AMC 8 · 2000 · #14
Easy mode Grade 5Problem
Consider the number 1919+9999.
This is a huge number — way too big to actually write out. But we only care about its very last digit (the ones place).
What is the units digit of 1919+9999?
Pick an answer.
AMC 8 2000 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 (ones place) of the sum $19^{19} + 99^{99}$.
Givens: Two huge numbers: $19^{19}$ and $99^{99}$; Both bases end in the digit $9$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $8$, (E) $9$
Unknowns: The units digit of the sum $19^{19} + 99^{99}$
Understand
Restated: Find the units digit (ones place) of the sum $19^{19} + 99^{99}$.
Givens: Two huge numbers: $19^{19}$ and $99^{99}$; Both bases end in the digit $9$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $8$, (E) $9$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem
We never need the full $19^{19}$ or $99^{99}$ — only their last digit. Tool #9 (Solve an Easier Related Problem) collapses both giants to the same easier question: "what is the units digit of $9^n$?" since both bases end in $9$. Tool #5 (Look for a Pattern) then takes over: compute $9^1, 9^2, 9^3, 9^4$ by hand, spot the length-$2$ cycle $9, 1, 9, 1, \dots$, and read off the units digit for any exponent. Add the two units digits at the end and take the units digit of that small sum.
Execute — Answer: D
5.NBT.B.5 Step 1 - Replace each big base with just its units digit.
- When you multiply two whole numbers, the units digit of the product is the units digit of (units digit of one) $\times$ (units digit of the other).
- Apply that rule repeatedly: the units digit of $19^{19}$ equals the units digit of $9^{19}$, and the units digit of $99^{99}$ equals the units digit of $9^{99}$.
💡 Long multiplication shows the ones-place digit of a product only ever depends on the ones-place digits of the factors — a Grade 5 multi-digit multiplication fact.
5.NBT.B.5 Step 2 - List the first few powers of $9$ and read off the units digits.
- Stop as soon as the pattern repeats.
💡 Hand-multiplying four small powers is well within Grade 5 fluency and exposes the cycle without algebra.
4.OA.C.5 Step 3 - Spot the cycle.
- The units digits of $9^1, 9^2, 9^3, 9^4$ are $9, 1, 9, 1$ — a length-$2$ pattern.
- Odd exponent gives units digit $9$; even exponent gives units digit $1$.
💡 Naming a repeating rule from a short list is the Grade 4 "generate a number pattern" move.
4.OA.C.5 Step 4 - Apply the rule to both exponents.
- $19$ is odd, so $\text{units}(9^{19}) = 9$.
- $99$ is also odd, so $\text{units}(9^{99}) = 9$.
💡 Once the cycle is known, every exponent question becomes a parity check — odd or even.
4.OA.A.3 Step 5 - Add the two units digits and take the units digit of that sum.
- $9 + 9 = 18$, whose units digit is $8$.
💡 The units digit of a sum depends only on the units digits of the addends — a Grade 4 place-value-aware addition.
5.NBT.B.5 Replace each big base with just its units digit. When you multiply two whole num 5.NBT.B.5 List the first few powers of $9$ and read off the units digits. Stop as soon as 4.OA.C.5 Spot the cycle. The units digits of $9^1, 9^2, 9^3, 9^4$ are $9, 1, 9, 1$ — a le 4.OA.C.5 Apply the rule to both exponents. $19$ is odd, so $\text{units}(9^{19}) = 9$. $9 4.OA.A.3 Add the two units digits and take the units digit of that sum. $9 + 9 = 18$, who Review
Reasonableness: Test the rule on a small case. $9^3 + 9^3 = 729 + 729 = 1458$, units digit $8$ — and indeed both exponents are odd, so each term ends in $9$ and the sum ends in $9 + 9 = 18 \to 8$. Same shape as our problem. Among the choices, only (D) $8$ matches the $9 + 9$ tail. Also, the units digit of $19^{19} + 99^{99}$ cannot be $9$ (choice E) because two odd-power-of-$9$ tails add to $18$, not $9$.
Alternative: Tool #2 (Make a Systematic List) for $99 \bmod 2$. Build a tiny table indexed by exponent parity: even $\to 1$, odd $\to 9$. Plug in $19 \to 9$ and $99 \to 9$; add to get $18$, units digit $8$. Same answer (D) with no multiplication beyond $9 \times 9 = 81$.
CCSS standards used (min grade 5)
5.NBT.B.5Fluently multiply multi-digit whole numbers (Reducing each base to its units digit (the multiplication algorithm shows the ones place of a product only depends on the ones places of the factors) and hand-computing $9^2 = 81, 9^3 = 729, 9^4 = 6561$ to expose the cycle.)4.OA.C.5Generate a number or shape pattern following a given rule (Naming the length-$2$ cycle $9, 1, 9, 1, \dots$ in the units digits of $9^n$ and turning it into the odd/even rule used to read off $\text{units}(9^{19})$ and $\text{units}(9^{99})$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding the two units digits $9 + 9 = 18$ and reading off the units digit $8$ as the final answer.)
⭐ Only the last digit matters: powers of $9$ flip between $9$ and $1$, both exponents are odd so each piece ends in $9$, and $9 + 9 = 18$ ends in $8$.
⭐ Only the last digit matters: powers of $9$ flip between $9$ and $1$, both exponents are odd so each piece ends in $9$, and $9 + 9 = 18$ ends in $8$.
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