Smallest Divisible Digit Product I
<p>You are given two integers <code>n</code> and <code>t</code>. Return the <strong>smallest</strong> number greater than or equal to <code>n</code> such that the <strong>product of its digits</strong> is divisible by <code>t</code>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 10, t = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">10</span></p>

<p><strong>Explanation:</strong></p>

<p>The digit product of 10 is 0, which is divisible by 2, making it the smallest number greater than or equal to 10 that satisfies the condition.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 15, t = 3</span></p>

<p><strong>Output:</strong> <span class="example-io">16</span></p>

<p><strong>Explanation:</strong></p>

<p>The digit product of 16 is 6, which is divisible by 3, making it the smallest number greater than or equal to 15 that satisfies the condition.</p>
</div>

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

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= t &lt;= 10</code></li>
</ul>
Math; Enumeration
class Solution {
    public int smallestNumber(int n, int t) {
        for (int i = n;; ++i) {
            int p = 1;
            for (int x = i; x > 0; x /= 10) {
                p *= (x % 10);
            }
            if (p % t == 0) {
                return i;
            }
        }
    }
}
