which of the following numbers are in x

Define a set X recursively as follows.

B. 3 and 7 are in X.
R. If x and y are in X, so is x+y. (Here it is possible that x=y.)

Decide which of the following numbers are in X. Explain each decision.

a. 24
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Commented:
24 is in X because 7 and 17 are in X
0

Experts Exchange Solution brought to you by

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Author Commented:
The answer which i have between 3 and 7 is
6,9,10,13,14,17

I dont know how 2 get these numbers
0
Commented:
if 3 and 14 are in X, so is 3+14
0
Author Commented:
can you please tell me in more simple way how to get each number
0
Commented:
The only ways to get numbers are by application of B. and R.
if "more simple" means "fewer applications of R" then you may prefer to use larger
x and y when you apply R
Another "simple" observation might be that since
12,13, and 14 are all in X, then all integers >= 12 are in X
0
Commented:
If 3 in x
then  3 + 3 = 6    in x
then  6 + 6 = 12  in x
then  12 + 12 = 24  in x

The set basically comprises of numbers of the form

3n + 7m  where n,m integer.

eg for 29

3*5 + 2*7 =29

0
Author Commented:
if i understand correctly
3+3=6
6+3=9
9+3=12
12+3=15

than
3+7=10
7+3=13

Another answer is 6, 10 and 14, 15 I copy it from the board but i dont get the question.
0
Author Commented:
The actual question parts are
a. 24
b. 11
0
Author Commented:
3 and 3 are in X ==> 3+3 = 6 is in X
6 and 3 are in X ==> 3+6 = 9 is in X
3 and 7 are in X ==> 3+7 = 10 is in X
10 and 3 are in X ==> 3+10 = 13 is in X
7 and 7 are in X ==> 7+7 = 14 is in X
10 and 7 are in X ==> 10+7=17 is in X
0
Commented:
Those are all correct applications of R.
0
Commented:
... I did not see that 11 was part of the question as well.

To show 11 in the set or not we must find integer m & n st  (see my answer above)

3n + 7m = 11

Clearly m=1 (or higher) won't work as we have 4 left over

Hence we need to find

3n = 11

Clearly integer n does not exist as 3 does not divide 11

hence 11 not the set

0
Commented:
and 1000000 because all powers of 10 are in.
0
Commented:
A 'simple' way to find all the numbers would be to build a table. Pull all the numbers you know are in X in the rows and columns put all the sums in the middle. Then add the new numbers do the row and column headers and go again. Keep doing this until you get three consecutive numbers all in X.

Since 3 is in the set, if you can find three consecutive numbers that are all in X, you can easily prove that all the integers greater than them are in X.
0
Commented:
You've already found that 13 and 14 are in the set, add 9 and 3 to put 12 in the set and then you know all the numbers > 14 are also in the set because every number is either 12+3x or 13+3x or 14+3x
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.