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The greatest four digit number which is divisible by 18, 25, 30, and 48 is: Options : A] 9000 B] 9200 C] 7200 D] 9729

Question

The greatest four digit number which is divisible by 18, 25, 30, and 48 is:

Options : A] 9000 B] 9200 C] 7200 D] 9729

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Solution 1

To find the greatest four-digit number divisible by 18, 25, 30, and 48, we need to find the least common multiple (LCM) of these numbers.

Step 1: Prime factorization of the numbers 18 = 2 * 3^2 25 = 5^2 30 = 2 * 3 * 5 48 = 2^4 * 3

Step 2: Find the LCM The LCM is the product of the highest powers of all the factors. LCM = 2^4 * 3^2 * 5^2 = 1800

Step 3: Find the greatest four-digit number divisible by the LCM The greatest four-digit number is 9999. Divide 9999 by the LCM (1800) and find the remainder.

9999 ÷ 1800 = 5 remainder 999

Subtract the remainder from 9999 to find the largest four-digit number divisible by 1800.

9999 - 999 = 9000

So, the greatest four-digit number which is divisible by 18, 25, 30, and 48 is 9000.

Therefore, the answer is option A] 9000.

This problem has been solved

Solution 2

To find the greatest four-digit number divisible by 18, 25, 30, and 48, we need to find the least common multiple (LCM) of these numbers.

Step 1: Prime factorization of the numbers 18 = 2 * 3^2 25 = 5^2 30 = 2 * 3 * 5 48 = 2^4 * 3

Step 2: Find the LCM The LCM is the product of the highest powers of all the factors. LCM = 2^4 * 3^2 * 5^2 = 3600

Step 3: Find the greatest four-digit number divisible by the LCM The greatest four-digit number is 9999. Divide 9999 by the LCM (3600) and find the remainder.

9999 ÷ 3600 = 2 remainder 2799

Subtract the remainder from 9999 to find the largest four-digit number that is divisible by the LCM.

9999 - 2799 = 7200

So, the greatest four-digit number which is divisible by 18, 25, 30, and 48 is 7200.

Therefore, the correct option is C] 7200.

This problem has been solved

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