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 $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$ ?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Let k = 2008^2 + 2^2008. Find the units digit (the last digit) of the number k^2 + 2^k.

Givens: k is defined as 2008^2 + 2^2008.; The target expression is k^2 + 2^k.

Unknowns: The units digit of k^2 + 2^k.

Understand

Restated: Let k = 2008^2 + 2^2008. Find the units digit (the last digit) of the number k^2 + 2^k.

Givens: k is defined as 2008^2 + 2^2008.; The target expression is k^2 + 2^k.

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

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 k^2, then the last digit of 2^k, and finally add.

Execute — Answer: D

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 1

Only last digits matter

When you add or multiply whole numbers, the last digit of the answer is decided entirely by the last digits of the pieces.
So to get the last digit of k^2 + 2^k, we never need the full giant numbers.
We only need the last digit of k (which controls k^2) and the last digit of 2^k.
This shrinks a monstrous problem down to single-digit bookkeeping.

$$\text{units}(a+b)\ \text{and}\ \text{units}(a\times b)\ \text{depend only on}\ \text{units}(a),\ \text{units}(b)$$

💡 Carrying only moves left, so the far-right digit is sealed off from everything above it.

#5 Look for a Pattern 4.OA.C.5 Step 2

Last digit of the powers of 2

List the last digits of powers of 2: 2^1 ends in 2, 2^2 ends in 4, 2^3 ends in 8, 2^4 ends in 6, then 2^5 ends in 2 again.
The last digit repeats in the block 2, 4, 8, 6 with length 4.
To read off any power, you just need where its exponent lands inside this length-4 cycle, which is the exponent's remainder when divided by 4.
An exponent that is a multiple of 4 sits at the end of the block, giving last digit 6.

$$2^1,2^2,2^3,2^4,2^5,\dots \to 2,4,8,6,2,\dots \quad(\text{cycle length }4)$$

💡 Each new power just multiplies the last digit by 2, and that little machine loops back after four steps.

#7 Identify Subproblems 4.OA.B.4 Step 3

Find the last digit of k

Split k = 2008^2 + 2^2008 into two last digits.
For 2008^2, only the last digit 8 matters, and 8^2 = 64 ends in 4.
For 2^2008, use the cycle: 2008 = 4 x 502 is a multiple of 4, so 2^2008 sits at the end of the block and ends in 6.
Add the last digits: 4 + 6 = 10, which ends in 0.
So k ends in 0.

$$2008^2 \to 8^2=64 \to 4;\quad 2^{2008}\ (2008=4\cdot 502) \to 6;\quad 4+6=10 \to 0$$

💡 A number's last digit is the leftover after knocking out full tens, so adding two last digits and keeping the ones place is enough.

#7 Identify Subproblems 4.NBT.B.5 Step 4

Last digit of k squared

We just found that k ends in 0.
Squaring a number that ends in 0 gives a number that ends in 0, since 0 x 0 = 0 and nothing else reaches the ones place.
So k^2 ends in 0.

$$k \equiv 0 \pmod{10} \;\Rightarrow\; k^2 \;\text{ends in}\; 0$$

💡 A trailing zero, when squared, only makes more trailing zeros.

#5 Look for a Pattern 4.OA.B.4 Step 5

Last digit of 2 to the k

For 2^k we need where k lands in the length-4 cycle, so we need k's remainder mod 4, not mod 10.
Check both parts of k: 2008 is a multiple of 4, so 2008^2 is a multiple of 4; and 2^2008 has plenty of factors of 2, so it is a multiple of 4.
A sum of two multiples of 4 is a multiple of 4, so k is a multiple of 4.
That puts 2^k at the end of the block, so 2^k ends in 6.

$$2008^2\equiv 0,\; 2^{2008}\equiv 0 \pmod 4 \;\Rightarrow\; k\equiv 0 \pmod 4 \;\Rightarrow\; 2^k \;\text{ends in}\; 6$$

💡 The power cycle is read by the exponent's remainder after fours, not by its last digit.

#7 Identify Subproblems 4.NBT.B.5 Step 6

Add the two last digits

The last digit of k^2 is 0 and the last digit of 2^k is 6.
Add them: 0 + 6 = 6.
So k^2 + 2^k ends in 6, which is choice (D).

$$0 + 6 = 6$$

💡 With both last digits in hand, the final last digit is just their ones-place sum.

[1] #9 4.NBT.B.5 When you add or multiply whole numbers, the last digit of the answer is decided

[2] #5 4.OA.C.5 List the last digits of powers of 2: 2^1 ends in 2, 2^2 ends in 4, 2^3 ends in 8

[3] #7 4.OA.B.4 Split k = 2008^2 + 2^2008 into two last digits. For 2008^2, only the last digit

[4] #7 4.NBT.B.5 We just found that k ends in 0. Squaring a number that ends in 0 gives a number

[5] #5 4.OA.B.4 For 2^k we need where k lands in the length-4 cycle, so we need k's remainder mo

[6] #7 4.NBT.B.5 The last digit of k^2 is 0 and the last digit of 2^k is 6. Add them: 0 + 6 = 6.

Review

Reasonableness: 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.

Alternative: Work purely with remainders mod 10 and mod 4. Compute k mod 10 = (8^2 + 2^2008) mod 10 = (4 + 6) mod 10 = 0, so k^2 mod 10 = 0. Compute k mod 4 = 0 to read 2^k mod 10 = 6 from the cycle. Then (k^2 + 2^k) mod 10 = (0 + 6) mod 10 = 6.

CCSS standards used (min grade 6)

4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Using the fact that the last digit of a product or sum comes only from the last digits, then squaring the last digit of k and adding the two final last digits.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Finding and using the repeating cycle 2, 4, 8, 6 in the last digits of the powers of 2.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Checking that 2008 and k are multiples of 4 to locate the exponent's spot in the length-4 cycle.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Reading and reasoning about the exponent notation 2008^2, 2^2008, k^2, and 2^k.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

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

Review

CCSS standards used (min grade 6)

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