AMC 10 · 2008 · #24

Grade 6 number-theory

Archetype Units-Digit Tracking └ Sub-archetype Power Cycle Lookup

units-digit-trackingmodular-arithmetic-mod-10pattern-recognition pattern-recognitioneasier-related-problem ↑ Prerequisites: units-digit-tracking

📏 Long solution 💡 3 insights

Problem

Let k = 2008² + 2²⁰⁰⁸, a whole number far too big to write out. Find the units digit (the last digit) of k² + 2^k.

Pick an answer.

(A)

$\ 0$

(B)

$\ 2$

(C)

$\ 4$

(D)

$\ 6$

(E)

$\ 8$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

The numbers are far too big to compute, but a units digit only depends on units digits, so the real problem is much smaller than it looks. The last digit of a power of a fixed base repeats in a short cycle, so finding and using that repeating pattern is the key move. Break the work into pieces: first the last digit of k, then the last digit of k², then the last digit of 2^k, and finally add.

1STEP 1

Only last digits matter

When you add or multiply whole numbers, the answer's last digit comes only from the pieces' last digits — no giant numbers needed.

units(a+b) and units(a × b) depend only on units(a), units(b)

2STEP 2

Last digit of the powers of 2

The last digits of 2¹, 2², 2³, 2⁴ run 2, 4, 8, 6 and then repeat, so powers of 2 have a cycle of length 4.

2¹,2²,2³,2⁴,2⁵,… → 2,4,8,6,2,… (cycle length 4)

3STEP 3

Find the last digit of k

8² = 64 ends in 4, and 2008 is a multiple of 4 so 2²⁰⁰⁸ ends in 6; since 4 + 6 = 10, k ends in 0.

2008² → 8²=64 → 4; 2²⁰⁰⁸ (2008=4 · 502) → 6; 4+6=10 → 0

4STEP 4

Last digit of k squared

A number ending in 0 squares to a number ending in 0, since nothing else reaches the ones place, so k² ends in 0.

k ≡ 0 (mod 10) → k² ends in 0

5STEP 5

Last digit of 2 to the k

For 2^k read k's remainder after fours, not its last digit: 2008² and 2²⁰⁰⁸ are both multiples of 4, so 2^k ends in 6.

2008²≡ 0, 2²⁰⁰⁸≡ 0 (mod 4) → k≡ 0 (mod 4) → 2^k ends in 6

6STEP 6

Add the two last digits

k² ends in 0 and 2^k ends in 6, and 0 + 6 = 6 needs no carry, so the units digit is 6 — choice (D).

0 + 6 = 6

Answer

Every step stayed inside single digits, and the two independent pieces behaved as expected: a number ending in 0 squares to something ending in 0, and a power of 2 whose exponent is a multiple of 4 ends in 6. Their sum 0 + 6 = 6 needs no carry, so the last digit is cleanly 6, matching (D). As a sanity check, the last digit of 2^k had to be one of 2, 4, 8, 6, and adding 0 to any of those can never produce an odd digit, so an even answer like 6 is exactly the right shape.

💡Key takeaway

To get a last digit of a huge power, list the last digits until they repeat, then jump to your exponent's spot in that short cycle.

Only last digits matter
Last digit of the powers of 2
Find the last digit of k
Last digit of k squared
Last digit of 2 to the k
Add the two last digits