AMC 10 · 2020 · #12
Grade 8 arithmeticProblem
The decimal representation of 20201 consists of a string of zeros after the decimal point, followed by a 9 and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
Pick an answer.
AMC 10 2020 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: Write $\dfrac{1}{20^{20}}$ as a decimal. After the decimal point comes a run of $0$'s, then a $9$, then more digits. How many $0$'s are in that opening run?
Givens: Target number: $\dfrac{1}{20^{20}}$; Its decimal expansion is $0.\underbrace{00\ldots 0}_{?}9\ldots$; Answer choices: $23,\; 24,\; 25,\; 26,\; 27$
Unknowns: Number of $0$'s between the decimal point and the first nonzero digit ($9$)
Understand
Restated: Write $\dfrac{1}{20^{20}}$ as a decimal. After the decimal point comes a run of $0$'s, then a $9$, then more digits. How many $0$'s are in that opening run?
Givens: Target number: $\dfrac{1}{20^{20}}$; Its decimal expansion is $0.\underbrace{00\ldots 0}_{?}9\ldots$; Answer choices: $23,\; 24,\; 25,\; 26,\; 27$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Tool #7 (Subproblems): split the denominator $20^{20}$ into $2^{20} \cdot 10^{20}$ and handle each piece separately — the $10^{20}$ part contributes exactly $20$ zeros, the $2^{20}$ part is what we still need to convert to a power-of-$10$ shift. Tool #9 (Easier Problem): replace the unwieldy $2^{20}$ with the small fact $2^{10} = 1024$ to find the digit count of $2^{20}$. Tool #3 (Eliminate): only one answer choice matches once we count digits.
Execute — Answer: D
6.EE.A.1 Step 1 - Rewrite the denominator using $20 = 2 \cdot 10$.
- This splits the problem into a clean power of $10$ (which just shifts the decimal point) and a power of $2$ (which controls the first nonzero digit).
💡 Grade 6 exponents: $(ab)^n = a^n b^n$ separates the two effects.
6.EE.A.1 Step 2 - Find how many digits $2^{20}$ has, using the easier fact $2^{10} = 1024$.
- Squaring, $2^{20} = 1024^2 = 1{,}048{,}576$ — a $7$-digit number.
💡 Grade 6: square the small known $2^{10}$ to get $2^{20}$ without a calculator.
8.EE.A.3 Step 3 - Combine.
- $\dfrac{1}{20^{20}} = \dfrac{1}{2^{20} \cdot 10^{20}} = \dfrac{1}{1{,}048{,}576} \cdot 10^{-20}$.
- Now $\dfrac{1}{1{,}048{,}576}$ is just under $\dfrac{1}{10^6}$, so it equals $0.000000\,9536\ldots$ — leading $9$ sits at the $7$th decimal place.
💡 Grade 8 scientific notation: $1/2^{20}$ starts with $9$ at the $10^{-7}$ place.
8.EE.A.4 Step 4 - Multiply by $10^{-20}$: this shifts every digit $20$ more places to the right.
- The $9$ that was at the $7$th decimal place lands at the $(7 + 20) = 27$th decimal place.
💡 Grade 8 scientific notation: multiplying by $10^{-20}$ shifts the exponent by $-20$.
5.NBT.A.2 Step 5 If the first nonzero digit is at decimal place $27$, then there are $27 - 1 = 26$ zeros after the decimal point before that $9$.
💡 Grade 5 place value: positions $1$ through $26$ are zeros; position $27$ is the $9$.
2.NBT.A.4 Step 6 - Match $26$ to the choices: option (D).
- The other choices ($23, 24, 25, 27$) miss the digit count of $2^{20}$ by one or two.
💡 Grade 2 number comparison: only one option equals $26$.
6.EE.A.1 Rewrite the denominator using $20 = 2 \cdot 10$. This splits the problem into a 6.EE.A.1 Find how many digits $2^{20}$ has, using the easier fact $2^{10} = 1024$. Squari 8.EE.A.3 Combine. $\dfrac{1}{20^{20}} = \dfrac{1}{2^{20} \cdot 10^{20}} = \dfrac{1}{1{,}0 8.EE.A.4 Multiply by $10^{-20}$: this shifts every digit $20$ more places to the right. T 5.NBT.A.2 If the first nonzero digit is at decimal place $27$, then there are $27 - 1 = 26 2.NBT.A.4 Match $26$ to the choices: option (D). The other choices ($23, 24, 25, 27$) miss Review
Reasonableness: Quick estimate: $20^{20} = (2 \cdot 10)^{20} = 2^{20} \cdot 10^{20} \approx 10^6 \cdot 10^{20} = 10^{26}$, so $\dfrac{1}{20^{20}} \approx 10^{-26}$. That puts the first nonzero digit near the $26$th or $27$th decimal place — which means $25$, $26$, or $27$ zeros. Refining the estimate ($2^{20}$ is just over $10^6$, not equal), the first nonzero digit slides to the $27$th place, giving $26$ zeros. Consistent with (D).
Alternative: Tool #6 (Guess and Check) on the digit count rule: for any positive integer $N$ with $d$ digits that is not a power of $10$, $\dfrac{1}{N}$ has $d - 1$ leading zeros before the first nonzero digit (because $10^{d-1} \le N < 10^d$ means $10^{-d} < \dfrac{1}{N} \le 10^{-(d-1)}$). Here $N = 20^{20}$ has $7 + 20 = 27$ digits, so $27 - 1 = 26$ zeros. Same answer, faster path.
CCSS standards used (min grade 8)
2.NBT.A.4Compare two three-digit numbers using symbols (Matching $26$ to the five answer choices.)5.NBT.A.2Explain patterns in number of zeros and placement of decimal point (Reading off that $27$ as the place of the first nonzero digit means $26$ zeros before it.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Splitting $20^{20} = 2^{20} \cdot 10^{20}$ and computing $2^{20} = 1024^2 = 1{,}048{,}576$.)8.EE.A.3Use numbers expressed in the form of a single digit times a power of 10 (Writing $\dfrac{1}{2^{20}} \approx 9.537 \cdot 10^{-7}$.)8.EE.A.4Perform operations with numbers expressed in scientific notation (Multiplying by $10^{-20}$ to shift the first nonzero digit to the $27$th decimal place.)
⭐ This AMC 10 problem only needs Grade 8 scientific notation you already know! Split $20^{20} = 2^{20} \cdot 10^{20}$. The $10^{20}$ part contributes $20$ zeros for free. The $2^{20} = 1{,}048{,}576$ part has $7$ digits, so $\dfrac{1}{2^{20}} \approx 9.5 \cdot 10^{-7}$ — $9$ sits at the $7$th decimal place. Shift $20$ more places right and the $9$ lands at the $27$th place, leaving $27 - 1 = \mathbf{26}$ leading zeros, answer (D).
⭐ This AMC 10 problem only needs Grade 8 scientific notation you already know! Split $20^{20} = 2^{20} \cdot 10^{20}$. The $10^{20}$ part contributes $20$ zeros for free. The $2^{20} = 1{,}048{,}576$ part has $7$ digits, so $\dfrac{1}{2^{20}} \approx 9.5 \cdot 10^{-7}$ — $9$ sits at the $7$th decimal place. Shift $20$ more places right and the $9$ lands at the $27$th place, leaving $27 - 1 = \mathbf{26}$ leading zeros, answer (D).
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