Question

If \(N = 2^3 \times 3^7 \times 5^7 \times 7^9 \times 10!\), then how many factors of N are there which are perfect squares as well as multiples of 420?

• A. 500
• B. 1080
• C. 9000
• D. 15840

Hint:

When dealing with factors that must satisfy multiple conditions (like being a square and a multiple of another number), analyze the constraints on the exponents of each prime factor separately and then multiply the number of possibilities.

Answer 1

The Correct Option is A

Step 1: Understanding the Question: 
The question asks for the number of factors of a given number N that satisfy two conditions simultaneously:
1. The factor must be a perfect square.
2. The factor must be a multiple of 420.
To solve this, we first need the complete prime factorization of N. Then, we need to analyze the exponents of the prime factors of any such factor based on the given conditions. 
Step 2: Key Formula or Approach: 
1. Prime Factorization: We will first express N in its canonical prime factorized form. This involves finding the prime factorization of 10!.
2. Condition for Perfect Squares: A number is a perfect square if and only if all the exponents in its prime factorization are even.
3. Condition for Multiples: A number F is a multiple of K if and only if for every prime p, the exponent of p in the prime factorization of F is greater than or equal to the exponent of p in the prime factorization of K.
4. Counting Factors: The total number of such factors is found by multiplying the number of choices available for each prime's exponent. 
Step 3: Detailed Explanation: 
Part A: Complete Prime Factorization of N 
First, let's find the prime factorization of 10!: \[ 10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \] \[ 10! = 2 \times 3 \times (2^2) \times 5 \times (2 \times 3) \times 7 \times (2^3) \times (3^2) \times (2 \times 5) \] The powers of the primes are:
- Power of 2: \(1 + 2 + 1 + 3 + 1 = 8\)
- Power of 3: \(1 + 1 + 2 = 4\)
- Power of 5: \(1 + 1 = 2\)
- Power of 7: \(1\)
So, \(10! = 2^8 \times 3^4 \times 5^2 \times 7^1\). Now, we combine this with the given expression for 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 B: 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 C: Applying the Conditions to a Factor 
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 within the range: \(0 \le a \le 11\), \(0 \le b \le 11\), \(0 \le c \le 9\), \(0 \le d \le 10\). Now we apply the two conditions to these exponents: 

F is a perfect square: This means \(a, b, c, d\) must all be even integers. 
F is a multiple of 420: This means the exponents of F must be greater than or equal to the exponents of 420. \[ a \ge 2, \quad b \ge 1, \quad c \ge 1, \quad d \ge 1 \] 
Part D: Counting the Possible Exponents 
We combine all constraints for each exponent:
- For exponent a (power of 2):
\(a \le 11\), \(a\) is even, and \(a \ge 2\). The possible values are \(\{2, 4, 6, 8, 10\}.\) 5 choices.
- For exponent b (power of 3):
\(b \le 11\), \(b\) is even, and \(b \ge 1\). The possible values are \(\{2, 4, 6, 8, 10\}.\) 5 choices.
- For exponent c (power of 5):
\(c \le 9\), \(c\) is even, and \(c \ge 1\). The possible values are \(\{2, 4, 6, 8\}\). 4 choices.
- For exponent d (power of 7):
\(d \le 10\), \(d\) is even, and \(d \ge 1\). The possible values are \(\{2, 4, 6, 8, 10\}\). 5 choices.
Part E: Final Calculation 
The total number of such factors is the product of the number of choices for each exponent. Total number of factors = \(5 \times 5 \times 4 \times 5 = 500\). 
Step 4: Final Answer: 
There are 500 factors of N that are perfect squares and multiples of 420. This corresponds to option (A).