The function f is defined for each positive three-digit integer n by \(\mathrm{f(n)} = \mathrm{2}^\mathrm{x} \cdot \mathrm{3}^\mathrm{y} \cdot \mathrm...

1. Translate the problem requirements: We need to find two three-digit numbers \(\mathrm{m}\) and \(\mathrm{v}\) where \(\mathrm{f(m) = 9 \times f(v)}\), given that \(\mathrm{f(n) = 2^x \times 3^y \times 5^z}\) for a number \(\mathrm{n}\) with digits \(\mathrm{x, y, z}\). Since \(\mathrm{9 = 3^2}\), this means the tens digit of \(\mathrm{m}\) must be exactly 2 more than the tens digit of \(\mathrm{v}\), while hundreds and units digits remain the same.
2. Express the constraint algebraically: Convert \(\mathrm{f(m) = 9 \times f(v)}\) into \(\mathrm{2^a \times 3^b \times 5^c = 3^2 \times 2^a \times 3^d \times 5^c}\), which simplifies to the requirement that \(\mathrm{b = d + 2}\).
3. Apply digit constraints: Since both \(\mathrm{m}\) and \(\mathrm{v}\) must be three-digit numbers, their digits must be between 1-9 for hundreds place and 0-9 for tens/units places. The constraint \(\mathrm{b = d + 2}\) limits our options significantly.
4. Find valid digit combinations: Test the possible values where \(\mathrm{d}\) can range from 0 to 7 (so that \(\mathrm{d + 2 ≤ 9}\)), and calculate \(\mathrm{m - v = 20}\) for any valid combination.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this function \(\mathrm{f}\) actually does in plain English. When we have a three-digit number like 234, the function \(\mathrm{f}\) takes each digit and uses it as an exponent for specific prime numbers. So for \(\mathrm{n = 234}\), we get \(\mathrm{f(234) = 2^2 \times 3^3 \times 5^4}\).

Now, we're told that \(\mathrm{f(m) = 9 \times f(v)}\). This means the output of function \(\mathrm{f}\) for number \(\mathrm{m}\) is exactly 9 times larger than the output for number \(\mathrm{v}\). The key insight here is that \(\mathrm{9 = 3^2}\), so we're essentially multiplying by an extra factor of \(\mathrm{3^2}\).

Since our function only uses powers of 2, 3, and 5, and we're multiplying by \(\mathrm{3^2}\), this extra factor must come from the \(\mathrm{3^y}\) term in our function. This means the tens digit of \(\mathrm{m}\) must be exactly 2 more than the tens digit of \(\mathrm{v}\), while the hundreds and units digits stay the same.

Process Skill: TRANSLATE - Converting the functional relationship into digit constraints

2. Express the constraint algebraically

Let's say \(\mathrm{m}\) has digits \(\mathrm{abc}\) (so \(\mathrm{m = 100a + 10b + c}\)) and \(\mathrm{v}\) has digits \(\mathrm{def}\) (so \(\mathrm{v = 100d + 10e + f}\)).

Then:

• \(\mathrm{f(m) = 2^a \times 3^b \times 5^c}\)
• \(\mathrm{f(v) = 2^d \times 3^e \times 5^f}\)
• \(\mathrm{f(m) = 9 \times f(v)}\) means \(\mathrm{2^a \times 3^b \times 5^c = 9 \times 2^d \times 3^e \times 5^f}\)

Since \(\mathrm{9 = 3^2}\), we get:
\(\mathrm{2^a \times 3^b \times 5^c = 3^2 \times 2^d \times 3^e \times 5^f = 2^d \times 3^{(e+2)} \times 5^f}\)

For these expressions to be equal, each prime factor must have the same exponent on both sides:

• \(\mathrm{a = d}\) (hundreds digits are the same)
• \(\mathrm{b = e + 2}\) (tens digit of \(\mathrm{m}\) is 2 more than tens digit of \(\mathrm{v}\))
• \(\mathrm{c = f}\) (units digits are the same)

3. Apply digit constraints

Since both \(\mathrm{m}\) and \(\mathrm{v}\) are three-digit positive integers, we have important restrictions:

• Hundreds digits: \(\mathrm{a, d ∈ \{1, 2, 3, 4, 5, 6, 7, 8, 9\}}\)
• Tens digits: \(\mathrm{b, e ∈ \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}}\)
• Units digits: \(\mathrm{c, f ∈ \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}}\)

From our constraint \(\mathrm{b = e + 2}\), and knowing that both \(\mathrm{b}\) and \(\mathrm{e}\) must be single digits (0-9), we need:

• \(\mathrm{e + 2 ≤ 9}\), so \(\mathrm{e ≤ 7}\)
• \(\mathrm{e ≥ 0}\) (since it's a digit)

This means \(\mathrm{e}\) can be 0, 1, 2, 3, 4, 5, 6, or 7, and correspondingly \(\mathrm{b}\) will be 2, 3, 4, 5, 6, 7, 8, or 9.

Process Skill: APPLY CONSTRAINTS - Using the three-digit requirement to limit possible values

4. Find valid digit combinations

Let's think about what \(\mathrm{m - v}\) equals. Since:

• \(\mathrm{m = 100a + 10b + c}\)
• \(\mathrm{v = 100d + 10e + f}\)
• \(\mathrm{a = d}\) and \(\mathrm{c = f}\) and \(\mathrm{b = e + 2}\)

We get:
\(\mathrm{m - v = 100a + 10b + c - (100d + 10e + f)}\)
\(\mathrm{= 100a + 10b + c - 100a - 10e - c}\)
\(\mathrm{= 10b - 10e}\)
\(\mathrm{= 10(b - e)}\)
\(\mathrm{= 10(e + 2 - e)}\)
\(\mathrm{= 10(2)}\)
\(\mathrm{= 20}\)

Let's verify with a concrete example: If \(\mathrm{v = 123}\) (so \(\mathrm{d=1, e=2, f=3}\)), then \(\mathrm{m}\) must have \(\mathrm{a=1, b=4, c=3}\), giving us \(\mathrm{m = 143}\).
Check: \(\mathrm{m - v = 143 - 123 = 20}\) ✓

Another example: If \(\mathrm{v = 507}\) (so \(\mathrm{d=5, e=0, f=7}\)), then \(\mathrm{m = 527}\).
Check: \(\mathrm{m - v = 527 - 507 = 20}\) ✓

4. Final Answer

Regardless of the specific values of the hundreds and units digits, \(\mathrm{m - v}\) will always equal 20.

The answer is (D) 20.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the function definition

Students often confuse which digit corresponds to which exponent in \(\mathrm{f(n) = 2^x3^y5^z}\). They might think \(\mathrm{x}\) is the units digit, \(\mathrm{y}\) is the tens digit, and \(\mathrm{z}\) is the hundreds digit, rather than correctly identifying \(\mathrm{x}\) as hundreds, \(\mathrm{y}\) as tens, and \(\mathrm{z}\) as units. This fundamental misunderstanding makes the entire solution incorrect from the start.

2. Not recognizing that 9 = 3²

When seeing \(\mathrm{f(m) = 9 \times f(v)}\), students might try to distribute the factor of 9 across all three prime bases (2, 3, and 5) instead of recognizing that \(\mathrm{9 = 3^2}\) and therefore can only affect the \(\mathrm{3^y}\) term. This leads to incorrect constraint equations involving multiple digits rather than focusing on just the tens digits.

3. Missing the digit constraint implications

Students may set up the correct equation \(\mathrm{b = e + 2}\) but fail to consider that both \(\mathrm{b}\) and \(\mathrm{e}\) must be single digits (0-9). They might not realize this constrains the possible values of \(\mathrm{e}\) to be at most 7, or they might not consider what happens when this constraint cannot be satisfied.

Errors while executing the approach

1. Algebraic manipulation errors when equating exponents

When equating \(\mathrm{2^a \times 3^b \times 5^c = 2^d \times 3^{(e+2)} \times 5^f}\), students might incorrectly conclude that the exponents can be different as long as the overall expressions are equal, rather than understanding that for expressions with different prime bases to be equal, each corresponding exponent must be identical.

2. Arithmetic errors in calculating m - v

Even with correct digit relationships (\(\mathrm{a = d, b = e + 2, c = f}\)), students might make errors when substituting into \(\mathrm{m - v = (100a + 10b + c) - (100d + 10e + f)}\). Common mistakes include forgetting to distribute the negative sign or miscalculating \(\mathrm{10(e + 2) - 10e}\).

Errors while selecting the answer

1. Choosing a specific example rather than the general result

Students might work through one specific example (like \(\mathrm{m = 143, v = 123}\)) and see that \(\mathrm{m - v = 20}\), but then doubt whether this holds for all valid pairs. They might instead choose an answer that represents just the difference in tens digits (like 2) or get confused by their specific example and select the wrong value.