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What is the least value of n such that the value of \((8^{\mathrm{n}})(25^{24})\) has 49 or more digits?
Let's start by understanding what it means for a number to have 49 digits. Think about familiar numbers: 1 has 1 digit, 10 has 2 digits, 100 has 3 digits, and so on. We can see that a number has exactly n digits when it's between \(10^{(n-1)}\) and \(10^n - 1\).
For example:
So for a number to have 49 or more digits, it must be at least \(10^{48}\).
We need: \((8^n)(25^{24}) \geq 10^{48}\)
Process Skill: TRANSLATE - Converting the digit requirement into a mathematical inequality
Our goal is clear now. We need to find the smallest value of n where:
\((8^n)(25^{24}) \geq 10^{48}\)
This gives us a concrete target to work toward - we need our expression to be at least as large as \(10^{48}\).
Here's where we use a key insight. Let's break down our numbers into their prime factors:
So our expression becomes:
\((8^n)(25^{24}) = (2^3)^n \times (5^2)^{24} = 2^{(3n)} \times 5^{48}\)
Now here's the clever part: since \(10 = 2 \times 5\), we can create factors of 10 by pairing up 2's and 5's. We have \(5^{48}\), which means we have 48 factors of 5. If we have at least 48 factors of 2, we can make 48 factors of 10.
Process Skill: SIMPLIFY - Breaking complex expressions into manageable prime factors
We have \(2^{(3n)} \times 5^{48}\). Let's see how many complete factors of 10 we can make:
If \(3n \geq 48\), then we can write:
\(2^{(3n)} \times 5^{48} = 2^{48} \times 2^{(3n-48)} \times 5^{48} = (2 \times 5)^{48} \times 2^{(3n-48)} = 10^{48} \times 2^{(3n-48)}\)
For our inequality \((8^n)(25^{24}) \geq 10^{48}\) to hold, we need:
\(10^{48} \times 2^{(3n-48)} \geq 10^{48}\)
Dividing both sides by \(10^{48}\):
\(2^{(3n-48)} \geq 1\)
Since \(2^0 = 1\), we need:
\(3n - 48 \geq 0\)
\(3n \geq 48\)
\(n \geq 16\)
From our work above, we found that \(n \geq 16\).
Let's verify this works:
Let's check \(n = 15\) to confirm it's too small:
Therefore, the minimum value is \(n = 16\).
The least value of n such that \((8^n)(25^{24})\) has 49 or more digits is 16.
The answer is C.
1. Misunderstanding the digit count requirement: Students often confuse "49 or more digits" with "exactly 49 digits" or "more than 49 digits." This leads them to set up the wrong inequality - using > instead of ≥, or aiming for \(10^{49}\) instead of \(10^{48}\) as the threshold.
2. Incorrect digit-to-power relationship: Students may incorrectly think that a number with n digits must be at least \(10^n\) instead of \(10^{(n-1)}\). For example, thinking 100 (3 digits) corresponds to \(10^3\) rather than being \(\geq 10^2\). This shifts their entire target by one power of 10.
3. Missing the prime factorization insight: Students might try to compute \((8^n)(25^{24})\) directly or convert everything to the same base without recognizing that factoring into 2's and 5's allows efficient pairing to create powers of 10. This leads to much more complex calculations.
1. Incorrect exponent manipulation: When converting \(8^n\) to \((2^3)^n\), students often write \(2^{(n+3)}\) instead of \(2^{(3n)}\), or when converting \(25^{24}\) to \((5^2)^{24}\), they write \(5^{(24+2)}\) instead of \(5^{48}\). These exponent rule errors completely derail the solution.
2. Mishandling the pairing of 2's and 5's: Students may incorrectly assume they can always make the maximum number of 10's without checking which prime factor is limiting. For instance, if they have \(2^{45} \times 5^{48}\), they might incorrectly think they can make 48 factors of 10 instead of only 45.
3. Algebraic errors in the inequality: When solving \(3n \geq 48\), students might make basic algebraic mistakes like forgetting to divide by 3, or incorrectly handling the inequality direction, leading to wrong values like \(n \geq 48\) or \(n \geq 144\).
1. Boundary condition confusion: Students might find \(n \geq 16\) but then select \(n = 15\), thinking they need the largest value that's less than 16, rather than understanding they need the smallest value that satisfies the inequality.
2. Verification errors: Students may correctly find \(n = 16\) but then incorrectly verify their answer, making computational errors that cause them to doubt their result and select a different answer choice instead.