# How to check if a given number is a Disarium number using Python?

What is a Disarium number?
A Disarium number is a number where the sum of digits raised to powers equal to their position gives back the number.
Example
135 has the 3 digits 1, 3 and 5. 1 is at position 1,  3 at position 2 and 5 at position 3.
So, the sum will be 11 + 32 + 53=1 + 9 + 125 = 135. So, this is a Disarium number.

Let us try with 123
sum =11 + 22 + 33=1 + 4 + 9=14
14 is not equal to 123 so this is not a Disarium number

The first thing to do is to separate the digits of the number.
We will use the two operators // and %
// is integer division. For example 123 divide by 10 is 12.3, 123 integer divide by 10 will give 12 since .3 will be ignored.
123/10 =12.3
123//10 =12

What does % do. It gives the remainder. Thus 123 % 10=3,

How does it help in finding the digits. Consider this sequence.
n=123
r= n % 10  = 3
n=n//10      = 12
r=n % 10   = 2
n=n//10      =1
r=n % 10   =1
n=n//10      =0

This gives us the following program.

n=123
while n!=0:
r=n%10
n=n//10
print(r)

Next step, we need to get the position of each digit. How shall we get that?
Let us add a counter to the loop.
n=123
count=0
while n!=0:
r=n%10
n=n//10
count+=1
print("Digit = ",r, ", Position = ", count)

The positions are in reverse order. To get it in proper order let us subtract the positions from 4, that is from 1 more than the position.

n=123
count=0
length=len(str(n)) + 1
while n!=0:
r=n%10
n=n//10
count+=1
position=length-count
print("Digit = ",r, ", Position = ", position)

We can complete the program now.
n=123
copy = n
count=0
sum=0
length=len(str(n)) + 1
while n!=0:
r=n%10
n=n//10
count+=1
position=length-count
sum = sum + r ** position
print("Digit = ",r, ", Position = ", position)
if sum==copy:
print("Disarium")
else:
print("Not Disarium")