AMC 8 · 1999 · #24
Grade 5 number-theoryProblem
When 19992000 is divided by 5, the remainder is
Pick an answer.
AMC 8 1999 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 remainder when $1999^{2000}$ is divided by $5$.
Givens: The dividend is $1999^{2000}$ — a $2000$-fold product of $1999$'s; The divisor is $5$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The remainder $1999^{2000} \bmod 5$
Understand
Restated: Find the remainder when $1999^{2000}$ is divided by $5$.
Givens: The dividend is $1999^{2000}$ — a $2000$-fold product of $1999$'s; The divisor is $5$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve a Related, Easier Problem
We can't compute $1999^{2000}$, so Tool #9 (Easier Problem) trades it for something we can handle: the units digit of $9^{2000}$ — which depends only on the units digit of the base ($1999$ ends in $9$) and gives the same remainder mod $5$. With the easier problem in hand, Tool #5 (Look for a Pattern) finishes it: $9^1, 9^2, 9^3, \ldots$ have units digits $9, 1, 9, 1, \ldots$, a length-$2$ cycle. Even exponents land on $1$, and $1 \bmod 5 = 1$. We deliberately avoid Tool #13 (Algebra) and modular-arithmetic notation; the pattern is enough.
Execute — Answer: B
4.NBT.A.1 Step 1 - Reduce to the units digit.
- Any number ending in the digit $d$ is $d$ more than a multiple of $10$, and $10$ is a multiple of $5$, so the remainder mod $5$ depends only on the units digit.
- So we only need the units digit of $1999^{2000}$.
💡 Place value (Grade 4): tens, hundreds, thousands are all multiples of $5$, so they leave the same remainder as the ones place alone.
5.NBT.B.5 Step 2 - Replace the base by its units digit.
- When you multiply, the units digit of the product depends only on the units digits of the factors.
- Since $1999$ ends in $9$, every factor of $1999$ in $1999 \times 1999 \times \cdots$ contributes the same way as a factor of $9$ does.
- So the units digit of $1999^{2000}$ equals the units digit of $9^{2000}$.
💡 Grade 5 multiplication: only the ones digits of the factors affect the ones digit of the product.
4.OA.C.5 Step 3 List the units digits of small powers of $9$ and look for the cycle.
💡 Grade 4 "generate and analyze a pattern": the units digits of powers of $9$ alternate $9, 1, 9, 1, \ldots$ with period $2$.
4.OA.C.5 Step 4 - Read the cycle at exponent $2000$.
- Odd exponents give units digit $9$; even exponents give units digit $1$.
- Since $2000$ is even, the units digit of $9^{2000}$ is $1$.
💡 Just match the exponent's parity to the right slot of the length-$2$ pattern.
4.NBT.B.6 Step 5 - Divide the units digit by $5$ to get the remainder.
- A number ending in $1$ leaves remainder $1$ when divided by $5$.
💡 Grade 4 division with remainder: any number whose ones digit is $1$ is $5k + 1$ for some whole number $k$, so the remainder is $1$.
4.NBT.A.1 Reduce to the units digit. Any number ending in the digit $d$ is $d$ more than a 5.NBT.B.5 Replace the base by its units digit. When you multiply, the units digit of the p 4.OA.C.5 List the units digits of small powers of $9$ and look for the cycle. 4.OA.C.5 Read the cycle at exponent $2000$. Odd exponents give units digit $9$; even expo 4.NBT.B.6 Divide the units digit by $5$ to get the remainder. A number ending in $1$ leave Review
Reasonableness: Spot-check the pattern with the exponents we can actually compute. $9^2 = 81$ ends in $1$, and $81 = 16 \times 5 + 1$ — remainder $1$. $9^4 = 6561$ ends in $1$, and $6561 = 1312 \times 5 + 1$ — remainder $1$. Every even power of $9$ keeps giving remainder $1$, so the answer (B) $1$ is consistent. Choices (A) $0$ and (E) $4$ would require the units digit to be $0$ or $5$ (for (A)) or $4$ or $9$ (for (E)) — neither matches the cycle.
Alternative: Tool #2 (Systematic List) on remainders directly. $1999 \div 5$ leaves remainder $4$, so the remainder of $1999^{2000}$ matches that of $4^{2000}$. List $4^1, 4^2, 4^3, 4^4$: remainders mod $5$ are $4, 1, 4, 1, \ldots$ — period $2$ again. Even exponent $\Rightarrow$ remainder $1$, confirming (B).
CCSS standards used (min grade 5)
4.NBT.A.1Recognize place value — a digit in one place represents ten times what it represents in the place to its right (Justifying why the remainder of any whole number when divided by $5$ depends only on its units digit (tens and above are all multiples of $5$).)5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Justifying that the units digit of a product depends only on the units digits of the factors, so $1999^{2000}$ and $9^{2000}$ share a units digit.)4.OA.C.5Generate a number or shape pattern that follows a given rule; identify apparent features of the pattern (Listing the units digits of $9^1, 9^2, 9^3, 9^4, \ldots$ as $9, 1, 9, 1, \ldots$ and reading off the parity rule (even exponent $\Rightarrow$ units digit $1$).)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors (Converting the units digit $1$ into the final remainder: $1 \div 5$ gives quotient $0$ and remainder $1$.)
⭐ You don't need to compute $1999^{2000}$ — just track the units digit. Powers of $9$ end in $9, 1, 9, 1, \ldots$, so an even exponent always ends in $1$, and $1$ leaves remainder $1$ when divided by $5$.
⭐ You don't need to compute $1999^{2000}$ — just track the units digit. Powers of $9$ end in $9, 1, 9, 1, \ldots$, so an even exponent always ends in $1$, and $1$ leaves remainder $1$ when divided by $5$.
More like this
Same archetype — closest grade level first.