AMC 10 · 2022 · #11

Grade 8 arithmetic

exponentspolynomial-roots convert-to-algebrapattern-recognition ↑ Prerequisites: exponents

📏 Medium solution 💡 2 insights

Problem

Ted should have written 2^m · √(

\frac{1}{4096}

), but he wrote 2 · sqrt[m]{

\frac{1}{4096}

} instead. For which real numbers m do these two expressions happen to give the same value? Find all such m and report their sum.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

The problem hands us an equation in disguise: "two expressions are equal". Tool #5 (Pattern) spots that 4096 is a power of 2, so every quantity in sight lives on a single base-2 ladder — turning radicals into clean exponents. Tool #13 (Convert to Algebra) then equates the two exponents (same base, so exponents match) and produces a tidy quadratic in m. We could find each root and add, but Tool #11 (Work Backwards from "sum of roots") plus Vieta's shortcut - $\frac{b}{a}$ skips solving and answers directly. Tool #3 (Eliminate Possibilities) at the end matches the integer sum against the five choices.

1STEP 1

Spot that 4096 = 2¹², so $\frac{1}{4096}$ = 2⁻¹² — every radical and power now sits on one base-2 ladder.

4096 = 2¹² →

\frac{1}{4096}

= 2⁻¹²

2STEP 2

Rewrite radicals as fractional exponents (√x = x¹/2, m-th root = x¹/m); both sides become clean powers of 2.

2^m · 2⁻¹2 ·

\frac{1}{2}

= 2¹ · 2⁻¹2 ·

\frac{1}{m}

3STEP 3

Combine exponents on each side (a^b·a^c = a^b+c); equal bases 2^m-6 = 2^1- $\frac{12}{m}$ force m - 6 = 1 - $\frac{12}{m}$ .

2^m-6 = 2¹ -

\frac{12}{m}

→ m - 6 = 1 -

\frac{12}{m}

4STEP 4

Multiply through by m (valid since m ≠ 0) and collect terms: m² - 7m + 12 = 0.

m(m - 6) = m - 12 → m² - 6m - m + 12 = 0 → m² - 7m + 12 = 0

5STEP 5

We need the sum of the roots, not each root: Vieta gives sum = - $\frac{b}{a}$ = 7; check (m-3)(m-4)=0 → 3+4 = 7.

sum of roots = -

\frac{b}{a}

= --

\frac{7}{1}

= 7 [(m-3)(m-4)=0, 3+4=7]

6STEP 6

Match the sum 7 to the options: 7 is choice (C); the others (5, 6, 8, 9) don't fit.

sum = 7 → (C)

Answer

Plug both roots back into the original equation. For m = 3: left = 2³ · √(2⁻¹²) = 8 · 2⁻⁶ =

\frac{8}{64}

\frac{1}{8}

; right = 2 · sqrt[3]{2⁻¹²} = 2 · 2⁻⁴ =

\frac{2}{16}

\frac{1}{8}

. Match. For m = 4: left = 2⁴ · 2⁻⁶ =

\frac{16}{64}

\frac{1}{4}

; right = 2 · sqrt[4]{2⁻¹²} = 2 · 2⁻³ =

\frac{2}{8}

\frac{1}{4}

. Match. Both roots are real and valid, so the sum 3 + 4 = 7 is correct and the answer is (C).

💡Key takeaway

This AMC 10 problem only needs Grade 8 integer-exponent rules you already know — once you turn every 4096 into 2¹² and equate the two exponents, you get a tidy quadratic m² - 7m + 12 = 0, and the sum of its roots is just - $\frac{b}{a}$ = 7.