AMC 10 · 2005 · #25
Easy mode Grade 5Choose some of the whole numbers from 1 to 100 to put into a group B. The only rule is that no two numbers in the group may add up to 125. You want the group to hold as many numbers as possible. What is the largest number of numbers the group B can have?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: From the whole numbers $1$ through $100$, choose as large a collection $B$ as possible so that no two chosen numbers add up to $125$. Find the greatest number of members $B$ can have.
Givens: The pool of numbers is every integer from $1$ to $100$; $B$ is a subset chosen from that pool; The forbidden condition: no two members of $B$ may sum to $125$; Answer choices: (A) $50$, (B) $51$, (C) $62$, (D) $65$, (E) $68$
Unknowns: The maximum possible number of elements in $B$
Understand
Restated: From the whole numbers $1$ through $100$, choose as large a collection $B$ as possible so that no two chosen numbers add up to $125$. Find the greatest number of members $B$ can have.
Givens: The pool of numbers is every integer from $1$ to $100$; $B$ is a subset chosen from that pool; The forbidden condition: no two members of $B$ may sum to $125$; Answer choices: (A) $50$, (B) $51$, (C) $62$, (D) $65$, (E) $68$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #2 Make a Systematic List
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 $x\ge 25$ 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.
Execute — Answer: C
3.NBT.A.2 Step 1 Find each number's forbidden partner
- Two members clash only when they add to $125$.
- So for any number $x$, the single number it may not sit beside is $125-x$.
- A number is dangerous only if that partner $125-x$ is itself in the pool $1$ to $100$.
- Solve $125-x\le 100$: this gives $x\ge 25$.
- So a number needs a partner inside the pool exactly when it is $25$ or larger.
💡 A number is only risky if the exact number that would complete the sum of $125$ actually exists in the pool.
4.OA.A.3 Step 2 Split the pool into safe and paired
- Numbers $1$ through $24$ are completely safe: for each, the partner $125-x$ is $101$ or more, which is outside the pool, so these numbers can never help form a sum of $125$.
- All $24$ of them may be kept with no risk.
- That leaves the numbers $25$ through $100$ — the $76$ numbers that do come in banned couples — as the only place a choice has to be made.
💡 Handle the risk-free numbers first so only the tricky ones are left to reason about.
5.OA.B.3 Step 3 List and count the banned couples
- Pair each number in $25$ through $100$ with its partner that sums to $125$: $(25,100),\,(26,99),\,(27,98),\,\ldots,\,(62,63)$.
- Every one of the $76$ numbers appears in exactly one couple.
- The smaller members run $25,26,\ldots,62$, so the number of couples is $62-25+1=38$.
💡 Every risky number belongs to one and only one couple, so lining them up turns the ban into a simple count of couples.
4.OA.A.3 Step 4 Apply the ceiling: at most one per couple
- From each banned couple, keeping both numbers would make a sum of $125$, so at most one number from each couple can be in $B$.
- That caps the paired region at $38$ numbers.
- Adding the $24$ safe numbers, no set can exceed $24+38=62$.
- This is a firm ceiling: $63$ numbers from a total of $24$ singles plus $38$ couples would force two picks out of some couple, breaking the rule.
💡 Two numbers can share a couple but only one seat in $B$, so the couples set the hard ceiling.
3.NBT.A.2 Step 5 Build a set that reaches 62
- Take $B=\{1,2,3,\ldots,62\}$.
- The two largest members are $61$ and $62$, whose sum is $61+62=123$, which is less than $125$, so no two members of this set can reach $125$.
- This set has exactly $62$ members and breaks no rule, so the ceiling of $62$ is actually reached.
- The maximum possible number of elements is $62$, 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.2 Two members clash only when they add to $125$. So for any number $x$, the single 4.OA.A.3 Numbers $1$ through $24$ are completely safe: for each, the partner $125-x$ is $ 5.OA.B.3 Pair each number in $25$ through $100$ with its partner that sums to $125$: $(25 4.OA.A.3 From each banned couple, keeping both numbers would make a sum of $125$, so at m 3.NBT.A.2 Take $B=\{1,2,3,\ldots,62\}$. The two largest members are $61$ and $62$, whose s Review
Reasonableness: The two views agree: the couples argument says no set can beat $62$, and the concrete set $\{1,\ldots,62\}$ hits $62$, so $62$ is both the ceiling and reachable. A quick check of the boundary couple $(62,63)$ confirms it: $62+63=125$, so $62$ and $63$ may not both appear — keeping $62$ and dropping $63$ is exactly what $\{1,\ldots,62\}$ does. The trap answer (A) $50$ comes from only splitting into odd/even or stopping at half of $100$; it undercounts because the $24$ small safe numbers plus one from each couple beat any even split. Choices (D) $65$ and (E) $68$ exceed the $62$ ceiling and would force a banned pair, so they are impossible.
Alternative: Think of it as filling seats greedily from the bottom up. Add $1,2,3,\ldots$ in order; each new number $n$ is safe as long as its partner $125-n$ has not already been taken. Partners only start landing inside the range once $n$ reaches $63$ (whose partner $62$ is already in), so every number from $1$ to $62$ goes in freely and $63$ is the first that must be refused. That again gives $62$ members without ever listing all the couples.
CCSS standards used (min grade 5)
3.NBT.A.2Fluently add and subtract within 1000 (Finding each number's forbidden partner $125-x$ and checking sums such as $61+62=123$ and $62+63=125$ to test which numbers can safely sit together.)5.OA.B.3Generate two numerical patterns using two given rules and identify relationships (Generating the couples $(25,100),(26,99),\ldots,(62,63)$ as two paired sequences that each sum to $125$, then counting that there are $38$ of them.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Splitting the pool into $24$ safe numbers and $38$ couples, capping the couples at one pick each, and combining $24+38=62$ for the maximum.)
⭐ Match 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.
⭐ Match 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.
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