AMC 10 · 2008 · #15

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

A number is built from a large square plus a large power of two. Find the last digit of that number squared plus two raised to it.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Only last digits matter for sums and products.

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

2STEP 2

Last digit of the powers of 2

The powers of two cycle with length four.

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

3STEP 3

Find the last digit of k

The number itself ends in zero.

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

4STEP 4

Last digit of k squared

So its square ends in zero too.

k ≡ 0 (mod 10) → k² ends in 0

5STEP 5

Last digit of 2 to the k

The exponent's remainder mod four gives the other last digit as 6.

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

6STEP 6

Add the two last digits

Adding gives 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