AMC 10 · 2011 · #9
Easy mode Grade 5A room holds 9 pairs of twins and 6 groups of triplets. Every pair and every group comes from a different family. Each twin shakes hands with every twin who is not in his or her own family, and with half of the triplets. Each triplet shakes hands with every triplet who is not in his or her own family, and with half of the twins. How many handshakes happen in all?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A convention holds $9$ sets of twins and $6$ sets of triplets, every set from a different family. Each twin shakes hands with every twin outside his or her own family, plus exactly half of the triplets. Each triplet shakes hands with every triplet outside his or her own family, plus exactly half of the twins. Count how many handshakes happened in all.
Givens: $9$ sets of twins, $2$ people per set; $6$ sets of triplets, $3$ people per set; All $15$ sets come from different families, so siblings only ever sit inside one set; Each twin shakes hands with every non-sibling twin and with half of the triplets; Each triplet shakes hands with every non-sibling triplet and with half of the twins
Unknowns: The total number of handshakes that took place
Understand
Restated: A convention holds $9$ sets of twins and $6$ sets of triplets, every set from a different family. Each twin shakes hands with every twin outside his or her own family, plus exactly half of the triplets. Each triplet shakes hands with every triplet outside his or her own family, plus exactly half of the twins. Count how many handshakes happened in all.
Givens: $9$ sets of twins, $2$ people per set; $6$ sets of triplets, $3$ people per set; All $15$ sets come from different families, so siblings only ever sit inside one set; Each twin shakes hands with every non-sibling twin and with half of the triplets; Each triplet shakes hands with every non-sibling triplet and with half of the twins
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #15 Organize Information in More Ways, #2 Make a Systematic List
Every handshake has two people in it, so it is twin-with-twin, triplet-with-triplet, or twin-with-triplet, and never two of those at once. That splits one tangled count into three clean ones. Inside each single group it is easier to count all pairs and subtract the sibling pairs than to count directly. The twin-with-triplet group needs more care: the problem describes it twice, once from each side, so I read it both ways and check the two readings give the same number, then build one pattern that actually obeys both rules.
Execute — Answer: B
3.OA.A.1 Step 1 Count the people in each group
- Sets of twins and sets of triplets are equal groups, so the head counts are products.
- $9$ sets of $2$ give $9 \times 2 = 18$ twins, and $6$ sets of $3$ give $6 \times 3 = 18$ triplets, for $36$ people in all.
- The two groups turn out to be the same size, which matters later.
- A twin has $1$ sibling; a triplet has $2$.
💡 Equal groups of people mean the head count is a multiplication, not a list.
4.OA.A.3 Step 2 Sort handshakes into three kinds
- Pick any handshake.
- Its two people are either both twins, both triplets, or one of each.
- Those three cases cover every handshake and no handshake fits two cases, so counting each kind on its own and adding the three counts gives every handshake exactly once.
- Now I have three smaller problems instead of one large one.
💡 Buckets that never overlap and never leave anything out let me add the bucket counts safely.
5.NBT.B.5 Step 3 Twin-twin: all pairs minus siblings
- Among the $18$ twins, everyone shakes everyone except siblings.
- First count all pairs: each of the $18$ can pair with the other $17$, giving $18 \times 17 = 306$ person-to-person records, but each pair shows up twice that way, so there are $\frac{306}{2} = 153$ pairs.
- Exactly one pair per set of twins is a sibling pair, so $9$ pairs never shake.
- That leaves $153 - 9 = 144$ twin-twin handshakes.
💡 Counting everything and removing the few forbidden pairs beats chasing who shakes whom.
5.NBT.B.6 Step 4 Triplet-triplet: subtract more sibling pairs
- The $18$ triplets also make $\frac{18 \times 17}{2} = 153$ pairs.
- This time each set has $3$ siblings, and $3$ people make $\frac{3 \times 2}{2} = 3$ sibling pairs, so each of the $6$ sets blocks $3$ pairs: $6 \times 3 = 18$ blocked pairs in all.
- That gives $153 - 18 = 135$ triplet-triplet handshakes.
- Same group size as the twins, but more handshakes are lost because triplet families are bigger.
💡 A family of three hides three sibling pairs, not one, so bigger families remove more handshakes.
5.NF.B.4 Step 5 Cross handshakes, read from both sides
- Read the twin-with-triplet handshakes from the twin side first.
- Half of $18$ triplets is $\frac{1}{2} \times 18 = 9$, so each of the $18$ twins is in $9$ of them, giving $18 \times 9 = 162$ records.
- Every such handshake has exactly one twin in it, so it makes exactly one record here: there are $162$ cross handshakes.
- Now read the same handshakes from the triplet side: each of the $18$ triplets is in $\frac{1}{2} \times 18 = 9$ of them, again $18 \times 9 = 162$.
- The two readings match.
- That agreement is what keeps the problem from contradicting itself, since the twin rule and the triplet rule are two descriptions of one and the same set of handshakes.
💡 Counting one pile from two directions is only safe when the two totals agree, so checking them is the point.
4.OA.C.5 Step 6 Check such a convention can exist
- The rules say each twin picks some half of the triplets but never say which half, so I should check a legal pattern exists at all, and that my count does not depend on the choice.
- Number the twins $1$ to $18$ and the triplets $1$ to $18$, and let twin $k$ shake hands with triplets $k, k+1, \ldots, k+8$, wrapping back to $1$ after $18$.
- Each twin meets $9$ triplets by construction, and each triplet $m$ is met by exactly the $9$ twins $m-8, \ldots, m$, so both rules hold.
- Any other legal pattern still gives $162$ cross handshakes, because that number came from the twin side alone, not from which triplets got chosen.
💡 A sliding window of nine wraps around evenly, so every triplet gets chosen by exactly nine twins.
4.NBT.B.4 Step 7 Add the three kinds
- The three buckets hold $144$ twin-twin handshakes, $135$ triplet-triplet handshakes, and $162$ twin-triplet handshakes, and every handshake sits in exactly one bucket.
- Adding gives $144 + 135 + 162 = 441$ handshakes, which is choice (B).
💡 Once the buckets do not overlap, the total is just their sum.
3.OA.A.1 Sets of twins and sets of triplets are equal groups, so the head counts are prod 4.OA.A.3 Pick any handshake. Its two people are either both twins, both triplets, or one 5.NBT.B.5 Among the $18$ twins, everyone shakes everyone except siblings. First count all 5.NBT.B.6 The $18$ triplets also make $\frac{18 \times 17}{2} = 153$ pairs. This time each 5.NF.B.4 Read the twin-with-triplet handshakes from the twin side first. Half of $18$ tri 4.OA.C.5 The rules say each twin picks some half of the triplets but never say which half 4.NBT.B.4 The three buckets hold $144$ twin-twin handshakes, $135$ triplet-triplet handsha Review
Reasonableness: Bound the answer from above. With $36$ people, the most handshakes possible is $\frac{36 \times 35}{2} = 630$, which is exactly choice (C) and would mean nobody sits any handshake out. Since $9 + 18 = 27$ sibling pairs never shake and half of the $18 \times 18 = 324$ twin-triplet pairs never meet, the answer has to be $630 - 27 - 162 = 441$, matching the main count. That upper bound of $630$ also kills two options outright: $648$ and $882$ are both larger than the number of pairs that exist. And $324$ is just the count of all twin-triplet pairs, the number you get by looking only at the cross handshakes.
Alternative: Count from people instead of from pair types. Each twin shakes $16$ non-sibling twins plus $9$ triplets, so $25$ handshakes; each triplet shakes $15$ non-sibling triplets plus $9$ twins, so $24$. Adding over all $36$ people counts every handshake once for each of its two people, so the total is $\frac{18 \times 25 + 18 \times 24}{2} = \frac{450 + 432}{2} = \frac{882}{2} = 441$. This route sums over people while the main route sums over kinds of pair, and they agree. Note that $882$, the sum before halving, is one of the choices, so forgetting the halving lands on a listed wrong answer.
CCSS standards used (min grade 5)
3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Turning $9$ sets of $2$ and $6$ sets of $3$ into head counts of $18$ twins and $18$ triplets.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Splitting every handshake into twin-twin, triplet-triplet, or twin-triplet, and planning to add the three counts.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing $18 \times 17 = 306$ pairs among the twins and $18 \times 9 = 162$ cross handshakes.)5.NBT.B.6Find whole-number quotients with up to four-digit dividends and two-digit divisors (Halving $18 \times 17$ to get $153$ pairs and halving $3 \times 2$ to get the $3$ sibling pairs inside a triplet set.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Reading 'half the triplets' as $\frac{1}{2} \times 18 = 9$ from each side of the cross handshakes.)4.OA.C.5Generate a number or shape pattern following a given rule (Building the wrap-around pattern that shows a handshake plan obeying both halves-rules really exists.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Removing sibling pairs from the pair counts and adding $144 + 135 + 162$ at the end.)
⭐ Sort the handshakes by who is in them, count all pairs and subtract the ones that never happen, and remember each handshake belongs to two people but happens only once.
⭐ Sort the handshakes by who is in them, count all pairs and subtract the ones that never happen, and remember each handshake belongs to two people but happens only once.
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