Armstrong Number
<p>Given an integer <code>n</code>, return <code>true</code> <em>if and only if it is an <strong>Armstrong number</strong></em>.</p>

<p>The <code>k</code>-digit number <code>n</code> is an Armstrong number if and only if the <code>k<sup>th</sup></code> power of each digit sums to <code>n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 153
<strong>Output:</strong> true
<strong>Explanation:</strong> 153 is a 3-digit number, and 153 = 1<sup>3</sup> + 5<sup>3</sup> + 3<sup>3</sup>.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 123
<strong>Output:</strong> false
<strong>Explanation:</strong> 123 is a 3-digit number, and 123 != 1<sup>3</sup> + 2<sup>3</sup> + 3<sup>3</sup> = 36.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>8</sup></code></li>
</ul>
Math
class Solution:
    def isArmstrong(self, n: int) -> bool:
        k = len(str(n))
        s, x = 0, n
        while x:
            s += (x % 10) ** k
            x //= 10
        return s == n
