AMC 10 · 2022 · #11
Grade 8 arithmeticProblem
Ted mistakenly wrote 2m⋅40961 as 2⋅m40961. What is the sum of all real numbers m for which these two expressions have the same value?
Pick an answer.
AMC 10 2022 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.
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
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) $9$
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
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.
💡 Grade-8 integer-exponent rules let us trade an awkward fraction for a clean $2^{-12}$.
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.
💡 Radicals and fractional exponents are the same idea written two ways — Grade 8 makes the conversion.
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.
💡 Equal powers of the same base mean equal exponents — that is exactly how exponent equations get solved in Grade 8.
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 $m$.
💡 Multiplying both sides by $m$ is a Grade-8 linear-equation move; rearranging gives a polynomial we can read off.
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.
💡 Going backwards from the quadratic's coefficients straight to the root-sum is faster than solving.
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.
💡 Comparing our computed answer against the five options is the standard multiple-choice closeout.
8.EE.A.1 Spot the pattern: $4096 = 2^{12}$ is a power of $2$. Rewrite the reciprocal as $ 8.EE.A.2 Turn each radical into a fractional exponent: $\sqrt{x} = x^{1/2}$ and $\sqrt[m] 8.EE.A.1 Combine exponents on each side using $a^{b} \cdot a^{c} = a^{b+c}$. The left sid 8.EE.C.7 Clear the fraction by multiplying through by $m$ (allowed since $m \neq 0$), the 8.EE.C.7 Tool #11 (Work Backwards): we want the SUM of the roots, not each root. For $am^ 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.1Know 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.2Use 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.7Solve 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.5Understand 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 $-b/a = 7$.
⭐ 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 $-b/a = 7$.