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Posted on 2011-10-20

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

14 Comments

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

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.

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