Special Programs in C− Check If The Number Is Armstrong Number 

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alternative int main() { int num,count=0,sum=0,rem,result=0; printf("enter a number "); scanf("%d",&num); int kela=num; int lela=num; while(num!=0){ num=num/10; count++; } while(lela!=0){ rem=lela%10; sum=sum+pow(rem,count); lela/=10; } if(kela==sum) { printf("its a armstrong number i.e %d",kela); } else{ printf("its not a armstrong number coz %d",sum); } return 0; }

Are u btech cse student?

Yar how can you love it...I'm unable to understand c language..can you help me

@@PoojaMishra-xc6lp start from the first lecture of c programming of neso,eventually u will understand everything

So true in mine case

@@ramarajujaya2480 yes it is true in my case also ❤️❤️ Neso is the best ❤️❤️

see the value changed inside while is permanent...not temprary

@@its_SR07 if its permanent then again mul value is changed into 343 right then is that 3 want to multiple with 343?

count=n; //count n again, for next digit q/=10; //to omit the used digit :) omitting starts from ones place :) //mul=1; like it if u like :) so that others can see :)

Baawli gaand😂

this code only works with a 3 digits number

header lines are obvious in every programs. i guess thats why he didnt mention

You can use that but that will not actually give u the feeling of the logic of making a number raise to some power

@Darshan patil did you find out ?

In scanf remove that gap between % and d then it's working. I checked.

@@paraprannoy3732 yeah thank you

yep

!! Well if you include math.h (By which you can include pow) in the header files it can be easier !! :) #include #include int main(){ int num,temp,rem,result=0,count=0; printf("enter the number that you want to check "); scanf("%d",&num); temp=num; while(temp!=0){ temp=temp/10; count++; } temp=num; while(num>0){ rem=num%10; result=result+pow(rem,count); num=num/10; } if(temp==result) { printf("It is an armstrong number "); } else{ printf("It is not an armstrong number "); } return 0; }#include #include int main(){ int num,temp,rem,result=0,count=0; printf("enter the number that you want to check "); scanf("%d",&num); temp=num; while(temp!=0){ temp=temp/10; count++; } temp=num; while(num>0){ rem=num%10; result=result+pow(rem,count); num=num/10; } if(temp==result) { printf("It is an armstrong number "); } else{ printf("It is not an armstrong number "); } return 0; }

We are storing the value of 37/10 in 'q' not in count and in next step we are simply incrementing the count variable to find the order of number.

37/10 = 3

@Rajesh Sanagala We have initialize mul=1 as 1 is the only number after multiply with we get the same number and moreover we have to use the mul variable each time the loop run and If we don't initialize it with 1 then we will be multiplying the new digit with the previous result Like Nu=153 mul=mul*rem // (1*3=3**3=9) For next digit we need to have the mul variable as 1 only mul=mul*rem //(1*5=5) Otherwise we will get 9*5 but we have to add the digit power 3..

Which cnt for first one we have given the value of count to cnt variable so that changing anything in cnt doesn't affect count For second one Count is = 3 we initialised cnt to 3 by equating it to count as while loop mein cnt ki value humne 0 kardi thi last process mein

Same with q while counting the no. Of digits for our given no. Humne q ki value 0 kardi thi we initialised it back to our original no.

Because we've use ‘count' to count the digit of the number at first step and in second step we're using cnt instead of count as count variable was used earlier

But, you are not allowed in all cases

instead of x-- use temp1=temp1/10; and in the condition in while use : temp1!=0.

@@aakashmudigonda3375 thank you

as we got the value of cnt from previous function and before entering the while loop the condition will be checked and if the condition will be true so code will be executed. and mul is initialised in main function recheck the full snippet and ya we again initialised mul as 1 so that we can again multiply the left digits with their respective order . thanku and watch this video again.

@@harmankaur3938 Thanks for help and info mate! Appreciated 👏

pratice makes a man prefect you havent head that. dont use your memory here. experience is the mother of all wisdom

Numbers from 0-9 are Armstrong numbers

Because if its 3. it reduces to 0 and it can not be looped again for the remaining values we need to multiply

so after it becomes 0 you basically need to set it to 3 again, so it continues that process of Looping. I hope I could help

Suppose n is the number you want to count the digits then let q= n then follow the code given, we don't want the number n which is given as input to change so we are using another variable q. Is my explanation clear ?

Yes if the variable in which you store it is integer

remainder is 0

@@saikiran5345 a small number % a bigger one is the smal one ==> so 3%10 is 3 bro (modulo property)

As clearly the left-hand side number is left than right hand side number....the result would be left hand side number ...without doubt

no its remainder so it would be 3