Question 499201: determine the smallest 6 digit number which is exactly divisible by 8,15,and21
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! What are the factors of 8?
1, 2, 4
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What are the factors of 15?
1, 3, 5
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What are the factors of 21?
1, 3, 7
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The domain set is: {1, 2, 4, 3, 5, 7}
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What is the least common multiple that is equal to or greater than 21 (the largest number)?
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1*2*4*3*5*7 = 840
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840/8 = 105
840/15 = 56
840/21 = 40
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So, 840 is looking good!
We can check 420, but we'll find it does not work for 8.
420/8 is not an integer solution.
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But what does the question ask? The smallest 6-digit number that is exactly divisible by 8, 15, and 21.
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How about 840,000?
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840,000/8 = 105,000
840,000/15 = 56,000
840,000/21 = 40,000
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That works, but is it the smallest 6-digit number?
Not likely.
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Looking back, 8*15*21*x, where x is an integer, will always be divisible exactly by 8, 15 & 21.
We want to find an x such that the product is > 100,000 by the smallest margin.
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8*15*21=2520
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2520x >= 100,000
x >= 39.6824397
The next integer is 40.
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40*2520 = 100,800
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100,800/8 = 12,600
100,800/15 = 6,720
100,800/21 = 4,800
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The smallest 6-digit number (that is, a number >= 100,000) that is exactly divisible by 8, 15, and 21 is 100,800.
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Done.
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