AMC 8 · 2010 · #24
Grade 8 arithmeticProblem
What is the correct ordering of the three numbers, 108, 512, and 224?
Pick an answer.
AMC 8 2010 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: Put the three numbers $10^8$, $5^{12}$, and $2^{24}$ in order from least to greatest, and pick the matching answer choice.
Givens: Three powers: $10^8$, $5^{12}$, and $2^{24}$; All three exponents are multiples of $4$: $8 = 4 \times 2$, $12 = 4 \times 3$, $24 = 4 \times 6$; Answer choices: (A) $2^{24}<10^8<5^{12}$, (B) $2^{24}<5^{12}<10^8$, (C) $5^{12}<2^{24}<10^8$, (D) $10^8<5^{12}<2^{24}$, (E) $10^8<2^{24}<5^{12}$
Unknowns: The correct ordering of $10^8$, $5^{12}$, and $2^{24}$
Understand
Restated: Put the three numbers $10^8$, $5^{12}$, and $2^{24}$ in order from least to greatest, and pick the matching answer choice.
Givens: Three powers: $10^8$, $5^{12}$, and $2^{24}$; All three exponents are multiples of $4$: $8 = 4 \times 2$, $12 = 4 \times 3$, $24 = 4 \times 6$; Answer choices: (A) $2^{24}<10^8<5^{12}$, (B) $2^{24}<5^{12}<10^8$, (C) $5^{12}<2^{24}<10^8$, (D) $10^8<5^{12}<2^{24}$, (E) $10^8<2^{24}<5^{12}$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems
Comparing $9$-digit numbers head-on is messy, but the exponents $8$, $12$, $24$ all share the factor $4$. So we replace the original problem with the easier related problem of comparing the fourth roots — Tool #9. Taking a fourth root preserves order (the function $x \mapsto x^{1/4}$ is increasing on positive numbers), so the ordering of the fourth roots is exactly the ordering of the originals. Tool #7 then turns the three-way comparison into easy pairwise subproblems on the small numbers $10^2$, $5^3$, $2^6$, which any 6th grader can evaluate.
Execute — Answer: A
8.EE.A.1 Step 1 - Use the exponent law $(a^m)^n = a^{mn}$ to rewrite each number as something raised to the $4$th power.
- Since $8 = 2 \cdot 4$, $12 = 3 \cdot 4$, and $24 = 6 \cdot 4$, we get a common outer exponent of $4$.
💡 Sharing a common exponent ($4$) is the simplification that makes the comparison easy — only the bases $10^2$, $5^3$, $2^6$ will matter.
6.EE.A.1 Step 2 - Because $x \mapsto x^4$ is increasing on positive numbers, comparing $A^4$ with $B^4$ is the same as comparing $A$ with $B$.
- So ordering $10^8$, $5^{12}$, $2^{24}$ is the same as ordering the bases $10^2$, $5^3$, $2^6$ — a much smaller problem.
💡 This is the Tool #9 move: replace a hard problem with an order-equivalent easier one whose numbers fit on one line.
6.EE.A.1 Step 3 - Evaluate each of the three small powers.
- These are basic exponent computations a 6th grader can do in their head.
💡 Computing $10^2$, $5^3$, $2^6$ separately is the Tool #7 subproblems move — three tiny independent calculations.
5.NBT.A.2 Step 4 Sort the three small results from least to greatest.
💡 Once everything fits inside three-digit numbers, ordering them is just a quick number-line comparison.
8.EE.A.1 Step 5 - Lift the ordering back to the originals: since each small number was a $4$th root of the matching big number, the same inequality holds for the original powers.
- That matches choice (A).
💡 The fourth-power function preserves order, so the easier comparison transfers cleanly to the original one.
8.EE.A.1 Use the exponent law $(a^m)^n = a^{mn}$ to rewrite each number as something rais 6.EE.A.1 Because $x \mapsto x^4$ is increasing on positive numbers, comparing $A^4$ with 6.EE.A.1 Evaluate each of the three small powers. These are basic exponent computations a 5.NBT.A.2 Sort the three small results from least to greatest. 8.EE.A.1 Lift the ordering back to the originals: since each small number was a $4$th roo Review
Reasonableness: Sanity check with rough magnitudes. $2^{10} \approx 1000$, so $2^{24} = 2^{20} \cdot 2^4 \approx 10^6 \cdot 16 \approx 1.6 \times 10^7$, which is well under $10^8$. And $5^{12} = 5^{12}$ versus $10^8 = (5 \cdot 2)^8 = 5^8 \cdot 2^8 = 5^8 \cdot 256$, while $5^{12} = 5^8 \cdot 5^4 = 5^8 \cdot 625$. Since $625 > 256$, $5^{12} > 10^8$. Both spot checks agree with $2^{24} < 10^8 < 5^{12}$.
Alternative: Tool #5 (Look for a Pattern) on a shared exponent of $8$ instead of a $4$th root. Rewrite $2^{24} = (2^3)^8 = 8^8$, leave $10^8$ as is, and rewrite $5^{12} = (5^{3/2})^8 = (\sqrt{125})^8$. Since the exponent is now common, ordering reduces to comparing the bases $8$, $10$, $\sqrt{125} \approx 11.18$. So $8 < 10 < \sqrt{125}$, giving the same answer $2^{24} < 10^8 < 5^{12}$, choice (A).
CCSS standards used (min grade 8)
5.NBT.A.2Explain patterns in the number of zeros and powers of 10 (Reading $10^8$ as a power of $10$ and ordering the small whole numbers $64$, $100$, $125$ on the number line.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the easier related expressions $10^2 = 100$, $5^3 = 125$, $2^6 = 64$.)8.EE.A.1Know and apply the properties of integer exponents (Rewriting $10^8 = (10^2)^4$, $5^{12} = (5^3)^4$, $2^{24} = (2^6)^4$ via $(a^m)^n = a^{mn}$ so all three share a common exponent.)
⭐ When numbers are too big to compute, look for a shared exponent — taking the same root of all of them keeps the order but shrinks the numbers to something you can handle by hand.
⭐ When numbers are too big to compute, look for a shared exponent — taking the same root of all of them keeps the order but shrinks the numbers to something you can handle by hand.
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