Problem 87 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of positive numbers less than 1000 and divisible by 5 (no digit being repeated) is (A) 150 (B) 154 (C) 166 (D) None of these

Step-by-Step Solution

Answer

Calculating each case, the total is 162.

Step 1: Identify the range

First determine the range of numbers we are considering. We are looking for numbers less than 1000, which means we can have numbers with 1-digit, 2-digits, or 3-digits.

Step 2: Identify divisibility condition

For a number to be divisible by 5, it must end in either 0 or 5.

Step 3: Combine conditions

Consider the cases based on digits and divisibility condition: 1-digit, 2-digit, and 3-digit numbers ending with 0 or 5.

Step 4: Case 1: One-digit numbers divisible by 5

The only one-digit numbers that are divisible by 5 are 5 and 0. However, 0 is not positive. Therefore, the one 1-digit positive number is 5.

Step 5: Case 2: Two-digit numbers ending in 0

If a two-digit number ends in 0, the first digit can be any of the 1-9, i.e., 9 choices (1 excludes repetition with 0). Therefore, there are 9 two-digit numbers.

Step 6: Case 3: Two-digit numbers ending in 5

In this case, the first digit can be any of 0-9 except 5, meaning 9 choices (0 does not repeat). But we exclude 5 itself (reducing options if previously counted), resulting in 8 numbers.

Step 7: Case 4: Three-digit numbers ending in 0

If it ends in 0, the hundreds place can be filled by 1-9 (9 choices), tens place by 0-9 but not the hundredth place digit (8 choices), so the total is 9*8 = 72 such numbers.

Step 8: Case 5: Three-digit numbers ending in 5

Here, hundreds place can be 1-9 (9 choices), tens by 0-9 but not the hundreds or unit digit of 5 (8 choices), resulting in 9*8 = 72 numbers.

Step 9: Sum the totals

Add all calculated cases: 1 (one-digit) + 9 + 8 (two-digit) + 72 + 72 (three-digit) to get the total number of positive numbers.

Key Concepts

One-digit Numbers

One-digit numbers are the simplest type of numbers that we encounter. They range from 0 to 9. However, in problems like our original exercise, we're interested in certain conditions and characteristics they must satisfy. For example, when considering divisibility by 5, we know two vital facts:

• The number must end in either 0 or 5.
• It must be positive, which means 0 is not an option for positive divisibility.

In our exercise, we noticed that for one-digit numbers divisible by 5, only the number 5 meets these criteria, as it is the only positive one-digit number that ends in 5. Simple, right? The one-digit count for our problem is thus just 1 number: 5. This foundational step sets the stage for tackling more complex, multi-digit scenarios.

Two-digit Numbers

Two-digit numbers introduce more complexity. They range from 10 to 99. Given the divisibility rule for 5, any valid two-digit number must end in either 0 or 5. Let's dissect this:

• Ending in 0: The first digit can be any number from 1 to 9, providing us with 9 possible numbers (10, 20, ..., 90).
• Ending in 5: Here, the first digit can be any number from 0 to 9, except 5, to avoid digit repetition, leaving us with 9 choices. However, we exclud 5 as the number itself, which results in numbers like 15, 25,..., 95.

Thus, two-digit numbers add richness and options to the list. In total, we find that we have a combined count of 17 two-digit numbers (9 ending in 0 and 8 ending in 5). This building block is crucial for expanding our list to three-digit divisors.

Three-digit Numbers

Three-digit numbers are the most involved category in our exercise, as they range from 100 to 999. The process involves more calculations, but it's just a logical extension of two-digit understanding. We still need the ending digit to be either 0 or 5:

• Ending in 0: The hundreds place offers 9 potential values (1 through 9). The tens place, however, excludes this hundreds digit and the fixed unit of 0, leaving us with 8 possibilities per hundreds choice. This results in a total of 72 numbers (from combinations like 100, 210, 320, and so forth).
• Ending in 5: Similar to zeros, options for hundreds start at 1-9, providing 9 choices again. For different tens digits, consider all except the hundreds digit and the last fixed digit of 5, again totaling to 8 valid outcomes per hundreds placement. Thus, we again calculate 72 possible numbers (like 105, 215, ..., 985).

This comprehensive approach yields a total of 144 three-digit numbers divisible by 5 without any repeated digits, underscoring the power of understanding basic divisibility and position rules.