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
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24 is in X because 7 and 17 are in X

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mustish1Author Commented:
The answer which i have between 3 and 7 is

I dont know how 2 get these numbers
if 3 and 14 are in X, so is 3+14
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mustish1Author Commented:
can you please tell me in more simple way how to get each number
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
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

So start with 3 then add 3 on 4 times,  then add 7 twice.
mustish1Author Commented:
if i understand correctly


Another answer is 6, 10 and 14, 15 I copy it from the board but i dont get the question.
mustish1Author Commented:
The actual question parts are
a. 24
b. 11
mustish1Author 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
Those are all correct applications of R.
... 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

and 1000000 because all powers of 10 are in.
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.
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
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