Loading...
If \(\mathrm{m}\), \(\mathrm{p}\), and \(\mathrm{t}\) are distinct positive prime numbers, then \(\mathrm{(m^3)(p)(t)}\) has how many different positive factors greater than \(\mathrm{1}\)?
Let's start by understanding what we're looking for. We have an expression \((\mathrm{m}^3)(\mathrm{p})(\mathrm{t})\) where m, p, and t are distinct positive prime numbers. We need to find how many positive factors this expression has that are greater than 1.
Think of it this way: if we have a number written as a product of prime powers, we want to count all the different ways we can multiply some (or all) of those prime factors together to create new numbers.
For example, if we had \(2^3 \times 3 \times 5\), we could form factors like 2, 4, 6, 10, 12, etc. - basically any combination of the available prime factors.
Process Skill: TRANSLATE - Converting the abstract expression into a concrete factor-counting problem
Since m, p, and t are distinct prime numbers, our expression \((\mathrm{m}^3)(\mathrm{p})(\mathrm{t})\) is already in prime factorization form. Let's think about what this means:
So our number looks like: \(\mathrm{m}^3 \times \mathrm{p}^1 \times \mathrm{t}^1\)
This is helpful because when we want to count factors, we need to consider all possible combinations of these prime powers.
Now here's the key insight: to form any factor of our number \(\mathrm{m}^3\mathrm{pt}\), we can choose:
Think of it like choosing toppings for a pizza - we can independently choose how much of each prime to include in our factor.
Since these choices are independent, we multiply the number of options:
Total factors = \(4 \times 2 \times 2 = 16\)
Technically, this follows the formula: if a number has prime factorization \(\mathrm{p}_1^{\mathrm{a}_1} \times \mathrm{p}_2^{\mathrm{a}_2} \times ... \times \mathrm{p}_\mathrm{k}^{\mathrm{a}_\mathrm{k}}\), then it has \((\mathrm{a}_1 + 1)(\mathrm{a}_2 + 1)...(\mathrm{a}_\mathrm{k} + 1)\) total factors.
We found that \(\mathrm{m}^3\mathrm{pt}\) has 16 total positive factors. However, the question asks for factors greater than 1.
One of our 16 factors is the number 1 (when we choose \(\mathrm{m}^0\mathrm{p}^0\mathrm{t}^0 = 1\)). Since we need factors greater than 1, we must exclude this.
Therefore: Number of factors greater than 1 = \(16 - 1 = 15\)
Process Skill: APPLY CONSTRAINTS - Remembering to exclude factors that don't meet the specified condition
The expression \((\mathrm{m}^3)(\mathrm{p})(\mathrm{t})\) has 15 different positive factors greater than 1.
This matches answer choice D: 15.
To verify: our systematic approach counted all possible factor combinations, properly applied the constraint "greater than 1," and used the fundamental principle that factors correspond to all possible combinations of prime powers up to their maximum exponents.
Students often confuse "factors" with "prime factors" or think they need to find only the prime factors of the expression. They may not realize that factors include ALL possible products that divide the number evenly, such as \(\mathrm{m}^2, \mathrm{mp}, \mathrm{m}^2\mathrm{p}, \mathrm{mpt}\), etc. This leads them to count only the 3 prime factors (m, p, t) instead of all possible factor combinations.
Many students correctly identify that they need to count all factors but miss the critical constraint that the factors must be "greater than 1." This causes them to either include the factor 1 in their final count or forget to subtract it, leading to an answer of 16 instead of 15.
Students may attempt to list out all factors manually instead of using the systematic approach with the formula (exponent + 1) for each prime. This manual approach becomes error-prone and time-consuming, especially when dealing with expressions like \(\mathrm{m}^3\mathrm{pt}\) that have multiple prime factors with different exponents.
Even when students correctly identify that they need to multiply \((3+1) \times (1+1) \times (1+1)\), they may make simple arithmetic mistakes, calculating \(4 \times 2 \times 2\) as something other than 16. This is particularly common under time pressure during the actual test.
Students might misread \(\mathrm{m}^3\mathrm{pt}\) and think all primes have the same exponent, leading them to calculate \((3+1)^3 = 64\) total factors instead of \((3+1)(1+1)(1+1) = 16\). They may also confuse which prime has which exponent when setting up their calculation.
Students who correctly calculate 16 total factors may forget to subtract 1 to exclude the factor "1" itself. They might look at the answer choices, see that 16 isn't available, but then second-guess their approach rather than remembering to apply the "greater than 1" constraint, leading them to select a different answer choice entirely.
Step 1: Choose specific prime values
Since m, p, and t are distinct positive prime numbers, let's use:
• m = 2 (smallest prime)
• p = 3 (next smallest prime)
• t = 5 (next smallest prime)
Step 2: Calculate the concrete expression
\((\mathrm{m}^3)(\mathrm{p})(\mathrm{t}) = (2^3)(3)(5) = (8)(3)(5) = 120\)
Step 3: Find prime factorization of our result
\(120 = 2^3 \times 3^1 \times 5^1\)
Step 4: Count all positive factors systematically
Using the prime factorization \(2^3 \times 3^1 \times 5^1\), each factor has the form \(2^\mathrm{a} \times 3^\mathrm{b} \times 5^\mathrm{c}\) where:
• a can be 0, 1, 2, or 3 (4 choices)
• b can be 0 or 1 (2 choices)
• c can be 0 or 1 (2 choices)
Total factors = \(4 \times 2 \times 2 = 16\)
Step 5: List all factors to verify
Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
That's 16 factors total.
Step 6: Apply the constraint
We need factors greater than 1, so we exclude the factor 1.
Factors greater than 1: 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Count: 15 factors
Step 7: Verify this pattern holds generally
For any distinct primes m, p, t, the expression \((\mathrm{m}^3)(\mathrm{p})(\mathrm{t})\) will have prime factorization \(\mathrm{m}^3\mathrm{p}^1\mathrm{t}^1\), giving us \((3+1)(1+1)(1+1) = 16\) total factors, and 15 factors greater than 1.
Answer: D. 15