We are asked to find the number of 3-digit integers greater than 430 that can be formed using the digits 0, 1, 2, 3, 4, and 5, where repetition of digits is allowed.

1. Problem Description

We are asked to find the number of 3-digit integers greater than 430 that can be formed using the digits 0, 1, 2, 3, 4, and 5, where repetition of digits is allowed.

2. Solution Steps

Let the 3-digit integer be represented as $abc$, where $a, b, c \in \{0, 1, 2, 3, 4, 5\}$.

Since the number must be greater than 430, we consider the possible values for $a$:

* Case 1: $a = 4$. Then $b > 3$, or $b = 3$ and $c > 0$.

* If $a=4$ and $b \in \{4, 5\}$, there are 2 choices for $b$. Then $c$ can be any of the 6 digits $\{0, 1, 2, 3, 4, 5\}$. So there are $2 \times 6 = 12$ possibilities.

* If $a = 4$ and $b = 3$, then $c$ must be greater than

0. $c \in \{1, 2, 3, 4, 5\}$. So there are 5 choices for $c$.

So, when $a = 4$, the number of possibilities is $12 + 5 = 17$.

* Case 2: $a \in \{5\}$. Then $b$ and $c$ can be any of the 6 digits.

* If $a = 5$, there are 6 choices for $b$ and 6 choices for $c$. So there are $1 \times 6 \times 6 = 36$ possibilities.

* Case 3: $a \in \{1, 2, 3\}$. These are not possible, because the number must be greater than

4

3

0.

* Case 4: If the number is a 3 digit number, the first digit cannot be zero. Hence, $a$ can be chosen from the set $\{1, 2, 3, 4, 5\}$.

The total number of 3-digit integers greater than 430 is the sum of the possibilities from the above cases.

Total possibilities = $17 + 36 = 53$.

3. Final Answer

53