AMC 10 · 2011 · #10
Grade 6 rate-ratioConsider the set of numbers {1,10,102,103,…,1010}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
Pick an answer.
AMC 10 2011 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 powers of ten $\{1, 10, 10^2, \ldots, 10^{10}\}$, take the biggest number, $10^{10}$, and divide it by the sum of all ten smaller numbers. Find which whole number that ratio is nearest to.
Givens: The set is $\{1,\,10,\,10^2,\,10^3,\,\ldots,\,10^{10}\}$, which is $11$ powers of ten; The largest element is $10^{10}$; The 'other ten elements' are $1,10,10^2,\ldots,10^9$; Answer choices: (A) $1$, (B) $9$, (C) $10$, (D) $11$, (E) $101$
Unknowns: The whole number closest to $\dfrac{10^{10}}{1+10+10^2+\cdots+10^9}$
Understand
Restated: From the powers of ten $\{1, 10, 10^2, \ldots, 10^{10}\}$, take the biggest number, $10^{10}$, and divide it by the sum of all ten smaller numbers. Find which whole number that ratio is nearest to.
Givens: The set is $\{1,\,10,\,10^2,\,10^3,\,\ldots,\,10^{10}\}$, which is $11$ powers of ten; The largest element is $10^{10}$; The 'other ten elements' are $1,10,10^2,\ldots,10^9$; Answer choices: (A) $1$, (B) $9$, (C) $10$, (D) $11$, (E) $101$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Eleven powers of ten look scary, so Tool #9 (Solve an Easier Related Problem) shrinks the set to two or three powers first, where the ratio is easy to compute by hand. Tool #5 (Look for a Pattern) then watches how that ratio behaves as the set grows, revealing that it always sits just above $9$. Tool #3 (Eliminate Possibilities) finishes: once we know the value is a hair over $9$, choices $1$, $10$, $11$, and $101$ are all clearly wrong.
Execute — Answer: B
6.RP.A.3 Step 1 Shrink the set and divide
- Replace the giant set with a tiny copy that works the same way: $\{1,10,100\}$.
- The largest is $100$ and the others add to $1+10=11$, so the ratio is $\dfrac{100}{11}=9.09\ldots$.
- Small numbers, but the same recipe as the real problem.
💡 A hard problem often has an easy twin with tiny numbers; solve the twin to see how the machine works.
4.OA.C.5 Step 2 Grow it one step and compare
- Add the next power to get $\{1,10,100,1000\}$.
- Now the largest is $1000$ and the rest add to $1+10+100=111$, giving $\dfrac{1000}{111}=9.009\ldots$.
- Line the results up: $9.09$, then $9.009$.
- Each time the set grows, the ratio stays just above $9$ and slides closer to it.
💡 Watching how the answer changes as the problem grows tells you where it is heading.
5.NBT.A.2 Step 3 Write the real denominator
- Back to the full set.
- The ten smaller elements are $1,10,10^2,\ldots,10^9$.
- Adding powers of ten just fills each place value with a single $1$, so the sum is $1{,}111{,}111{,}111$ (ten ones).
- The largest element $10^{10}$ is a $1$ followed by ten zeros: $10{,}000{,}000{,}000$.
💡 A sum of different powers of ten just writes a $1$ in each place, so the total is a neat string of ones.
5.NBT.B.5 Step 4 Nail down the exact ratio
- The ratio is $\dfrac{10{,}000{,}000{,}000}{1{,}111{,}111{,}111}$.
- Here is the clean trick: $9 \times 1{,}111{,}111{,}111 = 9{,}999{,}999{,}999$, which is exactly one less than $10^{10}$.
- So the top is a tiny bit more than $9$ times the bottom, making the ratio just barely over $9$ (about $9.000000009$).
💡 If nine copies of the bottom miss the top by only $1$, the ratio must be a whisker above $9$.
6.RP.A.3 Step 5 Round and pick the answer
- The value $9.000000009$ rounds to the whole number $9$.
- The other choices are nowhere close: $1$ is far too small, and $10$, $11$, $101$ are all too big.
- The ratio is closest to $9$, which is choice (B).
💡 A number that sits a hair above $9$ is closest to $9$, so every other choice drops away.
6.RP.A.3 Replace the giant set with a tiny copy that works the same way: $\{1,10,100\}$. 4.OA.C.5 Add the next power to get $\{1,10,100,1000\}$. Now the largest is $1000$ and the 5.NBT.A.2 Back to the full set. The ten smaller elements are $1,10,10^2,\ldots,10^9$. Addi 5.NBT.B.5 The ratio is $\dfrac{10{,}000{,}000{,}000}{1{,}111{,}111{,}111}$. Here is the cl 6.RP.A.3 The value $9.000000009$ rounds to the whole number $9$. The other choices are no Review
Reasonableness: The largest power of ten is about ten times the previous one, and the previous one already dwarfs everything before it. So the denominator $1{,}111{,}111{,}111$ is very close to $\tfrac{1}{9}$ of $10^{10}$ (since $\tfrac{1}{9}=0.111\ldots$), which makes the ratio close to $9$. The small-set experiments ($9.09$, then $9.009$) point to the same landing spot, so $9$ is exactly what we should expect.
Alternative: Use the geometric-series formula: $1+10+\cdots+10^9 = \dfrac{10^{10}-1}{9}$. Then the ratio is $\dfrac{10^{10}}{(10^{10}-1)/9} = \dfrac{9\cdot 10^{10}}{10^{10}-1}$, which is $9$ divided by $\left(1-10^{-10}\right)$ — just over $9$. Same answer (B), reached without listing digits.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Computing the ratio of largest element to the sum of the others and finding the nearest whole number.)4.OA.C.5Generate a number or shape pattern following a given rule (Comparing $9.09$ and $9.009$ from the shrunken sets to see the ratio settle just above $9$.)5.NBT.A.2Explain patterns in number of zeros and placement of decimal point (Recognizing that summing distinct powers of ten gives $1{,}111{,}111{,}111$ and that $10^{10}$ is a $1$ with ten zeros.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Checking $9 \times 1{,}111{,}111{,}111 = 9{,}999{,}999{,}999 = 10^{10}-1$ to pin the ratio just above $9$.)
⭐ When one number is a bit more than nine copies of another, their ratio is just over $9$ — so it rounds to $9$.
⭐ When one number is a bit more than nine copies of another, their ratio is just over $9$ — so it rounds to $9$.
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