How many three digit numbers divisible by 5 are there in which no digits are repeated?
Show Hint
Whenever the digit 0 is restricted from the first position and is also part of a divisibility condition, always split the counting into two scenarios: "ending in 0" and "ending in the non-zero option".
This prevents errors regarding the placement of 0 in the hundreds place.
Step 1 : Understanding the Question:
This question asks us to find the number of three-digit numbers divisible by 5 such that no digits are repeated. Step 2 : Key Formulas and Approach:
A three-digit number can be represented as $d_1d_2d_3$, where $d_1 \in \{1, 2, \dots, 9\}$ and $d_2, d_3 \in \{0, 1, \dots, 9\}$.
For a number to be divisible by 5, its units digit ($d_3$) must be either 0 or 5.
Since the digit 0 cannot occupy the hundreds place ($d_1$), we must analyze the two possible cases for $d_3$ separately to avoid double-counting or invalid placements. Step 3 : Detailed Explanation:
Let us divide the problem into two mutually exclusive cases:
• Case 1: The units digit $d_3$ is 0.
• The units place is fixed as 0, which can be done in $1$ way.
• Since repetition is not allowed and 0 is already used, the hundreds place $d_1$ can be filled with any of the remaining non-zero digits from $\{1, 2, \dots, 9\}$. This gives $9$ possible choices.
• The tens place $d_2$ can then be filled with any of the remaining $8$ digits.
• Number of ways for Case 1 = $9 \times 8 \times 1 = 72$ ways.
• Case 2: The units digit $d_3$ is 5.
• The units place is fixed as 5, which can be done in $1$ way.
• Since 5 is used, the hundreds place $d_1$ can be filled with any non-zero digit except 5.
The available digits for $d_1$ are $\{1, 2, 3, 4, 6, 7, 8, 9\}$, which gives $8$ choices.
• The tens place $d_2$ can be filled with any of the remaining digits, which now includes 0.
The total available digits are $\{0, 1, 2, \dots, 9\}$ excluding 5 and whichever digit was placed in $d_1$. This leaves $10 - 2 = 8$ choices.
• Number of ways for Case 2 = $8 \times 8 \times 1 = 64$ ways.
Adding the possibilities from both cases gives:
\[ \text{Total numbers} = 72 + 64 = 136 \] Step 4 : Final Answer:
The total number of such three-digit numbers is 136.
This corresponds to Option (A).
