AMC 10 · 2005 · #25
Grade 5 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question asks for the largest set, so Tool #14 (Extreme Principle) frames the whole attack: find a hard ceiling that no set can beat, then build one set that reaches it. Tool #4 (Introduce a Variable) names a general number x so its forbidden partner 125-x can be written down and the boundary x ≥ 25 solved once for all. The ban links numbers in couples (x, 125-x), so Tool #7 (Identify Subproblems) splits the pool into two clean groups — numbers whose forbidden partner is out of range (always safe) and numbers that come in banned couples. Tool #2 (Make a Systematic List) writes out those couples so they can be counted exactly, since from each couple at most one number may be kept.
Find each number's forbidden partner
Two numbers clash only if they add to 125, so x's sole forbidden partner is 125-x, which lies in 1 to 100 exactly when x ≥ 25.
A number is only risky if the exact number that would complete the sum of 125 actually exists in the pool.
3.NBT.A.2Introduce A VariableSplit the pool into safe and paired
So the 24 numbers 1 through 24 are always safe (their partners exceed 100), leaving only 25 through 100 to decide about.
Handle the risk-free numbers first so only the tricky ones are left to reason about.
4.OA.A.3Identify SubproblemsList and count the banned couples
Pair those up: (25,100), (26,99), …, (62,63). The smaller members run 25 to 62, so there are 38 couples.
Every risky number belongs to one and only one couple, so lining them up turns the ban into a simple count of couples.
Every risky number belongs to one and only one couple, so the ban becomes a simple count of couples.
▸ Why?
Each number has exactly one partner that would complete the forbidden sum, so the couples never overlap.
▸ Why?
Only one member of each couple can be kept, so the number of couples caps how many survive.
Apply the ceiling: at most one per couple
Each couple seats at most one member, so nothing beats 24 + 38 = 62; a 63rd pick would force some couple to send both.
Two numbers can share a couple but only one seat in B, so the couples set the hard ceiling.
4.OA.A.3Extreme PrincipleBuild a set that reaches 62
And B = {1, 2, …, 62} reaches it: its top two give 61 + 62 = 123, short of 125, so all 62 fit — choice (C).
The 62 smallest numbers are so small that even the top two fall short of 125, so all of them fit at once.
3.NBT.A.2Extreme PrincipleMatch up the numbers that would break the rule, keep just one from each pair plus all the numbers too small to ever pair up, and the smallest 62 numbers turn out to be a set that fits.
- Find each number's forbidden partner
- Split the pool into safe and paired
- List and count the banned couples
- Apply the ceiling: at most one per couple
- Build a set that reaches 62