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AMC 8 · 2012 · #12

Easy mode Grade 4
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Problem

Picture the giant number you get by multiplying 1313 by itself, 20122012 times in a row. That number is 13201213^{2012}.

You don't need to actually compute it. You only need its last digit — the digit in the ones place.

What is the ones digit of 13201213^{2012}?

Pick an answer.

(A)
$hspace{.05in}1$
(B)
$hspace{.05in}3$
(C)
$hspace{.05in}5$
(D)
$hspace{.05in}7$
(E)
$hspace{.05in}9$
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Toolkit + CCSS Solution

Understand

Restated: What single digit appears in the ones place of $13^{2012}$? We are not asked for the whole number, only its last digit.

Givens: Base = $13$; Exponent = $2012$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$

Unknowns: The units (ones) digit of $13^{2012}$

Understand

Restated: What single digit appears in the ones place of $13^{2012}$? We are not asked for the whole number, only its last digit.

Givens: Base = $13$; Exponent = $2012$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem

The exponent $2012$ is way too big to compute, so we use Tool #9 (Easier Problem): compute the first few powers $3^1, 3^2, 3^3, \dots$ and look only at their ones digits. Tool #5 (Look for a Pattern) then takes over — powers of $3$ cycle through ones digits $\{3, 9, 7, 1\}$ every $4$ steps. Once the cycle length is known, finding the $2012$th term becomes a single division-with-remainder problem. We avoid Tool #13 (Algebra) because a clean cycle pattern makes algebra unnecessary.

Execute — Answer: A

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 1
  • Strip the problem down.
  • The ones digit of a product depends only on the ones digits of its factors, so the ones digit of $13 \times 13 \times \dots$ matches the ones digit of $3 \times 3 \times \dots$.
  • Replace $13^{2012}$ with the easier $3^{2012}$.
$$\text{ones digit of } 13^{2012} = \text{ones digit of } 3^{2012}$$

💡 Multiplying multi-digit numbers and tracking how the ones place comes out of $(\text{tens} + \text{ones}) \times (\text{tens} + \text{ones})$ is a Grade 4 place-value insight.

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 2

Build a small table of powers of $3$ and record only the ones digit of each.

$3^1 = 3 \to \mathbf{3}$; $\; 3^2 = 9 \to \mathbf{9}$; $\; 3^3 = 27 \to \mathbf{7}$; $\; 3^4 = 81 \to \mathbf{1}$; $\; 3^5 = 243 \to \mathbf{3}$

💡 Computing $3 \times 3, 9 \times 3, 27 \times 3, \dots$ is straight Grade 4 multi-digit multiplication.

#5 Look for a Pattern 4.OA.C.5 Step 3
  • Spot the repeating pattern.
  • The ones digits read $3, 9, 7, 1, 3, 9, 7, 1, \dots$ — a block of four that repeats.
  • The cycle is $\{3, 9, 7, 1\}$ with length $4$.
$$\text{cycle} = (3, 9, 7, 1), \;\; \text{length} = 4$$

💡 Generating a sequence from a rule and noticing it repeats is exactly the Grade 4 "analyze patterns" standard.

#5 Look for a Pattern 4.NBT.B.6 Step 4
  • Use the cycle to jump to the $2012$th term.
  • Divide the exponent by the cycle length and look at the remainder: positions $1, 2, 3, 0$ in the cycle map to digits $3, 9, 7, 1$.
$$2012 \div 4 = 503 \text{ remainder } 0$$

💡 Finding a whole-number quotient with remainder for a $4$-digit number divided by a $1$-digit divisor is the Grade 4 division standard.

#5 Look for a Pattern 4.OA.C.5 Step 5
  • Read the result.
  • Remainder $0$ means $3^{2012}$ lands on the same spot in the cycle as $3^4$ — the last entry of the block, whose ones digit is $1$.
  • So the ones digit of $13^{2012}$ is $1$.
$$\text{ones digit of } 13^{2012} = 1 \;\Rightarrow\; \textbf{(A)}$$

💡 Mapping a position in a repeating cycle back to its value is the same pattern-analysis skill the table is built on.

[1] #9 4.NBT.B.5 Strip the problem down. The ones digit of a product depends only on the ones dig
[2] #9 4.NBT.B.5 Build a small table of powers of $3$ and record only the ones digit of each.
[3] #5 4.OA.C.5 Spot the repeating pattern. The ones digits read $3, 9, 7, 1, 3, 9, 7, 1, \dots$
[4] #5 4.NBT.B.6 Use the cycle to jump to the $2012$th term. Divide the exponent by the cycle len
[5] #5 4.OA.C.5 Read the result. Remainder $0$ means $3^{2012}$ lands on the same spot in the cy

Review

Reasonableness: Sanity-check the cycle on an exponent we can verify: $3^8 = 6561$, which ends in $1$. Our rule says $8 \div 4$ has remainder $0$, so position "last in cycle" $= 1$. Matches. Try $3^{10} = 59049$, ending in $9$: $10 \div 4 = 2$ remainder $2$, so position $2$ in the cycle $= 9$. Matches again. The cycle is reliable, so $3^{2012}$ ending in $1$ is trustworthy.

Alternative: Tool #2 (Systematic List) could replace Tool #9 here: list ones digits of $3^1, 3^2, \dots, 3^8$ in order until the repetition is obvious, then count by groups of $4$. Tool #3 (Eliminate) is weaker on this problem — no choice can be ruled out just by inspection, since all five digits $\{1, 3, 5, 7, 9\}$ are plausible ones digits of an odd power.

CCSS standards used (min grade 4)

  • 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Reducing $13^{2012}$ to $3^{2012}$ by place-value reasoning, and computing the small powers $3^1, 3^2, 3^3, 3^4, 3^5$ to expose the ones-digit pattern.)
  • 4.OA.C.5 Generate a number or shape pattern that follows a given rule and identify apparent features (Recognizing that the ones digits of powers of $3$ form the repeating cycle $(3, 9, 7, 1)$ and using that cycle to predict the $2012$th term.)
  • 4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors (Computing $2012 \div 4 = 503$ remainder $0$ to locate position $2012$ inside the length-$4$ cycle.)

⭐ This AMC 8 problem only needs Grade 4 pattern-spotting — find the cycle, divide to find the position, and read off the digit!

⭐ This AMC 8 problem only needs Grade 4 pattern-spotting — find the cycle, divide to find the position, and read off the digit!

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