AMC 10 · 2005 · #25
Grade 5 arithmeticA subset B of the set of integers from 1 to 100, inclusive, has the property that no two elements of B sum to 125. What is the maximum possible number of elements in B?
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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