AMC 8 · 2006 · #9
Easy mode Grade 5Problem
Look at this long string of fractions multiplied together:
23×34×45×⋯×20052006
The pattern continues all the way up: each fraction has a top number that is one bigger than its bottom number. The very last fraction is 20052006.
What is the value of this whole product?
Pick an answer.
AMC 8 2006 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: Find the value of the product $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$, where each factor is a fraction whose numerator is one more than its denominator and whose denominator matches the previous factor's numerator.
Givens: The product is $\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\cdots\times\dfrac{2006}{2005}$; Each factor has the form $\dfrac{k+1}{k}$ with $k$ running from $2$ to $2005$; There are $2004$ factors in total; Answer choices: (A) $1$, (B) $1002$, (C) $1003$, (D) $2005$, (E) $2006$
Unknowns: The value of the full product
Understand
Restated: Find the value of the product $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$, where each factor is a fraction whose numerator is one more than its denominator and whose denominator matches the previous factor's numerator.
Givens: The product is $\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\cdots\times\dfrac{2006}{2005}$; Each factor has the form $\dfrac{k+1}{k}$ with $k$ running from $2$ to $2005$; There are $2004$ factors in total; Answer choices: (A) $1$, (B) $1002$, (C) $1003$, (D) $2005$, (E) $2006$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Try an Easier Problem
Multiplying out $2004$ fractions directly is impossible, so the problem is asking us to spot structure. Tool #5 (Look for a Pattern) is the natural primary: each factor's numerator matches the next factor's denominator, which is the signature of a telescoping product where most terms cancel. To make the cancellation visible without writing all $2004$ factors, Tool #9 (Try an Easier Problem) is the perfect helper — first compute the same product for a few terms, see what survives, and trust the pattern to extend.
Execute — Answer: C
5.NF.B.4 Step 1 - Start with a shorter version of the product to expose the pattern.
- Take just the first three factors and write the result as one fraction.
💡 Trying a shorter version is the Grade 5 "multiply fractions" move — small enough to do by hand, big enough to see what happens.
4.OA.C.5 Step 2 - Notice what cancelled.
- The $3$ and $4$ each appeared once on top and once on bottom, so they disappeared.
- Only the very first denominator ($2$) and the very last numerator ($5$) survived.
- That is the pattern.
💡 Spotting that interior numbers appear once on top and once on bottom is the Grade 4 "generate and identify patterns" move.
5.NF.B.4 Step 3 - Apply the same pattern to the full product.
- Every interior number from $3$ to $2005$ shows up once as a numerator and once as a denominator, so it cancels.
- The only survivors are the first denominator $2$ and the last numerator $2006$.
💡 The cancellation rule extends from the short product to the long one because nothing structural changes — only the length.
5.NBT.B.6 Step 4 Divide to finish.
💡 A single Grade 5 division turns the telescoped fraction into the answer.
5.NF.B.4 Start with a shorter version of the product to expose the pattern. Take just the 4.OA.C.5 Notice what cancelled. The $3$ and $4$ each appeared once on top and once on bot 5.NF.B.4 Apply the same pattern to the full product. Every interior number from $3$ to $2 5.NBT.B.6 Divide to finish. Review
Reasonableness: Check the pattern against one more easy case. The product $\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}$ should leave $\dfrac{6}{2} = 3$. Multiplying directly: $\dfrac{3\cdot 4\cdot 5\cdot 6}{2\cdot 3\cdot 4\cdot 5} = \dfrac{360}{120} = 3$. Match. The rule "answer = last numerator $\div$ first denominator" is confirmed, and applying it to the contest product gives $2006/2 = 1003$. Also, the answer must be slightly larger than $1000$ since each factor is just above $1$ but there are $2004$ of them; $1003$ fits while $1$, $2005$, and $2006$ do not.
Alternative: Tool #4 (Introduce a Variable): write the product in compact form as $P = \prod_{k=2}^{2005}\dfrac{k+1}{k}$. The numerator is $3\cdot 4\cdots 2006 = \dfrac{2006!}{2!}$ and the denominator is $2\cdot 3\cdots 2005 = \dfrac{2005!}{1!}$, so $P = \dfrac{2006!/2!}{2005!/1!} = \dfrac{2006}{2} = 1003$. Same answer, just dressed up in factorial notation.
CCSS standards used (min grade 5)
5.NF.B.4Apply and extend previous understandings of multiplication to multiply fractions (Multiplying the short product $\tfrac{3}{2}\cdot\tfrac{4}{3}\cdot\tfrac{5}{4}$ by combining numerators and denominators, the same rule that extends to the full $2004$-factor product.)4.OA.C.5Generate a number or shape pattern that follows a given rule and identify apparent features of the pattern (Recognizing that every interior number appears once on top and once on bottom, so only the first denominator and last numerator survive.)5.NBT.B.6Find whole-number quotients of whole numbers (Computing $2006 \div 2 = 1003$ to turn the telescoped fraction into the final integer answer.)
⭐ When every fraction's top matches the next fraction's bottom, almost everything cancels. Only the first bottom and the last top survive — here that gives $2006 \div 2 = 1003$.
⭐ When every fraction's top matches the next fraction's bottom, almost everything cancels. Only the first bottom and the last top survive — here that gives $2006 \div 2 = 1003$.
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