
What is the number of numbers less than 1000 and divisible by 5, which can be formed with the digits 0 to 9, such that no digit gets repeated?
Answer
528.3k+ views
Hint: In order to solve this question, first we take the number of ways we can form a 1 digit number by considering the given condition, then we solve for the number of ways for 2 digit numbers can be formed by considering the given condition in the question and later we find the number of ways 3 digit number can be formed. Lastly add all the number of ways and we will get the required answer.
Complete step by step solution:
We have the given condition i.e.
Number < 1000
Now,
Suppose it is a 1 digit number.
The only way = 1.
Since the only number is 5 which is less than 1000 and divisible by 5.
Now,
Suppose it is a two digit number.
The divisibility rule of any number which is divisible by 5 is that the last digit of the number is 0 and 5 only.
So, if the second digit is 0, then the number of ways for the first digit is 9 i.e. 1, 2, 3, 4, 5, 6, 7, 8 and 9.
And if the second digit is 5, then the number of ways for the first digit is 8 because the first digit cannot be 5 as it is given in the question that no digits gets repeated.
Total ways = 9 + 8 = 17.
Now,
Suppose it is a three digit number.
If the last digit is 0, then the number of ways for the first digit = 9 and for the 2nd digit it will be 8 ways.
Total ways = \[9\times 8=72\]
If the last digit is 5, then the number of ways for the 1st digit = 8 and for the 2nd digit is 8 as this time 0 can be placed in the 2nd digit.
Total ways = \[8\times 8=64\]
Therefore, total ways = 72 + 64 = 136
Thus,
Total number of numbers = 1 + 17 + 136 = 154 ways.
Note: While solving these types of questions, students need to think explicitly about the number of ways. They also should not forget the given condition in the question as because by not considering the condition at any point of the solution, you will get the wrong answer. Here in this question when we formed a 3 digit number by considering the given condition we need to keep in mind that 0 cannot be placed in the first digit as it will become a two digit number.
Complete step by step solution:
We have the given condition i.e.
Number < 1000
Now,
Suppose it is a 1 digit number.
The only way = 1.
Since the only number is 5 which is less than 1000 and divisible by 5.
Now,
Suppose it is a two digit number.
The divisibility rule of any number which is divisible by 5 is that the last digit of the number is 0 and 5 only.
So, if the second digit is 0, then the number of ways for the first digit is 9 i.e. 1, 2, 3, 4, 5, 6, 7, 8 and 9.
And if the second digit is 5, then the number of ways for the first digit is 8 because the first digit cannot be 5 as it is given in the question that no digits gets repeated.
Total ways = 9 + 8 = 17.
Now,
Suppose it is a three digit number.
If the last digit is 0, then the number of ways for the first digit = 9 and for the 2nd digit it will be 8 ways.
Total ways = \[9\times 8=72\]
If the last digit is 5, then the number of ways for the 1st digit = 8 and for the 2nd digit is 8 as this time 0 can be placed in the 2nd digit.
Total ways = \[8\times 8=64\]
Therefore, total ways = 72 + 64 = 136
Thus,
Total number of numbers = 1 + 17 + 136 = 154 ways.
Note: While solving these types of questions, students need to think explicitly about the number of ways. They also should not forget the given condition in the question as because by not considering the condition at any point of the solution, you will get the wrong answer. Here in this question when we formed a 3 digit number by considering the given condition we need to keep in mind that 0 cannot be placed in the first digit as it will become a two digit number.
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