# Disarium Number

Given a number “n”, find if it is Disarium or not. A number is called Disarium if sum of its digits powered with their respective positions is equal to the number itself.

Examples:

Input : n = 135 Output : Yes 1^1 + 3^2 + 5^3 = 135 Therefore, 135 is a Disarium number Input : n = 89 Output : Yes 8^1+9^2 = 89 Therefore, 89 is a Disarium number Input : n = 80 Output : No 8^1 + 0^2 = 8

The idea is to fist count digits in given numbers. Once we have count, we traverse all digits from right most (using % operator), raise its power to digit count and decrement the digit count.

Below is implementation of above idea.

## C++

`// C++ program to check whether a number is Desoriam ` `// or not ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Finds count of digits in n ` `int` `countDigits(` `int` `n) ` `{ ` ` ` `int` `count_digits = 0; ` ` ` ` ` `// Count number of digits in n ` ` ` `int` `x = n; ` ` ` `while` `(x) ` ` ` `{ ` ` ` `x = x/10; ` ` ` ` ` `// Count the no. of digits ` ` ` `count_digits++; ` ` ` `} ` ` ` `return` `count_digits; ` `} ` ` ` `// Function to check whether a number is disarium or not ` `bool` `check(` `int` `n) ` `{ ` ` ` `// Count digits in n. ` ` ` `int` `count_digits = countDigits(n); ` ` ` ` ` `// Compute sum of terms like digit multiplied by ` ` ` `// power of position ` ` ` `int` `sum = 0; ` `// Initialize sum of terms ` ` ` `int` `x = n; ` ` ` `while` `(x) ` ` ` `{ ` ` ` `// Get the rightmost digit ` ` ` `int` `r = x%10; ` ` ` ` ` `// Sum the digits by powering according to ` ` ` `// the positions ` ` ` `sum = sum + ` `pow` `(r, count_digits--); ` ` ` `x = x/10; ` ` ` `} ` ` ` ` ` `// If sum is same as number, then number is ` ` ` `return` `(sum == n); ` `} ` ` ` `//Driver code to check if number is disarium or not ` `int` `main() ` `{ ` ` ` `int` `n = 135; ` ` ` `if` `( check(n)) ` ` ` `cout << ` `"Disarium Number"` `; ` ` ` `else` ` ` `cout << ` `"Not a Disarium Number"` `; ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to check whether a number is Desoriam ` `// or not ` ` ` `class` `Test ` `{ ` ` ` `// Method to check whether a number is disarium or not ` ` ` `static` `boolean` `check(` `int` `n) ` ` ` `{ ` ` ` `// Count digits in n. ` ` ` `int` `count_digits = Integer.toString(n).length(); ` ` ` ` ` `// Compute sum of terms like digit multiplied by ` ` ` `// power of position ` ` ` `int` `sum = ` `0` `; ` `// Initialize sum of terms ` ` ` `int` `x = n; ` ` ` `while` `(x!=` `0` `) ` ` ` `{ ` ` ` `// Get the rightmost digit ` ` ` `int` `r = x%` `10` `; ` ` ` ` ` `// Sum the digits by powering according to ` ` ` `// the positions ` ` ` `sum = (` `int` `) (sum + Math.pow(r, count_digits--)); ` ` ` `x = x/` `10` `; ` ` ` `} ` ` ` ` ` `// If sum is same as number, then number is ` ` ` `return` `(sum == n); ` ` ` `} ` ` ` ` ` `// Driver method ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `n = ` `135` `; ` ` ` ` ` `System.out.println(check(n) ? ` `"Disarium Number"` `: ` `"Not a Disarium Number"` `); ` ` ` `} ` `} ` |

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## Python

`# Python program to check whether a number is Desarium ` `# or not ` `import` `math ` ` ` `# Method to check whether a number is disarium or not ` `def` `check(n) : ` ` ` ` ` `# Count digits in n. ` ` ` `count_digits ` `=` `len` `(` `str` `(n)) ` ` ` ` ` `# Compute sum of terms like digit multiplied by ` ` ` `# power of position ` ` ` `sum` `=` `0` `# Initialize sum of terms ` ` ` `x ` `=` `n ` ` ` `while` `(x!` `=` `0` `) : ` ` ` ` ` `# Get the rightmost digit ` ` ` `r ` `=` `x ` `%` `10` ` ` ` ` `# Sum the digits by powering according to ` ` ` `# the positions ` ` ` `sum` `=` `(` `int` `) (` `sum` `+` `math.` `pow` `(r, count_digits)) ` ` ` `count_digits ` `=` `count_digits ` `-` `1` ` ` `x ` `=` `x` `/` `10` ` ` ` ` `# If sum is same as number, then number is ` ` ` `if` `sum` `=` `=` `n : ` ` ` `return` `1` ` ` `else` `: ` ` ` `return` `0` ` ` `# Driver method ` `n ` `=` `135` `if` `(check(n) ` `=` `=` `1` `) : ` ` ` `print` `"Disarium Number"` `else` `: ` ` ` `print` `"Not a Disarium Number"` ` ` `# This code is contributed by Nikita Tiwari. ` |

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Output:

Disarium Number

This article is contributed by **Sahil Chhabra(KILLER)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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