AMC 10 · 2005 · #13
Grade 8 arithmeticHow many positive integers n satisfy the following condition:
(130n)50>n100>2200 ?
Pick an answer.
AMC 10 2005 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: Count the positive integers $n$ for which the chain of inequalities $(130n)^{50} > n^{100} > 2^{200}$ holds. The exponents are enormous, so the real work is turning them into a plain range of allowed $n$.
Givens: The chain $(130n)^{50} > n^{100} > 2^{200}$ must hold; $n$ is a positive integer; Answer choices: (A) $0$, (B) $7$, (C) $12$, (D) $65$, (E) $125$
Unknowns: The range of positive integers $n$ that satisfy both inequalities; How many such integers there are
Understand
Restated: Count the positive integers $n$ for which the chain of inequalities $(130n)^{50} > n^{100} > 2^{200}$ holds. The exponents are enormous, so the real work is turning them into a plain range of allowed $n$.
Givens: The chain $(130n)^{50} > n^{100} > 2^{200}$ must hold; $n$ is a positive integer; Answer choices: (A) $0$, (B) $7$, (C) $12$, (D) $65$, (E) $125$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Numbers like $2^{200}$ are impossible to compare directly, so Tool #9 (Solve an Easier Related Problem) is the whole game: rewrite each power so both sides share one exponent, then compare the bases instead of the giant numbers. Tool #7 (Identify Subproblems) splits the double inequality into two independent inequalities — one giving an upper bound on $n$, one a lower bound. Tool #3 (Eliminate Possibilities) finishes by counting the integers in the surviving range and matching to the answer list.
Execute — Answer: E
8.EE.A.1 Step 1 Match powers in the left inequality
- Look at $(130n)^{50} > n^{100}$.
- The right side is already a $50$th power in disguise: $n^{100}=(n^2)^{50}$.
- So the inequality reads $(130n)^{50} > (n^2)^{50}$ — the same exponent $50$ on both sides.
💡 $100$ is $2\times 50$, so $n^{100}$ is really $(n^2)$ raised to the $50$th — that lets both sides wear the same exponent.
7.EE.B.4 Step 2 Compare the bases, not the giants
- Both sides are positive numbers raised to the same power $50$.
- Raising to a positive power keeps order, so the bigger base wins: $(130n)^{50} > (n^2)^{50}$ is true exactly when $130n > n^2$.
- Since $n>0$, divide both sides by $n$ to get $130 > n$.
- That means $n$ can be at most $129$.
💡 If two positive numbers to the same power compare a certain way, their bases compare the same way, so drop the exponent and read the small inequality.
8.EE.A.1 Step 3 Match powers in the right inequality
- Now the other half: $n^{100} > 2^{200}$.
- Rewrite the giant on the right so it becomes a $100$th power: $2^{200}=(2^2)^{100}=4^{100}$.
- The inequality becomes $n^{100} > 4^{100}$, again with one shared exponent, $100$.
💡 $200$ is $2\times 100$, so $2^{200}$ folds into $4^{100}$ and lines up with $n^{100}$.
7.EE.B.4 Step 4 Get the lower bound on n
- Both sides are positive numbers to the same power $100$, so compare bases: $n^{100} > 4^{100}$ holds exactly when $n > 4$.
- As a positive integer, $n$ must be at least $5$.
💡 Same power on both sides means the base $n$ just has to beat $4$.
4.OA.A.3 Step 5 Count the surviving integers
- Both conditions must hold at once, so $n$ satisfies $5 \le n \le 129$.
- The count of whole numbers from $5$ up to $129$ is $129-5+1=125$.
- Scanning the answer list, $125$ is choice (E).
💡 To count a run of integers, subtract the ends and add one for the fencepost you would otherwise miss.
8.EE.A.1 Look at $(130n)^{50} > n^{100}$. The right side is already a $50$th power in dis 7.EE.B.4 Both sides are positive numbers raised to the same power $50$. Raising to a posi 8.EE.A.1 Now the other half: $n^{100} > 2^{200}$. Rewrite the giant on the right so it be 7.EE.B.4 Both sides are positive numbers to the same power $100$, so compare bases: $n^{1 4.OA.A.3 Both conditions must hold at once, so $n$ satisfies $5 \le n \le 129$. The count Review
Reasonableness: The two bounds come from independent, honest steps: $n^2 < 130n$ caps $n$ at $129$, and $n>4$ floors it at $5$. Spot-check the ends. At $n=5$: $n^{100}=5^{100}>4^{100}=2^{200}$ (true) and $(130\cdot 5)^{50}=650^{50}>25^{50}=5^{100}$ (true), so $5$ works. At $n=129$: $129<130$ keeps the left inequality, and $129>4$ keeps the right, so $129$ works. Just outside, $n=4$ fails ($4^{100}=2^{200}$ is not strictly greater than itself) and $n=130$ fails ($130\cdot 130=130^2$ is not strictly greater than $130^2$). So the range is exactly $5$ through $129$, giving $125$ integers, matching choice (E).
Alternative: Take the $50$th root of the whole chain at once. Since everything is positive, $(130n)^{50} > n^{100} > 2^{200}$ becomes $130n > n^2 > 2^4$, i.e. $130n > n^2$ and $n^2 > 16$. The first gives $n<130$; the second gives $n>4$ (as $n$ is a positive integer). Same window $5\le n\le 129$, same count $125$.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Rewriting $n^{100}=(n^2)^{50}$ and $2^{200}=4^{100}$ so each inequality has one shared exponent and the bases can be compared.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Solving $130n>n^2$ down to $n\le 129$ and $n^{100}>4^{100}$ down to $n\ge 5$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Counting the integers in $5\le n\le 129$ as $129-5+1=125$.)
⭐ When powers are too big to compare, rewrite them to share one exponent, then just compare the bases.
⭐ When powers are too big to compare, rewrite them to share one exponent, then just compare the bases.
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