# Java Program to Check Armstrong Number between Two Integers

A positive integer with digits p, q, r, sâ€¦, is known as an Armstrong number of order n if the following condition is fulfilled.

`pqrs... = pn + qn + rn + sn +....`

Example:

```370 = 3*3*3 + 7*7*7 + 0
=  27 + 343 + 0
=  370```

Therefore, 370 is an Armstrong number.

Examples:

```Input : 100 200
Output :153
Explanation : 100 and 200 are given
two integers.
153 = 1*1*1 + 5*5*5 + 3*3*3
= 1 + 125 + 27
=  153
Therefore, only 153 is an Armstrong number between 100 and 200.```

Approach:

Firstly, we will traverse through all the numbers in the given range. Then, for every number, we have to count the number of digits in it. If the number of digits in the current number is n then, find the sum of (n-th) power of all the digits in the number stated. And if the sum is equal to the current number i, then print the number.

Example:

## Java

 `// JAVA program to find Armstrong ` `// numbers between two integers ` `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `gfg { ` ` `  `    ``// Function to print Armstrong ` `    ``// Numbers between two integers ` `    ``static` `void` `ArmstrongNum(``int` `l, ``int` `h) ` `    ``{ ` `        ``for` `(``int` `j = l + ``1``; j < h; ++j) { ` ` `  `            ``// Calculating number of digits ` `            ``int` `y = j; ` `            ``int` `N = ``0``; ` `            ``while` `(y != ``0``) { ` `                ``y /= ``10``; ` `                ``++N; ` `            ``} ` ` `  `            ``// Calculating the sum of nth ` `            ``// power of all the digits ` `            ``int` `sum_power = ``0``; ` `            ``y = j; ` `            ``while` `(y != ``0``) { ` `                ``int` `d = y % ``10``; ` `                ``sum_power += Math.pow(d, N); ` `                ``y /= ``10``; ` `            ``} ` ` `  `            ``// Checking if the current number ` `            ``// i is equal to the sum of nth ` `            ``// power of all the digits ` `            ``if` `(sum_power == j) ` `                ``System.out.print(j + ``" "``); ` `        ``} ` `    ``} ` ` `  `    ``// The Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n1 = ``50``; ` `        ``int` `n2 = ``500``; ` `        ``ArmstrongNum(n1, n2); ` `        ``System.out.println(); ` `    ``} ` `}`

Output

```153 370 371 407
```

Similarly, we can find Armstrong numbers within any given range.

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