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Tom [10]
3 years ago
13

Determine the smallest 6 digit number which is exactly divisible by 8,15,and21

Mathematics
1 answer:
ch4aika [34]3 years ago
3 0
Find the least common multiple of 8, 15, and 21. . . . . . It's 280 .

So all the numbers divisible by 8, 15, and 21 are all the multiples of 280.

The smallest 6-digit number is 100,000 .

100,000 / 280 = 357.1 , so the next whole-number multiple is the 358th one.

280 x 358 = <u>100,240</u> .

You can tell right away that this is the smallest one that qualifies,
because if you take away 280 and go to the next smaller multiple,
it won't have 6 digits anymore.
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Now remember we have our foci at the point (0, ± 3). Our major axis length is 10, so if covers 5 units on either side of the x - axis. Therefore, c = 3, and a = 5. But remember that c is not part of an ellipse equation. Take a look at the formula below,

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<u>Our solution is hence option number 1</u> : \frac{x^2}{25}+\frac{y^2}{16}=1

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