while loop:
Output:
enter number153
153 is a ArmStrong Number
§ If we want to execute a group of statements multiple
times until some condition false, then we should go for while loop.
Syntax:
while condition:
body
§
Write a program to display first n natural
numbers.
n=int(input('enter number'))
i=1
while i<=n:
print(i)
i=i+1
Output:
enter number5
1
2
3
4
5
Note-
·
In
implementation when we need to extract last digit from the given data than go
for %10. If
we require to remove last digit from the given number than go for divide by /10
a=
158%10 à 8
a=123%10 à 3
a=97%10 à 7
a=5%10 à 5
Note-
If the numerator value is less than of
denominator value than if it is modular operator returns numerator value.
|
a= 158//10
à 15
a=123//10 à 12
a=97//10 à 9
a=5//10 à 0
Note-
If
the numerator value is less than of denominator value than if it is division
operator returns zero.
|
§
Write a program to display the sum of all digits to
given number.
n=int(input('enter number'))
sum=0
while n>0:
d=n%10
sum=sum+d
n=n//10
print('Sum of Digits = ',sum)
Output:
enter number123
Sum of Digits = 6
§
Write a program to display the reverse of a given number.
n=int(input('enter number'))
rev=0
while n>0:
d=n%10
rev=rev*10+d
n=n//10
print('Reverse of Number = ',rev)
Output:
enter number123
Reverse of Number
= 321
§
Write a program to check whether given number is an
Armstrong or not.
Armstrong number is a
number that is equal to the sum of cubes of its digits.
For example 0,
1, 153, 370, 371 and 407 are the Armstrong numbers.
Let's try to understand why 153 is
an Armstrong number.
153 = (1*1*1)+(5*5*5)+(3*3*3)
where:
(1*1*1)=1
(5*5*5)=125
(3*3*3)=27
So:
1+125+27=153
n=int(input('enter
number'))
sum=0
temp=n
while n>0:
#Extracting the last digit
from given number
d=n%10
sum=sum+(d**3)
n=n//10
if
temp==sum:
print(temp,"
is a ArmStrong Number")
else:
print(temp,"
is a ArmStrong Number")
Output:
enter number153
153 is a ArmStrong Number
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