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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

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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Start your 7-day free trialif "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

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.

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.

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

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

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.

Math / Science

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