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Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15, and 21.

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105200

104200

106200

109200

answer is D.

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Detailed Solution

To be divisible by 8,15 and 21 a number should be divided by lowest common factor of these numbers.
Here, factors of the given numbers,

8 = 2 \times 2 \times 2

15 = 3 \times 5

21 = 3 \times 7

So, the lowest common factor,

LCM = 2 \times 2 \times 2 \times 3 \times 5 \times 7

LCM = 840

Now, we get 800 as reminder when 110000 is divided by 840.
So, the number near 110000 but greater than 100000 and divisible by 8,15 and

21

= 110000 - 840

=109200
Hence option 4 is correct.

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