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# Armstrong Numbers between two integers

• Difficulty Level : Basic
• Last Updated : 27 Apr, 2021

A positive integer with digits a, b, c, d… is called an Armstrong number of order n if following condition is satisfied.

`abcd... = an + bn + cn + dn +...`

```153 = 1*1*1 + 5*5*5 + 3*3*3
=  1 + 125 + 27
=  153
Therefore, 153 is an Armstrong number.```

Examples:

```Input : 100 400
Output :153 370 371
Explanation : 100 and 400 are given
two integers.(interval)
153 = 1*1*1 + 5*5*5 + 3*3*3
= 1 + 125 + 27
=  153
370 = 3*3*3 + 7*7*7 + 0
= 27 + 343
= 370
371 = 3*3*3 + 7*7*7 + 1*1*1
= 27 + 343 +1
= 371```

The approach implemented below is simple. We traverse through all numbers in given range. For every number, we first count number of digits in it. Let the number of digits in current number be n. Them we find sum of n-th power of all digits. If sum is equal to i, we print the number.

## C++

 `// CPP program to find Armstrong numbers in a range``#include ``using` `namespace` `std;` `// Prints Armstrong Numbers in given range``void` `findArmstrong(``int` `low, ``int` `high)``{``    ``for` `(``int` `i = low+1; i < high; ++i) {` `        ``// number of digits calculation``        ``int` `x = i;``        ``int` `n = 0;``        ``while` `(x != 0) {``            ``x /= 10;``            ``++n;``        ``}` `        ``// compute sum of nth power of``        ``// its digits``        ``int` `pow_sum = 0;``        ``x = i;``        ``while` `(x != 0) {``            ``int` `digit = x % 10;``            ``pow_sum += ``pow``(digit, n);``            ``x /= 10;``        ``}` `        ``// checks if number i is equal to the``        ``// sum of nth power of its digits``        ``if` `(pow_sum == i)``            ``cout << i << ``" "``;    ``    ``}``}` `// Driver code``int` `main()``{``    ``int` `num1 = 100;``    ``int` `num2 = 400;``    ``findArmstrong(num1, num2);``    ``cout << ``'\n'``;``    ``return` `0;``}`

## Java

 `// JAVA program to find Armstrong``// numbers in a range``import` `java.io.*;``import` `java.math.*;` `class` `GFG {``    ` `    ``// Prints Armstrong Numbers in given range``    ``static` `void` `findArmstrong(``int` `low, ``int` `high)``    ``{``        ``for` `(``int` `i = low + ``1``; i < high; ++i) {``     ` `            ``// number of digits calculation``            ``int` `x = i;``            ``int` `n = ``0``;``            ``while` `(x != ``0``) {``                ``x /= ``10``;``                ``++n;``            ``}``     ` `            ``// compute sum of nth power of``            ``// its digits``            ``int` `pow_sum = ``0``;``            ``x = i;``            ``while` `(x != ``0``) {``                ``int` `digit = x % ``10``;``                ``pow_sum += Math.pow(digit, n);``                ``x /= ``10``;``            ``}``     ` `            ``// checks if number i is equal``            ``// to the sum of nth power of``            ``// its digits``            ``if` `(pow_sum == i)``                ``System.out.print(i + ``" "``);    ``        ``}``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `num1 = ``100``;``        ``int` `num2 = ``400``;``        ``findArmstrong(num1, num2);``        ``System.out.println();``    ``}``}` `/*This code is contributed by Nikita Tiwari.*/`

## Python

 `# PYTHON program to find Armstrong``# numbers in a range``import` `math` `# Prints Armstrong Numbers in given range``def` `findArmstrong(low, high) :``    ` `    ``for` `i ``in` `range``(low ``+` `1``, high) :``        ` `        ``# number of digits calculation``        ``x ``=` `i``        ``n ``=` `0``        ``while` `(x !``=` `0``) :``            ``x ``=` `x ``/` `10``            ``n ``=` `n ``+` `1``            ` `        ``# compute sum of nth power of``        ``pow_sum ``=` `0``        ``x ``=` `i``        ``while` `(x !``=` `0``) :``            ``digit ``=` `x ``%` `10``            ``pow_sum ``=` `pow_sum ``+` `math.``pow``(digit, n)``            ``x ``=` `x ``/` `10``            ` `        ``# checks if number i is equal to``        ``# the sum of nth power of its digits``        ``if` `(pow_sum ``=``=` `i) :``            ``print``(``str``(i) ``+` `" "``),` `# Driver code``num1 ``=` `100``num2 ``=` `400``findArmstrong(num1, num2)``print``("")` `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find Armstrong``// numbers in a range``using` `System;` `class` `GFG {` `    ``// Prints Armstrong Numbers in given range``    ``static` `void` `findArmstrong(``int` `low, ``int` `high)``    ``{``        ``for` `(``int` `i = low + 1; i < high; ++i) {` `            ``// number of digits calculation``            ``int` `x = i;``            ``int` `n = 0;``            ``while` `(x != 0) {``                ``x /= 10;``                ``++n;``            ``}` `            ``// compute sum of nth power of``            ``// its digits``            ``int` `pow_sum = 0;``            ``x = i;``            ``while` `(x != 0) {``                ``int` `digit = x % 10;``                ``pow_sum += (``int``)Math.Pow(digit, n);``                ``x /= 10;``            ``}` `            ``// checks if number i is equal``            ``// to the sum of nth power of``            ``// its digits``            ``if` `(pow_sum == i)``                ``Console.Write(i + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `num1 = 100;``        ``int` `num2 = 400;``        ``findArmstrong(num1, num2);``        ``Console.WriteLine();``    ``}``}` `/*This code is contributed by vt_m.*/`

## PHP

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## Javascript

 ``

Output:

`153 370 371`

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