AMC 10 · 2022 · #11

Grade 8 arithmetic

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

📏 Medium solution 💡 2 insights

Problem

Ted mistakenly wrote $2^m\cdot\sqrt{\frac{1}{4096}}$ as $2\cdot\sqrt[m]{\frac{1}{4096}}.$ What is the sum of all real numbers $m$ for which these two expressions have the same value?

$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Ted should have written $2^{m} \cdot \sqrt{\frac{1}{4096}}$, but he wrote $2 \cdot \sqrt[m]{\frac{1}{4096}}$ instead. For which real numbers $m$ do these two expressions happen to give the same value? Find all such $m$ and report their sum.

Givens: Original expression: $2^{m} \cdot \sqrt{\dfrac{1}{4096}}$; Ted's mistake: $2 \cdot \sqrt[m]{\dfrac{1}{4096}}$; $4096 = 2^{12}$ (so $\dfrac{1}{4096} = 2^{-12}$); Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$

Unknowns: The sum of all real $m$ that make the two expressions equal

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #5 Look for a Pattern, #11 Work Backwards, #3 Eliminate Possibilities

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 $-b/a$ skips solving and answers directly. Tool #3 (Eliminate Possibilities) at the end matches the integer sum against the five choices.

Execute — Answer: C

#5 Look for a Pattern 8.EE.A.1 Step 1

Spot the pattern: $4096 = 2^{12}$ is a power of $2$.
Rewrite the reciprocal as $\dfrac{1}{4096} = 2^{-12}$.
This lets every radical and power live on the same base-$2$ ladder.

$$4096 = 2^{12} \;\Rightarrow\; \dfrac{1}{4096} = 2^{-12}$$

💡 Grade-8 integer-exponent rules let us trade an awkward fraction for a clean $2^{-12}$.

#5 Look for a Pattern 8.EE.A.2 Step 2

Turn each radical into a fractional exponent: $\sqrt{x} = x^{1/2}$ and $\sqrt[m]{x} = x^{1/m}$.
The two sides become $2^{m} \cdot (2^{-12})^{1/2}$ on the left and $2 \cdot (2^{-12})^{1/m}$ on the right.

$$2^{m} \cdot 2^{-12 \cdot \frac{1}{2}} = 2^{1} \cdot 2^{-12 \cdot \frac{1}{m}}$$

💡 Radicals and fractional exponents are the same idea written two ways — Grade 8 makes the conversion.

#13 Convert to Algebra 8.EE.A.1 Step 3

Combine exponents on each side using $a^{b} \cdot a^{c} = a^{b+c}$.
The left side becomes $2^{m-6}$, the right side $2^{1 - 12/m}$.
Same base on both sides forces the exponents to be equal.

$$2^{m-6} = 2^{1 - \frac{12}{m}} \;\Rightarrow\; m - 6 = 1 - \dfrac{12}{m}$$

💡 Equal powers of the same base mean equal exponents — that is exactly how exponent equations get solved in Grade 8.

#13 Convert to Algebra 8.EE.C.7 Step 4

Clear the fraction by multiplying through by $m$ (allowed since $m \neq 0$), then move every term to one side to get a standard quadratic in $m$.

$$m(m - 6) = m - 12 \;\Rightarrow\; m^{2} - 6m - m + 12 = 0 \;\Rightarrow\; m^{2} - 7m + 12 = 0$$

💡 Multiplying both sides by $m$ is a Grade-8 linear-equation move; rearranging gives a polynomial we can read off.

#11 Work Backwards 8.EE.C.7 Step 5

Tool #11 (Work Backwards): we want the SUM of the roots, not each root.
For $am^{2} + bm + c = 0$, the sum equals $-b/a$ (Vieta).
Here $a = 1$, $b = -7$, so sum $= 7$.
Quick sanity-factor: $m^{2} - 7m + 12 = (m-3)(m-4)$, roots $3$ and $4$, sum $3 + 4 = 7$.
Matches.

$$\text{sum of roots} = -\dfrac{b}{a} = -\dfrac{-7}{1} = 7 \quad\big[(m-3)(m-4)=0,\; 3+4=7\big]$$

💡 Going backwards from the quadratic's coefficients straight to the root-sum is faster than solving.

#3 Eliminate Possibilities 6.EE.B.5 Step 6

Match $7$ to the choice list: $7$ is choice (C).
The other choices ($5, 6, 8, 9$) would require different coefficients in the quadratic and don't fit.

$$\text{sum} = 7 \;\Rightarrow\; \textbf{(C)}$$

💡 Comparing our computed answer against the five options is the standard multiple-choice closeout.

[1] #5 8.EE.A.1 Spot the pattern: $4096 = 2^{12}$ is a power of $2$. Rewrite the reciprocal as $

[2] #5 8.EE.A.2 Turn each radical into a fractional exponent: $\sqrt{x} = x^{1/2}$ and $\sqrt[m]

[3] #13 8.EE.A.1 Combine exponents on each side using $a^{b} \cdot a^{c} = a^{b+c}$. The left sid

[4] #13 8.EE.C.7 Clear the fraction by multiplying through by $m$ (allowed since $m \neq 0$), the

[5] #11 8.EE.C.7 Tool #11 (Work Backwards): we want the SUM of the roots, not each root. For $am^

[6] #3 6.EE.B.5 Match $7$ to the choice list: $7$ is choice (C). The other choices ($5, 6, 8, 9$

Review

Reasonableness: Plug both roots back into the original equation. For $m = 3$: left $= 2^{3} \cdot \sqrt{2^{-12}} = 8 \cdot 2^{-6} = 8/64 = 1/8$; right $= 2 \cdot \sqrt[3]{2^{-12}} = 2 \cdot 2^{-4} = 2/16 = 1/8$. Match. For $m = 4$: left $= 2^{4} \cdot 2^{-6} = 16/64 = 1/4$; right $= 2 \cdot \sqrt[4]{2^{-12}} = 2 \cdot 2^{-3} = 2/8 = 1/4$. Match. Both roots are real and valid, so the sum $3 + 4 = 7$ is correct and the answer is (C).

Alternative: Tool #6 (Guess and Check) using small integer guesses for $m$: try $m = 3$ — both sides give $1/8$. Try $m = 4$ — both sides give $1/4$. Try $m = 5$ — left $= 32/64 = 1/2$, right $= 2 \cdot 2^{-12/5} \approx 0.379$, mismatch. So the integer roots are $3$ and $4$. Their sum is $7$ — same answer (C) without setting up the quadratic.

CCSS standards used (min grade 8)

8.EE.A.1 Know and apply the properties of integer exponents (Rewriting $4096 = 2^{12}$, $\dfrac{1}{4096} = 2^{-12}$, and combining $2^{m} \cdot 2^{-6} = 2^{m-6}$ on both sides of the equation.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Trading $\sqrt{\,\cdot\,}$ and $\sqrt[m]{\,\cdot\,}$ for the fractional exponents $\,\cdot\,^{1/2}$ and $\,\cdot\,^{1/m}$.)
8.EE.C.7 Solve linear equations in one variable (Multiplying $m - 6 = 1 - 12/m$ by $m$ to clear the fraction and reading the quadratic $m^{2} - 7m + 12 = 0$ off its coefficients to apply Vieta's sum $-b/a$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Substituting the candidate roots $m = 3$ and $m = 4$ back into the original equation as a verification step, and comparing the sum to the multiple-choice list.)

⭐ This AMC 10 problem only needs Grade 8 integer-exponent rules you already know — once you turn every $4096$ into $2^{12}$ and equate the two exponents, you get a tidy quadratic $m^{2} - 7m + 12 = 0$, and the sum of its roots is just $-b/a = 7$.