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determine the number nearest to 110000 but greater than 100000 which is exactly divisible by 8,15 and 21

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Answer:

Explanation:

To find the number nearest to 110,000 but greater than 100,000 that is exactly divisible by 8, 15, and 21, we need to find the least common multiple (LCM) of these three numbers and then round it up to the nearest whole number.

The LCM of 8, 15, and 21 is the smallest positive integer that is divisible by all three numbers without any remainder.

Prime factorization of the given numbers:

8 = 2^3

15 = 3 * 5

21 = 3 * 7

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

LCM = 2^3 * 3 * 5 * 7

LCM = 8 * 3 * 5 * 7

LCM = 840

Now, we need to find the number nearest to 110,000 but greater than 100,000, which is divisible by 840.

Dividing 100,000 by 840 gives us 119 with a remainder of 440.

To find the next multiple of 840 greater than 100,000, we add the remainder (440) to 100,000:

100,000 + 440 = 100,440

So, the number nearest to 110,000 but greater than 100,000, which is exactly divisible by 8, 15, and 21, is 100,440.

User Eggs McLaren
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