armstrong") else: print("Number is not Armstrong") the above code is also right or not at all conditions. please reply me.

[@Ansh_Rajpoot_]

Sir please make a whats app group for this course

[@atulsharma2490]

Ye muskil lag Raha he sir ji...koi or method ho to bataye

[@codeitup]

Armstrong ka program hota hi thora complex hai. Bas video, 2-3 baar dekhie with full concentration and I am sure aapko samajh aa jaega..

[@atulsharma2490]

Sir Aapke video achche se samaj aa rahe he....thanks.👍👍👍...Mera ek sawal he abi recently mere pas PC nhi he.. kya ye aap k samjhye hue program Android pr Chala sakta hu ? Kese? Or ye Jupiter note book par bhi aap k program run ho jayege?

[@codeitup]

@@atulsharma2490 You can use QPython app to run the python programs in Android. Yes, all these programs may run on Jupyter.

[@codeitup]

@@atulsharma2490 You may use QPython app to run all the python programs in mobile.

[@codeitup]

@@atulsharma2490 You can run all python programs in an App called QPython.

[@GarywinBhullar]

Complicated video

[@mayanksinghbhadoria5808]

thank you sir