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What is the largest number of 4-digit exactly divisible by 12, 15, 18 & 27?
- a)9900
- b)9909
- c)9990
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
What is the largest number of 4-digit exactly divisible by 12, 15, 18 ...
LCM of 12, 15, 18, 27 = 180
Largest number of 4-digit = 9999
Required number = 9999 – 99 = 9900
Largest number of 4-digit = 9999
Required number = 9999 – 99 = 9900
Most Upvoted Answer
What is the largest number of 4-digit exactly divisible by 12, 15, 18 ...
To find the largest number of 4 digits that is divisible by 12, 15, and 18, we need to find the least common multiple (LCM) of these three numbers.
The prime factorization of 12 is 2^2 * 3.
The prime factorization of 15 is 3 * 5.
The prime factorization of 18 is 2 * 3^2.
To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:
2^2 * 3^2 * 5 = 180
So, the LCM of 12, 15, and 18 is 180.
To find the largest 4-digit number divisible by 180, we divide 9999 by 180:
9999 ÷ 180 = 55.55 (repeating)
Since we are looking for the largest whole number, we round down to 55.
Therefore, the largest number of 4 digits exactly divisible by 12, 15, and 18 is 55 multiplied by 180:
55 * 180 = 9900
So, the largest number of 4 digits exactly divisible by 12, 15, and 18 is 9900.
The prime factorization of 12 is 2^2 * 3.
The prime factorization of 15 is 3 * 5.
The prime factorization of 18 is 2 * 3^2.
To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:
2^2 * 3^2 * 5 = 180
So, the LCM of 12, 15, and 18 is 180.
To find the largest 4-digit number divisible by 180, we divide 9999 by 180:
9999 ÷ 180 = 55.55 (repeating)
Since we are looking for the largest whole number, we round down to 55.
Therefore, the largest number of 4 digits exactly divisible by 12, 15, and 18 is 55 multiplied by 180:
55 * 180 = 9900
So, the largest number of 4 digits exactly divisible by 12, 15, and 18 is 9900.
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