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

Approach: A factor counts only if every prime's exponent is even (perfect square) AND at least as large as that prime's exponent in 420. So for each prime, just list the even exponents in the allowed range that clear the 420 floor, and multiply the counts.

Step 1: Build the full prime factorisation of N.
The only tricky piece is \(10!\). Counting prime powers in \(10!\): power of 2 \(=8\), power of 3 \(=4\), power of 5 \(=2\), power of 7 \(=1\), so \(10! = 2^8 \times 3^4 \times 5^2 \times 7^1\).

Now merge with the given part:
\[ N = 2^{3+8} \times 3^{7+4} \times 5^{7+2} \times 7^{9+1} = 2^{11} \times 3^{11} \times 5^{9} \times 7^{10}. \]

Step 2: Factorise the floor, 420.
\[ 420 = 2^2 \times 3^1 \times 5^1 \times 7^1. \]

Step 3: Set the rule for each exponent.
Write a qualifying factor as \(2^a 3^b 5^c 7^d\). Two conditions stack on every exponent: it must be even (perfect square) and it must meet the 420 floor. It also cannot exceed the exponent available in \(N\).

Step 4: Count the legal exponents prime by prime.
Prime 2: even, \(a \ge 2\), \(a \le 11\) \(\Rightarrow \{2,4,6,8,10\}\) = 5 choices.

Prime 3: even, \(b \ge 1\) (so effectively \(\ge 2\)), \(b \le 11\) \(\Rightarrow \{2,4,6,8,10\}\) = 5 choices.

Prime 5: even, \(c \ge 1\) (so \(\ge 2\)), \(c \le 9\) \(\Rightarrow \{2,4,6,8\}\) = 4 choices.

Prime 7: even, \(d \ge 1\) (so \(\ge 2\)), \(d \le 10\) \(\Rightarrow \{2,4,6,8,10\}\) = 5 choices.

Step 5: Multiply.
\[ 5 \times 5 \times 4 \times 5 = 500. \]

Final Answer: \(\boxed{500}\) such factors. (Option 3)

Answer 2

Approach (factor 420 out first): Every qualifying number is a multiple of 420, so write it as \( F = 420\times G \) and ask what \( G \) is allowed to be for \( F \) to stay a perfect square.

Step 1: \( N = 2^{11}3^{11}5^{9}7^{10} \) and \( 420 = 2^{2}3^{1}5^{1}7^{1} \), so write \( G = 2^{p}3^{q}5^{r}7^{s} \) with \( 0\le p\le9,\ 0\le q\le10,\ 0\le r\le8,\ 0\le s\le9 \).
Step 2: For \( F=420G \) to be a perfect square, each combined exponent must be even: exponent of 2 is \( 2+p \) (even iff \( p \) even); exponent of 3 is \( 1+q \) (even iff \( q \) odd); exponent of 5 is \( 1+r \) (even iff \( r \) odd); exponent of 7 is \( 1+s \) (even iff \( s \) odd).
Step 3: Count valid values: \( p\in\{0,2,4,6,8\} \) (5 values); \( q\in\{1,3,5,7,9\} \) (5 values); \( r\in\{1,3,5,7\} \) (4 values); \( s\in\{1,3,5,7,9\} \) (5 values).
Step 4: Multiply: \( 5\times5\times4\times5 = 500 \).

Final Answer: \( \boxed{500} \) factors. (Option 3)