AMC 8 · 2011 · #22

Grade 6 number-theory

modular-arithmeticpattern-recognitionexponents pattern-recognitionmodular-arithmetic ↑ Prerequisites: modular-arithmeticexponents

📏 Medium solution 💡 3 insights

Problem

What is the tens digit of $7^{2011}$ ?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Find the tens digit (the second-to-last digit) of the huge number $7^{2011}$.

Givens: The number is $7^{2011}$; Answer choices: (A) $0$, (B) $1$, (C) $3$, (D) $4$, (E) $7$

Unknowns: The tens digit of $7^{2011}$

Understand

Restated: Find the tens digit (the second-to-last digit) of the huge number $7^{2011}$.

Givens: The number is $7^{2011}$; Answer choices: (A) $0$, (B) $1$, (C) $3$, (D) $4$, (E) $7$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem

The exponent $2011$ is enormous, so direct computation is hopeless. Tool #9 (Easier Problem) says: try smaller exponents first — $7^1, 7^2, 7^3, \dots$ — and only keep the last two digits at each step (anything earlier than the tens place can never affect the tens digit). Tool #5 (Look for a Pattern) takes over once those last-two-digit values start repeating: a short cycle lets us replace exponent $2011$ with a much smaller equivalent exponent.

Execute — Answer: D

#9 Solve an Easier Related Problem 5.NBT.A.1 Step 1

Compute the last two digits of small powers of $7$.
Why only last two digits?
When you multiply by $7$, the new tens digit depends only on the previous two digits — anything further left gets carried into the hundreds and above, never down.
So at each step keep just the last two digits.

$$7^1 = 07,\; 7^2 = 49,\; 7^3 = 343 \to 43,\; 7^4 = 43 \times 7 = 301 \to 01$$

💡 Replacing a huge exponent with $1, 2, 3, 4$ is the Easier Problem move. Tracking only the last two digits uses the Grade 5 place-value idea that the tens and ones come from below the hundreds.

#5 Look for a Pattern 4.OA.C.5 Step 2

Spot the cycle.
The last two digits ran $07, 49, 43, 01$.
Multiplying by $7$ again gives $01 \times 7 = 07$, which is exactly where we started.
So the last two digits of $7^n$ repeat with period $4$: the pattern $07, 49, 43, 01$ loops forever.

$$7^5 \to 01 \times 7 = 07 \;(= 7^1),\; 7^6 \to 49 \;(= 7^2),\; \dots$$

💡 Detecting and stating a repeating cycle is the Grade 4 "generate and analyze patterns" skill — once $7^4$ returns to $01$, the cycle must restart.

#5 Look for a Pattern 6.NS.B.2 Step 3

Match $2011$ to a spot in the cycle.
The cycle has length $4$, so $7^n$ has the same last two digits as $7^r$, where $r$ is the remainder when $n$ is divided by $4$ (and $r = 4$ means "the last spot").
Divide: $2011 \div 4 = 502$ remainder $3$, so $2011$ is at position $3$ in the cycle.

$$2011 = 4 \times 502 + 3 \;\Rightarrow\; 7^{2011} \text{ ends like } 7^3$$

💡 Using division-with-remainder to locate a number inside a repeating cycle is the Grade 6 "divide multi-digit numbers" skill applied to pattern position.

#5 Look for a Pattern 5.NBT.A.1 Step 4

Read off the tens digit.
From step 1, $7^3$ ends in $43$, so $7^{2011}$ also ends in $43$.
The tens digit (the second-to-last digit) is $4$.

$$7^{2011} \to \dots 43 \;\Rightarrow\; \text{tens digit} = 4 \;\Rightarrow\; \textbf{(D)}$$

💡 Identifying which digit sits in the tens place is the Grade 5 place-value definition itself.

[1] #9 5.NBT.A.1 Compute the last two digits of small powers of $7$. Why only last two digits? Wh

[2] #5 4.OA.C.5 Spot the cycle. The last two digits ran $07, 49, 43, 01$. Multiplying by $7$ aga

[3] #5 6.NS.B.2 Match $2011$ to a spot in the cycle. The cycle has length $4$, so $7^n$ has the

[4] #5 5.NBT.A.1 Read off the tens digit. From step 1, $7^3$ ends in $43$, so $7^{2011}$ also end

Review

Reasonableness: The cycle $07, 49, 43, 01$ has four different tens digits: $0, 4, 4, 0$. So whatever the answer is, it has to be $0$ or $4$ — and the answer choices include exactly those two ((A) $0$ and (D) $4$). Since $2011 \bmod 4 = 3$ lands at $43$, the tens digit is $4$, choice (D). A second check: $7^{2012}$ would end in $01$ (tens digit $0$), so adjacent exponents really do alternate the way the cycle says.

Alternative: Tool #16 (Change Focus / Complement) cuts the work down further: only the last two digits ever matter, so all arithmetic can be done $\bmod{100}$. Then $7^2 = 49$, $7^4 = 49^2 = 2401 \equiv 1 \pmod{100}$, so $7^{2011} = 7^{4 \cdot 502 + 3} \equiv 1^{502} \cdot 7^3 = 343 \equiv 43 \pmod{100}$ — same answer (D).

CCSS standards used (min grade 6)

4.OA.C.5 Generate and analyze numerical patterns (Recognizing that the last two digits of $7^n$ cycle through $07, 49, 43, 01$ with period $4$.)
5.NBT.A.1 Understand the place value system (Justifying that only the last two digits matter for the tens digit, and reading $4$ off as the tens digit of $\dots 43$.)
6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm (Dividing $2011 \div 4$ to get quotient $502$ and remainder $3$, which locates $2011$ inside the length-$4$ cycle.)

⭐ Huge exponents look scary, but the last two digits cycle quickly — a Grade 6 division with remainder is all you need to land the answer!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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