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Q: Use a division ladder to find the gcf of each setof numbers 2730?

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The LCM of the given two numbers is 2730

Numbers that are divisable by 2730 include: 5460, 8190, 10920, 13650, 16380, 19110, 21840, 24570, 27300, 30030 etc. basicly the whole of the 2730 times table

6 x 13 x 35 = 2730

The GCF of the given two numbers is 273

2730 = 1 x 2730 2730 = 2 x 1365 2730 = 3 x 910 2730 = 5 x 546 2730 = 6 x 455 2730 = 7 x 390 2730 = 10 x 273 2730 = 13 x 210 2730 = 14 x 195 2730 = 15 x 182 2730 = 21 x 130 2730 = 26 x 105 2730 = 30 x 91 2730 = 35 x 78 2730 = 39 x 70 2730 = 42 x 65

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LCM(65, 42) = 2730.

Well for starters, you should do your own math homework, and perhaps phrase the question intelligibly. Assuming that you mean "what is the sum of three consecutive numbers whose product is 2730?": Take any of the numbers and call it X. In this example, we'll say X is our smallest number. We know that the numbers are consecutive, so if X is the smallest, then the other two numbers are X + 1 and X + 2. We also know that their product is 2730, so we can take those values and assemble them into an equation like this: (x)(x + 1)(x + 2) = 2730 At this point, there are two ways to solve this. The simplest way is somewhat intuitive though. We know that we are looking for a number that is slightly less than the cube root of 2730. The cube root of 2730 is approximately 13.976, so if we guess that our smallest number in that set is 13, then the other two numbers would be 14 and 15. If you multiply those three numbers together, you get 2730, so that is indeed the value you need. Their sum then, is 13 + 14 + 15, which equals 42. Of course, finding the answer that way is conveniently easy in homework land, but if they didn't come to such a beautifully clean answer, you'd have to solve it differently.

As a product of its prime factors: 2*3*5*7*13 = 2730

35x78=2730

It is a positive integer.

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