Nonprime numbers

[bdelaughter]

Write down 10 consecutive nonprime numbers in standard form such that the first has a factor of 2, the second a factor of 3, the third a factor of 4, and so on with the last having a factor of 11. Explain why you know that for any positive integer n, there exists n consecutive numbers that are composite.[/i]

 

[KaieraAi]

Nonprime numbers are any numbers that can be calculated (through multiplication) by more than just 1 and themselves. For example, 1 is a prime number, becasue it can only be multiplied by itself to get itself. So is 2, and 3. 4 isn't, since it can be made through 1 times 4, or 2 times 2. (If you get where I'm going...) So when factoring, (do you use a factor tree?) you are looking for numbers that use the numbers 2 through 11 somwhere in their line of multiplication. (So you could use 4 for the first one, becasue it uses 2.)

 

[lookagain]

For example, 1 is a prime number, becasue it can only be multiplied by itself to get itself.
So is 2, and 3. 4 isn't, since it can be made through 1 times 4, or 2 times 2.

KaieraAi,

1 is not a prime number.

A prime number is a positive integer, such that it has exactly two distinct positive factors, 1 and itself.


\(\displaystyle 2, 3 , 5 , 7 , 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 \\)

\(\displaystyle \ are \ the \ first \ 25\ prime \ numbers.\)

 

[KaieraAi]

Sorry... ^^; I haven't done this in a few years, my bad.

 

[soroban]

Hello, bdelaughter!

Does anyone remember the original question?

There is no need to ask whether someone remembers the question.
It's still right there for everyone to read. It hasn't gone away.

This is not an appropriate question to ask, *especially* with sarcasm as I was being
corrective concerning the topic. Feel free to delete the question anytime.

 

[lookagain]

Write down 10 consecutive nonprime numbers in standard form such that the first has a factor of 2, the second a factor of 3, the third a factor of 4, and so on with the last having a factor of 11. Explain why you know that for any positive integer n, there exists n consecutive numbers that are composite.[/i]

bdelaughter,

here are 10 consecutive composite numbers which have your properties:

11! + 2
11! + 3
11! + 4
.
.
.
11! + 11

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Suppose n is a positive integer.

Let n > 9

In general, the following are divisible 2, 3, 4, ..., n - 1, n, n + 1 . . . .

There are n numbers here, as well as here:

(n + 1)! + 2
(n + 1)! + 3
(n + 1)! + 4
.
.
.
(n + 1)! + (n - 1)
(n + 1)! + n

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2, 3, 4, ..., (n - 1), n, (n + 1) all divide (n + 1)!,
so they divide the sum of (n + 1)! + one of those factors