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Q: Is it possible that 3 consecutivewhole numbers can be primefactors?

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3 and 5

1, 3, 11, 33 3 and 11 are prime.

2 and 3

90 = 2 x 3 x 3 x 5 = 2 x 32 x 5

Not possible in whole numbers

3 and 36

-2

No.

YES

Divide the number of possible even numbers by all the possible numbers. So there are three possible even numbers; 2, 4 & 6. And there are six possible numbers; 1, 2, 3, 4, 5, & 6. So the probability of throwing an even number is 3/6 = .5 .

Total number of possible 3-digit numbers = 9!x10!10!

Winning number for tonight

From {1, 2, 3, 4} there are two prime numbers (2, 3} which can go in the hundreds position, which once chosen allows 3 possible choices for the tens digit and a further 2 possible choices for the units digit, giving 2 x 3 x 2 = 12 possible numbers.

All whole numbers are rational because they can be expressed as fractions as for example 3 is 3/1 as an improper fraction.

3, 5 and 23

3

No. Because 2 are consecutive to 3.

456, 546 and 564 are three of the four possible even numbers.

1, 2.5 and 24 is one possible set of numbers.

1.5, 2.25 and 572.25, when added together is one possible set of numbers.

Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.

Yes. The mean of 1, 3, and 5 is 3. The mean is the average. Add all numbers and then divide by how many numbers were added. In this case there were three numbers (1, 3, 5). 1+3+5=9. 9/3=3.

There are 6 numbers on a die. Therefore there are 6 possible numbers that can be rolled. "3" is one of those numbers. The odds are therefore 1/6.

If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.

The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5