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What is the smallest number which is divisible by 18, 24, 32.

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The smallest number divisible by 18, 24, 32 is 1152. Initially, determine the smallest number that can be evenly divided by 18, 24, and 32. Proceed by incrementing this number until reaching a multiple that is a four-digit number. This resulting four-digit number represents the smallest number divisible by 18, 24, and 32.

Smallest Number Divisible

To find the smallest number divisible by 18, 24, and 32, we need to determine their least common multiple (LCM).

The prime factorization of 18 is 2 * 32.

The prime factorization of 24 is 23 * 3.

The prime factorization of 32 is 25.

Steps to Find the Smallest Number Divisible

Solution:

To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:

25 * 32 = 32 * 9 = 288.

So, the smallest number divisible by 18, 24, and 32 is 288.

To find the smallest four-digit number divisible by 18, 24, and 32, we need to increase 288 to its multiple until we reach a four-digit number.

Starting with 288, we can multiply it by 2:

288 * 2 = 576 (not a four-digit number)

Next, we multiply it by 3:

288 * 3 = 864 (not a four-digit number)

Continuing, we multiply it by 4:

288 * 4 = 1152 (a four-digit number)

Therefore, the smallest four-digit number divisible by 18, 24, and 32 is 1152.

Summary:

What is the smallest number which is divisible by 18,24,32.

The smallest number divisible by 18, 24, 32 is 1152. This resulting four-digit number is the smallest one that can be evenly divided by 18, 24, and 32.

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