If m is a positive integer, then m^3 has how many digits? m has 3 digits. m^2 has 5 digits....

Understanding the Question

We need to find how many digits \(\mathrm{m}^3\) has, where \(\mathrm{m}\) is a positive integer.

In simpler terms: We're looking for the exact number of digits in \(\mathrm{m}^3\). For example, if \(\mathrm{m}^3 = 1{,}000{,}000\), it has 7 digits. If \(\mathrm{m}^3 = 10{,}000{,}000\), it has 8 digits.

What "sufficient" means here: We have sufficient information if we can determine exactly one possible value for the number of digits in \(\mathrm{m}^3\). If \(\mathrm{m}^3\) could have 7 digits OR 8 digits, that's not sufficient. [STOP - Not Sufficient!]

Given information:

• \(\mathrm{m}\) is a positive integer
• We need the digit count of \(\mathrm{m}^3\)

Key insight: When we cube a number, it grows exponentially. Even small changes in \(\mathrm{m}\) can lead to large changes in \(\mathrm{m}^3\), potentially changing the digit count.

Analyzing Statement 1

Statement 1 tells us: \(\mathrm{m}\) has 3 digits (meaning \(100 \leq \mathrm{m} \leq 999\))

Let's think about what happens when we cube numbers in this range:

• At the low end: \(\mathrm{m} = 100\), so \(\mathrm{m}^3 = 100^3 = 1{,}000{,}000\) (7 digits)
• At the high end: \(\mathrm{m} = 999\), so \(\mathrm{m}^3\) is nearly \(1000^3 = 1{,}000{,}000{,}000\) (9 digits)

This is a massive range! Going from \(100^3\) to \(1000^3\) represents a 1,000-fold increase (since \((10\mathrm{x})^3 = 1000\mathrm{x}^3\)).

To see this more clearly:

• When \(\mathrm{m} = 100\): \(\mathrm{m}^3 = 1{,}000{,}000\) (7 digits)
• When \(\mathrm{m} = 300\): \(\mathrm{m}^3 = 27{,}000{,}000\) (8 digits)
• When \(\mathrm{m} = 999\): \(\mathrm{m}^3\) approaches 1 billion (9 digits)

Since \(\mathrm{m}^3\) can have 7, 8, or 9 digits depending on the value of \(\mathrm{m}\), Statement 1 is NOT sufficient. [STOP - Not Sufficient!]

This eliminates choices A and D.

Analyzing Statement 2

Important: Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: \(\mathrm{m}^2\) has 5 digits (meaning \(10{,}000 \leq \mathrm{m}^2 < 100{,}000\))

This gives us a much narrower range for \(\mathrm{m}\). If \(\mathrm{m}^2\) has 5 digits:

• The smallest \(\mathrm{m}\) would be when \(\mathrm{m}^2 = 10{,}000\), so \(\mathrm{m} = 100\)
• The largest \(\mathrm{m}\) would be when \(\mathrm{m}^2 < 100{,}000\), so \(\mathrm{m} < \sqrt{100{,}000} \approx 316\)

Therefore: \(100 \leq \mathrm{m} < 317\) (approximately)

Now, what happens to \(\mathrm{m}^3\) in this narrower range?

• When \(\mathrm{m} = 100\): \(\mathrm{m}^3 = 1{,}000{,}000\) (just over 1 million) → 7 digits
• When \(\mathrm{m} = 316\): \(\mathrm{m}^3\) would be around 31 million → 8 digits

Here's the critical observation: Even in this narrower range, we jump from "just over 1 million" to "tens of millions." This represents a change from 7 digits to 8 digits!

To visualize the boundary:

• \(\mathrm{m} = 100\) to 215: \(\mathrm{m}^3\) stays in the millions (7 digits)
• \(\mathrm{m} = 216\) to 316: \(\mathrm{m}^3\) moves into the tens of millions (8 digits)

Since \(\mathrm{m}^3\) can have either 7 or 8 digits, Statement 2 is NOT sufficient. [STOP - Not Sufficient!]

This eliminates choice B.

Combining Statements

Let's see what happens when we use both statements together.

From both statements:

• Statement 1: \(100 \leq \mathrm{m} \leq 999\)
• Statement 2: \(100 \leq \mathrm{m} \leq 316\) (approximately)

When we combine these constraints, we take the intersection: \(100 \leq \mathrm{m} \leq 316\)

But wait—this is exactly the same range we had from Statement 2 alone! The combination doesn't give us any new restrictions beyond what Statement 2 already provided.

Since Statement 2 alone showed that \(\mathrm{m}^3\) could have either 7 or 8 digits in this range, the combined statements still allow \(\mathrm{m}^3\) to have either 7 or 8 digits.

Therefore, both statements together are NOT sufficient. [STOP - Not Sufficient!]

This eliminates choice C.

The Answer: E

The statements together are not sufficient to determine how many digits \(\mathrm{m}^3\) has, as \(\mathrm{m}^3\) can have either 7 or 8 digits even when both conditions are satisfied.

Key takeaway: When dealing with powers and digit counts, remember that exponential growth means even relatively narrow ranges (like 100 to 316) can produce outputs that span multiple digit counts. Always check the boundaries!

Answer Choice E: "The statements together are not sufficient."