Competition · AMC preparation · step 4 of 4

AMC 8 · 1999 · #24

Grade 5 number-theory
Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue
modular-arithmeticunits-digit-trackingexponentspattern-recognitionparity modular-arithmeticpattern-recognitionmodular-arithmetic-mod-10 ↑ Prerequisites: exponentsmodular-arithmetic
📏 Short solution 💡 3 insights
📘 View easy version →
Problem
Find the remainder when 1999²⁰⁰⁰ is divided by 5.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

We can't compute 1999²⁰⁰⁰, so Tool #9 (Easier Problem) trades it for something we can handle: the units digit of 9²⁰⁰⁰ — 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: 9¹, 9², 9³, … have units digits 9, 1, 9, 1, …, a length-2 cycle. Even exponents land on 1, and 1 mod 5 = 1. We deliberately avoid Tool #13 (Algebra) and modular-arithmetic notation; the pattern is enough.

1STEP 1

Reduce to the units digit

The remainder mod 5 depends only on the units digit, so we just need the units digit of 1999²⁰⁰⁰.

1999²⁰⁰⁰ mod 5 = (units digit of 1999²⁰⁰⁰) mod 5
2STEP 2

Swap in the base's units digit

Only the base's units digit matters, so the units digit of 1999²⁰⁰⁰ equals that of 9²⁰⁰⁰.

units digit of 1999²⁰⁰⁰ = units digit of 9²⁰⁰⁰
3STEP 3

Find the cycle of 9

List units digits of powers of 9: they cycle 9, 1, 9, 1, … with period 2.

9¹ = 9, 9² = 81, 9³ = 729, 9⁴ = 6561 — units digits 9, 1, 9, 1, …
4STEP 4

Read the cycle at 2000

Odd exponents give 9, even give 1; since 2000 is even, the units digit of 9²⁰⁰⁰ is 1.

2000 even → units digit of 9²⁰⁰⁰ = 1
5STEP 5

Divide by 5 for the remainder

A number ending in 1 leaves remainder 1 when divided by 5 — that's the answer.

1 ÷ 5 = 0 remainder 1 → (B)
Answer
1
Spot-check the pattern with the exponents we can actually compute. 9² = 81 ends in 1, and 81 = 16 × 5 + 1 — remainder 1. 9⁴ = 6561 ends in 1, and 6561 = 1312 × 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.
💡Key takeaway

You don't need to compute 1999²⁰⁰⁰ — just track the units digit. Powers of 9 end in 9, 1, 9, 1, …, so an even exponent always ends in 1, and 1 leaves remainder 1 when divided by 5.

  • Reduce to the units digit
  • Swap in the base's units digit
  • Find the cycle of 9
  • Read the cycle at 2000
  • Divide by 5 for the remainder

A parent dashboard for the family lives at sensimlab.com.

More like this

Same problem type. Tags show what each one shares with this problem.