Question:medium

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?

Show 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.
Updated On: Jul 4, 2026
  • 500
  • 1080
  • 9000
  • 15840
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Write each exponent as twice a whole number, \(a=2p,\ b=2q,\ c=2r,\ d=2s\), since every exponent must be even. Substituting \(N = 2^{11}3^{11}5^9 7^{10}\) and \(420=2^2\cdot3\cdot5\cdot7\) turns "even and at least 420's exponent" into simple bounds on \(p,q,r,s\).

Step 2: For \(2^{2p}\): \(2p\ge2\) and \(2p\le11\), so \(1\le p\le5\) (5 values). For \(3^{2q}\): \(2q\ge1\Rightarrow q\ge1\), and \(2q\le11\Rightarrow q\le5\) (5 values). For \(5^{2r}\): \(r\ge1,\ 2r\le9\Rightarrow r\le4\) (4 values). For \(7^{2s}\): \(s\ge1,\ 2s\le10\Rightarrow s\le5\) (5 values).

Step 3: Multiply the counts:
\[ 5\times5\times4\times5 = \boxed{500} \]

Final Answer: 500.
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