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What is the smallest square number which is divisible by each of the numbers 6,9 and 15 ?
- a)900
- b)810
- c)630
- d)730
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
What is the smallest square number which is divisible by each of the n...
LCM of 6,9 and 15 is 90. So any multiple of 90 will be divisible by 6,9 and 15. The only square number in the options is 900, so 900 is the smallest square number which is divisible by 6,9 and 15.
Most Upvoted Answer
What is the smallest square number which is divisible by each of the n...
Because its smaller cannot be divisible all 3 numbers and its bigger number also cannot be divisible by all 3
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What is the smallest square number which is divisible by each of the n...
Understanding the Problem
To find the smallest square number that is divisible by 6, 9, and 15, we first need to determine the least common multiple (LCM) of these numbers.
Finding the LCM
1. Prime Factorization:
- 6: \(2^1 \times 3^1\)
- 9: \(3^2\)
- 15: \(3^1 \times 5^1\)
2. Calculate the LCM:
- Take the highest power of each prime factor:
- \(2^1\) from 6
- \(3^2\) from 9
- \(5^1\) from 15
- Thus, the LCM is:
\[
LCM = 2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90
\]
Finding the Smallest Square Number
To find the smallest square number divisible by 90, we need to ensure that all prime factors in the square are raised to even powers.
1. Prime Factorization of 90:
\[
90 = 2^1 \times 3^2 \times 5^1
\]
2. Adjusting Exponents to Even Numbers:
- \(2^1\) needs one more \(2\) to become \(2^2\)
- \(3^2\) is already even
- \(5^1\) needs one more \(5\) to become \(5^2\)
3. Forming the Square Number:
- Thus, the smallest square number is:
\[
2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900
\]
Conclusion
The smallest square number that is divisible by 6, 9, and 15 is indeed 900, confirming option A.
To find the smallest square number that is divisible by 6, 9, and 15, we first need to determine the least common multiple (LCM) of these numbers.
Finding the LCM
1. Prime Factorization:
- 6: \(2^1 \times 3^1\)
- 9: \(3^2\)
- 15: \(3^1 \times 5^1\)
2. Calculate the LCM:
- Take the highest power of each prime factor:
- \(2^1\) from 6
- \(3^2\) from 9
- \(5^1\) from 15
- Thus, the LCM is:
\[
LCM = 2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90
\]
Finding the Smallest Square Number
To find the smallest square number divisible by 90, we need to ensure that all prime factors in the square are raised to even powers.
1. Prime Factorization of 90:
\[
90 = 2^1 \times 3^2 \times 5^1
\]
2. Adjusting Exponents to Even Numbers:
- \(2^1\) needs one more \(2\) to become \(2^2\)
- \(3^2\) is already even
- \(5^1\) needs one more \(5\) to become \(5^2\)
3. Forming the Square Number:
- Thus, the smallest square number is:
\[
2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900
\]
Conclusion
The smallest square number that is divisible by 6, 9, and 15 is indeed 900, confirming option A.
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What is the smallest square number which is divisible by each of the numbers 6,9 and 15 ?a)900b)810 c)630d)730Correct answer is option 'A'. Can you explain this answer?
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