AMC 10 · 2008 · #5
Easy mode Grade 5Multiply this long chain of fractions all the way to the end. Which answer choice does it equal?
48⋅812⋅1216⋅⋯⋅4n4n+4⋅⋯⋅20042008?
Pick an answer.
AMC 10 2008 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: Multiply the long chain of fractions $\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdots\frac{2008}{2004}$, where each fraction has the form $\frac{4n+4}{4n}$, and find which answer choice the product equals.
Givens: The product is $\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdots\frac{4n+4}{4n}\cdots\frac{2008}{2004}$; Every factor has the form $\frac{4n+4}{4n}$, so numerators and denominators both step up by $4$; The last factor is $\frac{2008}{2004}$; Answer choices: (A) $251$, (B) $502$, (C) $1004$, (D) $2008$, (E) $4016$
Unknowns: The single number the whole product simplifies to
Understand
Restated: Multiply the long chain of fractions $\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdots\frac{2008}{2004}$, where each fraction has the form $\frac{4n+4}{4n}$, and find which answer choice the product equals.
Givens: The product is $\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdots\frac{4n+4}{4n}\cdots\frac{2008}{2004}$; Every factor has the form $\frac{4n+4}{4n}$, so numerators and denominators both step up by $4$; The last factor is $\frac{2008}{2004}$; Answer choices: (A) $251$, (B) $502$, (C) $1004$, (D) $2008$, (E) $4016$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
Multiplying 501 fractions head-on is hopeless, so Tool #9 (Solve an Easier Related Problem) says try just the first two or three factors first. That reveals a repeating cancellation, and Tool #5 (Look for a Pattern) explains it: the top of each fraction is the bottom of the next one. Tool #7 (Identify Subproblems) then lets me write the whole thing as one big fraction, cancel every matching pair, and finish with a single easy division.
Execute — Answer: B
5.NF.B.4 Step 1 Multiply the first few factors
- Instead of all 501 fractions, start with two: $\frac{8}{4}\cdot\frac{12}{8}=\frac{8\cdot12}{4\cdot8}$.
- The $8$ on top cancels the $8$ on the bottom, leaving $\frac{12}{4}=3$.
- Bring in the next factor: $3\cdot\frac{16}{12}=\frac{12}{4}\cdot\frac{16}{12}$, and now the $12$ cancels, leaving $\frac{16}{4}=4$.
💡 Trying a tiny version of the product shows what happens without doing all the work.
4.OA.C.5 Step 2 See why a number always cancels
- Line up the tops and bottoms.
- Numerators: $8, 12, 16, \ldots, 2008$.
- Denominators: $4, 8, 12, \ldots, 2004$.
- Every numerator shows up again as the denominator of the very next fraction: the $8$ on top of the first is the $8$ on the bottom of the second, the $12$ on top of the second is the $12$ on the bottom of the third, and so on.
- So each middle number appears once on top and once on the bottom.
💡 Because each numerator equals the next denominator, matching pairs sit on top and bottom ready to cancel.
5.NF.B.4 Step 3 Cancel everything in the middle
- Write the whole product as one fraction: top $=8\cdot12\cdot16\cdots2008$ and bottom $=4\cdot8\cdot12\cdots2004$.
- Every number from $8$ up to $2004$ appears on both the top and the bottom, so all of them cancel.
- The only survivors are the very last numerator, $2008$, and the very first denominator, $4$.
- The product collapses to $\frac{2008}{4}$.
💡 A number on top and the same number on the bottom multiply to $1$, so only the two unmatched ends are left.
4.NBT.B.6 Step 4 Do the final division
- All that remains is $\frac{2008}{4}=2008\div4=502$.
- That matches answer choice (B).
💡 Once the chain telescopes, a single division finishes the problem.
5.NF.B.4 Instead of all 501 fractions, start with two: $\frac{8}{4}\cdot\frac{12}{8}=\fra 4.OA.C.5 Line up the tops and bottoms. Numerators: $8, 12, 16, \ldots, 2008$. Denominator 5.NF.B.4 Write the whole product as one fraction: top $=8\cdot12\cdot16\cdots2008$ and bo 4.NBT.B.6 All that remains is $\frac{2008}{4}=2008\div4=502$. That matches answer choice ( Review
Reasonableness: Check the division: $502\cdot4=2008$, so $\frac{2008}{4}=502$ is right. The size is sensible too — each factor is only a little bigger than $1$ (about $2, 1.5, 1.33, \ldots$), so multiplying about 500 of them lands well below the largest choices. The traps line up with common slips: $2008$ (D) forgets to divide by the first denominator, $4016$ (E) doubles instead, and $251$ (A) is $\frac{2008}{8}$ from cancelling one number too many.
Alternative: Simplify the general factor first: $\frac{4n+4}{4n}=\frac{4(n+1)}{4n}=\frac{n+1}{n}$. The product becomes $\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{502}{501}$, which telescopes to $\frac{502}{1}=502$. The last factor $\frac{2008}{2004}$ is $n=501$, so there are $501$ factors and the top of the last one, $502$, is the answer — again (B).
CCSS standards used (min grade 5)
5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying the fractions by combining all numerators over all denominators, so matching factors on top and bottom cancel.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that the numerators $8,12,16,\ldots$ and denominators $4,8,12,\ldots$ overlap, so each numerator equals the next denominator.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing the final quotient $2008\div4=502$.)
⭐ When the top of each fraction is the bottom of the next, they cancel down the line, so only the very last top and the very first bottom are left.
⭐ When the top of each fraction is the bottom of the next, they cancel down the line, so only the very last top and the very first bottom are left.
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