<p>The number of distinct natural numbers up to a maximum of four digits and divisible by 5, which can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, each digit not occurring more than once in each number, is</p>
Step-by-Step Solution
Key Concept: A number is divisible by 5 iff it ends in 0 or 5. Count 1-digit, 2-digit, 3-digit, and 4-digit numbers separately, treating cases where the last digit is 0 vs 5 differently (since 0 cannot be the leading digit).
<p><strong>Step 1: One-digit numbers</strong><br/>Numbers: 5 (only one, since 0 is not a natural number in standard context)<br/>Count = 1</p><p><strong>Step 2: Two-digit numbers</strong><br/><em>Case 1 - Ends in 0:</em> First digit from {1,2,3,4,5,6,7,8,9} = 9 choices<br/><em>Case 2 - Ends in 5:</em> First digit from {1,2,3,4,6,7,8,9} (not 0, not 5) = 8 choices<br/>Count = 9 + 8 = 17</p><p><strong>Step 3: Three-digit numbers</strong><br/><em>Case 1 - Ends in 0:</em> First digit has 9 choices {1-9}, second digit has 8 choices = 9 × 8 = 72<br/><em>Case 2 - Ends in 5:</em> First digit has 8 choices {1,2,3,4,6,7,8,9}, second digit has 8 choices {0,1,2,3,4,6,7,8,9} minus first = 8 × 8 = 64<br/>Count = 72 + 64 = 136</p><p><strong>Step 4: Four-digit numbers</strong><br/><em>Case 1 - Ends in 0:</em> First digit has 9 choices, second has 8, third has 7 = 9 × 8 × 7 = 504<br/><em>Case 2 - Ends in 5:</em> First digit has 8 choices {1,2,3,4,6,7,8,9}, remaining two digits filled from remaining 8 digits (including 0) in order = 8 × 8 × 7 = 448<br/>Count = 504 + 448 = 952</p><p><strong>Step 5: Total</strong><br/>1 + 17 + 136 + 952 = <strong>1106</strong></p><p>∴ Answer: C</p>
Correct Answer: C