Using digits 2, 3, 5, 6, 7 and 9, how many 3-digit numbers can be formed that are divisible by 5 such that the repetition of digits is not allowed?
herpity derp. If a number is divisible by 5, it must have five or 0 in its ones digit. Thus, five must be in the ones. 6 can be in 1 of 4 places. So can 5 and 4 and 3 and 2. So, the answer is....
Your number must be in the form
xx5, where the x can only be from 2,3,6,7 or 9
number of 3 digit numbers which are divisible by 5 are
5x4x1 = 20
To solve this problem, we can use the concept of permutation.
Since the number needs to be divisible by 5, the last digit must be either 5 or 0. However, since repetition of digits is not allowed, we cannot use the digit 5 as the last digit as it also appears in the set of digits given.
So, the last digit has to be 0. We are left with the digits 2, 3, 6, 7, and 9 to choose from for the first and second positions.
To find the number of 3-digit numbers that can be formed, we can break down the problem into two parts:
1. Choosing the first digit: We have 5 options (2, 3, 6, 7, and 9) to choose from for the first position.
2. Choosing the second digit: After choosing the first digit, we have 4 options left for the second position.
The total number of 3-digit numbers that can be formed is calculated by multiplying the number of options for each step. Therefore, the total number of 3-digit numbers is 5 * 4 = 20.
So, there are 20 different 3-digit numbers that can be formed using the given set of digits (2, 3, 5, 6, 7, and 9) that are divisible by 5 and do not allow repetition of digits.