Question

For the number \(N = 2^3 \times 3^7 \times 5^7 \times 7^9 \times 10!\), how many factors are perfect squares and also multiples of 420?

• A. 400
• B. 450
• C. 500
• D. 600

Hint:

To solve problems on factors with multiple constraints, first establish the full prime factorization of the base number. Then, for each prime, determine the possible range of its exponent based on all given conditions. The total number of factors is the product of the counts of valid exponents for each prime.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are asked to find the number of factors of a given composite number \(N\) that satisfy two conditions simultaneously: the factor must be a perfect square, and it must be a multiple of 420.
Step 2: Key Formula or Approach:
1. Find the complete prime factorization of the number \(N\).
2. Find the prime factorization of 420.
3. A factor is a perfect square if all the exponents in its prime factorization are even.
4. A factor is a multiple of 420 if the exponents of its prime factors are greater than or equal to the exponents of the prime factors of 420.
5. Combine these conditions to find the number of possible exponents for each prime factor, then multiply the counts.
Step 3: Detailed Explanation:
Part 1: Prime Factorization of N
First, we need the prime factorization of \(10!\).
\(10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10\)
\(10! = 1 \times 2 \times 3 \times (2^2) \times 5 \times (2 \times 3) \times 7 \times (2^3) \times (3^2) \times (2 \times 5)\)
Let's count the powers of each prime:
- Power of 2: \(1 + 2 + 1 + 3 + 1 = 8\) \(\implies 2^8\)
- Power of 3: \(1 + 1 + 2 = 4\) \(\implies 3^4\)
- Power of 5: \(1 + 1 = 2\) \(\implies 5^2\)
- Power of 7: \(1\) \(\implies 7^1\)
So, \(10! = 2^8 \times 3^4 \times 5^2 \times 7^1\).
Now, combine this with the rest of \(N\):
\(N = (2^3 \times 3^7 \times 5^7 \times 7^9) \times (2^8 \times 3^4 \times 5^2 \times 7^1)\)
\(N = 2^{3+8} \times 3^{7+4} \times 5^{7+2} \times 7^{9+1}\)
\(N = 2^{11} \times 3^{11} \times 5^9 \times 7^{10}\)
Part 2: Prime Factorization of 420
\(420 = 42 \times 10 = (6 \times 7) \times (2 \times 5) = (2 \times 3 \times 7) \times (2 \times 5) = 2^2 \times 3^1 \times 5^1 \times 7^1\).
Part 3: Finding the number of factors
Let a factor of \(N\) be \(F = 2^a \times 3^b \times 5^c \times 7^d\). For F to be a factor of N, the exponents must be in the ranges: \(0 \leq a \leq 11\), \(0 \leq b \leq 11\), \(0 \leq c \leq 9\), \(0 \leq d \leq 10\).
We need to apply two conditions on F:
Condition 1: F is a perfect square.
This means the exponents \(a, b, c, d\) must all be even numbers.
Condition 2: F is a multiple of 420.
This means the exponents of F must be greater than or equal to the exponents of 420.
\(a \geq 2, b \geq 1, c \geq 1, d \geq 1\).
Combining the conditions:
- For exponent 'a' (of prime 2): Must be even AND \(a \geq 2\). Within the range [0, 11], the possibilities are \{2, 4, 6, 8, 10\}. (5 choices)
- For exponent 'b' (of prime 3): Must be even AND \(b \geq 1\). Within the range [0, 11], the possibilities are \{2, 4, 6, 8, 10\}. (5 choices)
- For exponent 'c' (of prime 5): Must be even AND \(c \geq 1\). Within the range [0, 9], the possibilities are \{2, 4, 6, 8\}. (4 choices)
- For exponent 'd' (of prime 7): Must be even AND \(d \geq 1\). Within the range [0, 10], the possibilities are \{2, 4, 6, 8, 10\}. (5 choices)
The total number of such factors is the product of the number of choices for each exponent.
Total factors = \(5 \times 5 \times 4 \times 5 = 500\).
Step 4: Final Answer
There are 500 factors of \(N\) that are both perfect squares and multiples of 420.