AMC 10 · 2011 · #23
Grade 8 arithmeticWhat is the hundreds digit of 20112011?
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: Find the hundreds digit of the giant number $2011^{2011}$ — the digit in the third place from the right.
Givens: The number is $2011^{2011}$ (a power far too large to write out).; We want only ONE digit: the hundreds place.; Answer choices: (A) $1$, (B) $4$, (C) $5$, (D) $6$, (E) $9$.
Unknowns: The hundreds digit of $2011^{2011}$, i.e. the hundreds digit of $2011^{2011} \bmod 1000$.
Understand
Restated: Find the hundreds digit of the giant number $2011^{2011}$ — the digit in the third place from the right.
Givens: The number is $2011^{2011}$ (a power far too large to write out).; We want only ONE digit: the hundreds place.; Answer choices: (A) $1$, (B) $4$, (C) $5$, (D) $6$, (E) $9$.
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #5 Look for a Pattern
The number $2011^{2011}$ is astronomically large, but the question asks for a single digit near the right end. Tool #9 (Solve an Easier Related Problem) collapses the giant into something small: the hundreds digit is fixed by the last three digits, and last three digits of a product depend only on the last three digits of the factors — so $2011^{2011}$ and $11^{2011}$ have the same last three digits. Now we only need $11^{2011} \bmod 1000$. Tool #7 (Identify Subproblems) cracks that open: write $11 = 10 + 1$ and expand $(10+1)^{2011}$; almost every term is a multiple of $1000$ and vanishes, leaving just three pieces to compute. Tool #2 (Make a Systematic List) adds those three pieces. Tool #5 (Look for a Pattern) gives an independent cross-check.
Execute — Answer: D
5.NBT.A.1 Step 1 Shrink the giant to its last three digits
- The hundreds digit only cares about the last three digits of the number.
- When numbers are multiplied, each digit of the answer is built from digits at or below its own place, so the last three digits of a product depend only on the last three digits of the factors.
- Since $2011$ ends in $011$, the powers $2011^{2011}$ and $11^{2011}$ share the same last three digits.
- The problem becomes: find $11^{2011} \bmod 1000$.
💡 A carry can only travel leftward, so distant high digits never disturb the hundreds place.
8.EE.A.1 Step 2 Rewrite 11 as 10 + 1
- Write $11 = 10 + 1$, so $11^{2011} = (10 + 1)^{2011}$.
- Expanding this product multiplies out $2011$ copies of $(10 + 1)$.
- Every term is a product of some $10$'s and some $1$'s.
- A term that uses three or more $10$'s carries a factor of $10^3 = 1000$, so it is a multiple of $1000$ and contributes nothing to the last three digits.
- Only terms using zero, one, or two $10$'s can survive.
💡 Powers of $10$ from $1000$ up are invisible modulo $1000$, so only the low powers of $10$ matter.
6.EE.A.3 Step 3 Keep only the three surviving pieces
- Count the three kinds of surviving terms.
- Zero tens: the single all-$1$ term contributes $1$.
- Exactly one ten: choose which one of the $2011$ factors supplies the $10$ — that is $2011$ ways, giving $2011 \times 10 = 20110$, whose last three digits are $110$.
- Exactly two tens: choose $2$ of the $2011$ factors to supply the $10$'s — the number of pairs is $\tfrac{2011 \times 2010}{2} = 2011 \times 1005 = 2{,}021{,}055$, and this many hundreds is $2{,}021{,}055 \times 100$.
- Its last three digits come from the units digit of $2{,}021{,}055$, which is $5$, so it contributes $500$.
💡 Expanding a power is just organized counting: one term with no tens, $n$ terms with one ten, and 'pairs' of factors for two tens.
4.NBT.B.4 Step 4 Add the pieces and read the digit
- Add the three surviving contributions modulo $1000$: $1 + 110 + 500 = 611$.
- So the last three digits of $2011^{2011}$ are $611$.
- The hundreds digit is $6$, which is choice (D).
💡 Three small numbers add to $611$; the middle digit is the answer.
4.OA.C.5 Step 5 Cross-check with an independent method
- Verify $611$ a second way.
- First, the last two digits: $1 + (2011 \times 10 \bmod 100) = 1 + 10 = 11$, so $2011^{2011}$ ends in $\ldots11$ — consistent with $611$.
- Second, an independent modular check by repeated squaring of $11$ modulo $1000$ (keeping only the last three digits at every step) also lands on $611$: $11^2 \equiv 121$, $11^4 \equiv 641$, $11^8 \equiv 881$, and combining the powers whose exponents add to $2011$ returns $611$.
- Both routes agree.
💡 A different road reaching the same $611$ makes an arithmetic slip very unlikely.
5.NBT.A.1 The hundreds digit only cares about the last three digits of the number. When nu 8.EE.A.1 Write $11 = 10 + 1$, so $11^{2011} = (10 + 1)^{2011}$. Expanding this product mu 6.EE.A.3 Count the three kinds of surviving terms. Zero tens: the single all-$1$ term con 4.NBT.B.4 Add the three surviving contributions modulo $1000$: $1 + 110 + 500 = 611$. So t 4.OA.C.5 Verify $611$ a second way. First, the last two digits: $1 + (2011 \times 10 \bmo Review
Reasonableness: The last three digits $611$ end in $11$, exactly as expected since any power of a number ending in $11$ still ends in $\ldots1$ in the units and picks up $11$ in the last two places here. Two independent computations — the truncated $(10+1)^{2011}$ expansion and repeated squaring modulo $1000$ — both give $611$, so the hundreds digit $6$ is solid. Among the choices, only (D) $6$ matches; the distractors $1, 4, 5, 9$ correspond to common slips (dropping the two-tens term, mis-halving $2011 \times 2010$, or reading the wrong place).
Alternative: Use the Chinese Remainder Theorem. Split $1000 = 8 \times 125$. Modulo $8$: $11 \equiv 3$ and $3^2 \equiv 1$, so $11^{2011} \equiv 3^{2011} \equiv 3 \pmod 8$. Modulo $125$: by Euler ($\varphi(125) = 100$), $11^{2011} = 11^{2000} \cdot 11^{11} \equiv 11^{11} \pmod{125}$, which reduces to $611 \bmod 125 = 111$. The unique residue mod $1000$ that is $\equiv 3 \pmod 8$ and $\equiv 111 \pmod{125}$ is $611$ — same hundreds digit $6$.
CCSS standards used (min grade 8)
5.NBT.A.1Recognize that a digit in one place represents ten times as much as it represents in the place to its right and 1/10 of what it represents to its left (Seeing that the hundreds digit depends only on the last three digits, so the base $2011$ can be replaced by $11$.)8.EE.A.1Know and apply the properties of integer exponents to generate equivalent numerical expressions (Rewriting $11^{2011}$ as $(10+1)^{2011}$ and using $10^3 = 1000 \equiv 0$ to discard high powers of ten.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(10+1)^{2011}$ by the distributive property and counting the three surviving terms.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Adding the three surviving contributions $1 + 110 + 500 = 611$.)4.OA.C.5Generate a number or shape pattern that follows a given rule, and identify apparent features of the pattern (Cross-checking $611$ by tracking last-digit patterns and repeated squaring modulo $1000$.)
⭐ For the hundreds digit of $2011^{2011}$, throw away everything but the last three digits: $2011$ becomes $11$, expanding $(10+1)^{2011}$ leaves only three pieces, and $1 + 110 + 500 = 611$ — so the hundreds digit is $6$.
⭐ For the hundreds digit of $2011^{2011}$, throw away everything but the last three digits: $2011$ becomes $11$, expanding $(10+1)^{2011}$ leaves only three pieces, and $1 + 110 + 500 = 611$ — so the hundreds digit is $6$.
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