AMC 10 · 2018 · #12
Grade 4 number-theoryPick an answer.
The question asks for the least possible value of the smallest element, which is a minimize-the-boundary question — Tool #14 (Extreme Principle). The smart move is to test candidate smallest values from the bottom up (2, then 3, then 4) and stop at the first one that lets a full set of 6 be built. Tool #6 (Guess and Check): for each candidate smallest value, try to build a valid set of 6. Tool #2 (Make a Systematic List): after fixing the smallest element, list exactly which larger numbers are still allowed. Tool #3 (Eliminate Possibilities): the answer choices let us rule out 2 and 3, so the first choice that survives is the answer.
Search from the bottom up
One divides everything, so it is out immediately.
To find the smallest possible start, try the smallest values first and stop at the first that works.
4.OA.B.4Extreme PrincipleTry smallest =2
With two, the leftovers still contain a multiple pair.
Starting at 2 throws away every even number, leaving too few to reach six.
Starting at two throws away every even number, leaving too few to reach the required count.
▸ Why?
Every even number is a multiple of two, so keeping two forbids all of them at once.
▸ Why?
Too few candidates are left to fill the required slots, so the case dies before any arrangement is tried.
Try smallest =3
Three fails to reach six for the same reason.
Each divides-pair forces a sacrifice, so the allowed list keeps coming up one short.
4.OA.B.4Guess And CheckTry smallest =4 and finish
Four admits a real example, so the answer is 4.
Once 2 and 3 are ruled out, the first value that lets six fit is the answer.
4.OA.B.4Eliminate PossibilitiesTest the smallest start first: 2 and 3 leave too few non-multiples to reach six, but starting at 4 the set {4,5,6,7,9,11} fits, so the least element is 4, choice (C).
- Search from the bottom up
- Try smallest =2
- Try smallest =3
- Try smallest =4 and finish