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Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.?
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Find the smallest 5 digit no which is exactly divisible by 21. ,28 and...
Introduction:
In this problem, we need to find the smallest 5 digit number which is exactly divisible by 21, 28, and 36.
Approach:
To find the smallest 5 digit number which is exactly divisible by 21, 28, and 36, we need to follow the below steps:
1. Find the LCM of 21, 28, and 36.
2. Divide the LCM by each number given in the problem and find the highest quotient.
3. Multiply the highest quotient by the LCM to get the smallest 5 digit number which is exactly divisible by 21, 28, and 36.
Solution:
1. Finding the LCM of 21, 28, and 36.
- Prime factorization of 21 = 3 x 7
- Prime factorization of 28 = 2 x 2 x 7
- Prime factorization of 36 = 2 x 2 x 3 x 3
- LCM = 2 x 2 x 3 x 3 x 7 = 252
2. Divide the LCM by each number given in the problem and find the highest quotient.
- 252 ÷ 21 = 12
- 252 ÷ 28 = 9
- 252 ÷ 36 = 7
- Highest quotient = 12
3. Multiply the highest quotient by the LCM to get the smallest 5 digit number which is exactly divisible by 21, 28, and 36.
- Smallest 5 digit number = 12 x 252 = 3024
Conclusion:
The smallest 5 digit number which is exactly divisible by 21, 28, and 36 is 3024.
In this problem, we need to find the smallest 5 digit number which is exactly divisible by 21, 28, and 36.
Approach:
To find the smallest 5 digit number which is exactly divisible by 21, 28, and 36, we need to follow the below steps:
1. Find the LCM of 21, 28, and 36.
2. Divide the LCM by each number given in the problem and find the highest quotient.
3. Multiply the highest quotient by the LCM to get the smallest 5 digit number which is exactly divisible by 21, 28, and 36.
Solution:
1. Finding the LCM of 21, 28, and 36.
- Prime factorization of 21 = 3 x 7
- Prime factorization of 28 = 2 x 2 x 7
- Prime factorization of 36 = 2 x 2 x 3 x 3
- LCM = 2 x 2 x 3 x 3 x 7 = 252
2. Divide the LCM by each number given in the problem and find the highest quotient.
- 252 ÷ 21 = 12
- 252 ÷ 28 = 9
- 252 ÷ 36 = 7
- Highest quotient = 12
3. Multiply the highest quotient by the LCM to get the smallest 5 digit number which is exactly divisible by 21, 28, and 36.
- Smallest 5 digit number = 12 x 252 = 3024
Conclusion:
The smallest 5 digit number which is exactly divisible by 21, 28, and 36 is 3024.
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Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.?
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Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.? for Class 6 2024 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.? covers all topics & solutions for Class 6 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.?.
Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.? for Class 6 2024 is part of Class 6 preparation. The Question and answers have been prepared according to the Class 6 exam syllabus. Information about Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.? covers all topics & solutions for Class 6 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the smallest 5 digit no which is exactly divisible by 21. ,28 and 36.?.
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