Question

How many three-digit numbers that are divisible by 5, can be formed, using the digits \(0,2,3,5,7\), if no digit occurs more than once in each number? (1) 10 (2) 15 (3) 21 (4) 25

Short answer

Answer: 18

Step-by-step solution

Identify possible units digit

As mentioned, a number will only be divisible by \(5\) if it ends in either \(0\) or \(5\). Hence, the possible units digits for the required three-digit numbers are \(0\) and \(5\).

Calculate the possible digit combinations

Let's now find the count of all possible digit combinations based on the two choices of units digits: Case 1: Units digit = \(0\) - Here, we have four remaining digits \({2,3,5,7}\) to choose from for the tens and hundreds place. - We have \(3\) choices for the hundreds digit (we can't use \(0\) as the hundreds digit will result in a two-digit number), and \(3\) choices for the tens digit (since no digit should repeat). - Hence, \(3*3=9\) three-digit numbers can be formed when the units digit is \(0\). Case 2: Units digit = \(5\) - Here, we have four remaining digits \({0,2,3,7}\) to choose from for the tens and hundreds place. - We have \(3\) choices for the hundreds digit and \(3\) choices for the tens digit (as no digit should repeat). - Hence, \(3*3=9\) three-digit numbers can be formed when the units digit is \(5\).

Total count of three-digit numbers divisible by 5

Adding the count of numbers formed in both cases, we get the total count of three-digit numbers that are divisible by \(5\) and formed using the given digits: Total count = Count of numbers with units digit \(0\) + Count of numbers with units digit \(5\) Total count = \(9 + 9 = 18\) So, there are \(18\) such three-digit numbers. The correct answer is not given in the given options, which might be a mistake in the question.

Key concepts

Counting Three-Digit Numbers

Understanding how to count three-digit numbers that fit specific criteria is a fundamental skill in combinatorics, a branch of mathematics dealing with counting, arrangement, and combination. Consider the simple requirement that a number must be a multiple of 5 and formed from non-repeating chosen digits, such as in our given problem with digits \(0,2,3,5,7\).

When counting three-digit numbers, the first digit (hundreds place) cannot be zero, as that would result in a two-digit number. Hence, the choice for the first digit is restricted. When we have a set of digits and we need to form numbers with the condition that no digit repeats, the problem becomes one of permutations. For a three-digit number, we are effectively choosing and arranging 3 digits out of our set to different positions - hundreds, tens, and units place.

Permutations Without Repetition

Permutations without repetition are arrangements of elements where the order is important, and each element can be used only once. When forming our three-digit numbers, this principle is key. Starting with the units digit being 5, since the number must be divisible by 5, we then have three choices left for the hundreds place, as zero cannot be used. Finally, there are three choices remaining for the tens place.

In such permutations, the count of possible combinations for each digit's place is multiplied together to get the total count. This multiplication principle is a cornerstone of combinatorial counting and a powerful tool in solving such problems.

Multiples of 5

For a number to be a multiple of 5, its units digit must be either 0 or 5. This simple rule streamlines the process of creating divisible numbers and defines which digits we can use in the units place for our permutations. Bearing in mind this divisibility rule, we tailored our solution to include only those permutations ending in 0 or 5.

This rule is one of the basic divisibility rules in mathematics that can be quickly applied to questions involving multiples and can also help in checking the correctness of answers without having to perform the full division.