AMC 10 · 2003 · #16
Easy mode Grade 4Multiply 13 by itself 2003 times to make one huge number. You do not need the whole number — only its last digit, the one in the ones place. What is that last digit?
Pick an answer.
AMC 10 2003 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 units digit (the ones place) of the number $13^{2003}$ — that is, the last digit you would see if you actually multiplied thirteen by itself $2003$ times.
Givens: The base is $13$ and the exponent is $2003$.; Answer choices: (A) $1$, (B) $3$, (C) $7$, (D) $8$, (E) $9$.
Unknowns: The units digit of $13^{2003}$.
Understand
Restated: Find the units digit (the ones place) of the number $13^{2003}$ — that is, the last digit you would see if you actually multiplied thirteen by itself $2003$ times.
Givens: The base is $13$ and the exponent is $2003$.; Answer choices: (A) $1$, (B) $3$, (C) $7$, (D) $8$, (E) $9$.
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Computing $13^{2003}$ is impossible by hand, so the plan is to track only the one piece that matters — the units digit. Tool #9 (Solve an Easier Related Problem) makes the first cut: when you multiply, the units digit of the product depends only on the units digits of the factors, so $13^{2003}$ has the same units digit as $3^{2003}$. That replaces $13$ with the much smaller $3$. Tool #5 (Look for a Pattern) then handles the giant exponent: the units digits of $3^1, 3^2, 3^3, \dots$ repeat in a short cycle, so instead of $2003$ multiplications we just find where $2003$ lands in that cycle. Tool #3 (Eliminate Possibilities) confirms the result is one of the listed choices.
Execute — Answer: C
4.NBT.B.5 Step 1 Only the base's units digit matters
- When two numbers are multiplied, the units digit of the product is decided entirely by the units digits of the two numbers — the tens, hundreds, and everything higher never reach the ones place.
- So each time you multiply by $13$, the new units digit is controlled only by the $3$ in $13$.
- That means $13^{2003}$ ends in the same digit as $3^{2003}$, and we can throw the $10$ away.
💡 The ones place of a product only ever comes from the ones places of the factors, so the $1$ in $13$ is dead weight.
4.OA.C.5 Step 2 List units digits of powers of 3
- Compute the units digit of the first few powers of $3$: $3^1 = 3$, $3^2 = 9$, $3^3 = 27$ (units digit $7$), $3^4 = 81$ (units digit $1$), $3^5 = 243$ (units digit $3$ again).
- The units digits run $3, 9, 7, 1$ and then start over at $3$.
- So the pattern repeats in a block of length $4$.
💡 Multiplying by $3$ over and over cycles the units digit through the same four values forever.
4.NBT.B.6 Step 3 Locate 2003 in the length-4 cycle
- Because the cycle has length $4$, the units digit of $3^n$ depends only on where $n$ falls within a block of four: remainder $1$ gives $3$, remainder $2$ gives $9$, remainder $3$ gives $7$, and remainder $0$ (a multiple of $4$) gives $1$.
- Divide the exponent by $4$: $2003 = 4 \times 500 + 3$, so the remainder is $3$.
- A remainder of $3$ points to the third digit in the block.
💡 The full blocks of four each land back on $1$, so only the leftover $3$ steps into the new block decide the answer.
4.OA.C.5 Step 4 Read off the answer
- The third entry of the repeating block $3, 9, 7, 1$ is $7$.
- So the units digit of $3^{2003}$, and therefore of $13^{2003}$, is $7$.
- That matches choice (C).
💡 Remainder $3$ means stop on the third digit of the cycle, which is $7$.
4.NBT.B.5 When two numbers are multiplied, the units digit of the product is decided entir 4.OA.C.5 Compute the units digit of the first few powers of $3$: $3^1 = 3$, $3^2 = 9$, $3 4.NBT.B.6 Because the cycle has length $4$, the units digit of $3^n$ depends only on where 4.OA.C.5 The third entry of the repeating block $3, 9, 7, 1$ is $7$. So the units digit o Review
Reasonableness: The check is that the four possible units digits of a power of $3$ are exactly $3, 9, 7, 1$, so the answer must be one of those — this immediately rules out (D) $8$, which can never be the units digit of a power of $3$. Among the remaining choices, the exponent $2003$ leaves remainder $3$ when divided by $4$, and every exponent congruent to $3$ (like $3, 7, 11, \dots$) gives units digit $7$; you can confirm on the small case $3^3 = 27$, which indeed ends in $7$. So (C) is consistent, not a fluke of the large exponent.
Alternative: Reduce the exponent using the cycle directly with modular arithmetic: since the units digit of $3^n$ has period $4$, $3^{2003}$ has the same units digit as $3^{2003 \bmod 4} = 3^{3} = 27$, whose units digit is $7$. Same answer (C).
CCSS standards used (min grade 4)
4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Recognizing through place value that the units digit of a product comes only from the units digits of the factors, so $13^{2003}$ ends in the same digit as $3^{2003}$.)4.OA.C.5Generate a number or shape pattern following a given rule (Building the repeating block $3, 9, 7, 1$ of units digits of powers of $3$ and reading its third entry.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2003$ by $4$ to get remainder $3$, which pins down the position inside the length-4 cycle.)
⭐ Only the last digit of the base matters, and last digits of powers of $3$ cycle $3, 9, 7, 1$; since $2003$ leaves remainder $3$ after dividing by $4$, the last digit is the third one, $7$.
⭐ Only the last digit of the base matters, and last digits of powers of $3$ cycle $3, 9, 7, 1$; since $2003$ leaves remainder $3$ after dividing by $4$, the last digit is the third one, $7$.
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