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

Approach: Build the prime factorisation of \(N\), then treat each prime's exponent independently — a perfect-square-and-multiple-of-420 factor just means each exponent is even AND at least as big as it is in 420. Count valid exponents per prime and multiply.

Step 1: Factorise \(10!\).
Count each prime in \(1\cdot2\cdots10\):
Power of 2: from 2,4,6,8,10 \(\to 1+2+1+3+1=8\).
Power of 3: from 3,6,9 \(\to 1+1+2=4\).
Power of 5: from 5,10 \(\to 1+1=2\). Power of 7: 1.
So \(10! = 2^8\times3^4\times5^2\times7^1\).

Step 2: Build \(N\).
\[ N = 2^{3}3^{7}5^{7}7^{9}\times 2^{8}3^{4}5^{2}7^{1} = 2^{11}\times3^{11}\times5^{9}\times7^{10}. \]

Step 3: Factorise 420.
\[ 420 = 2^2\times3\times5\times7. \]

Step 4: Set up the exponent constraints.
Write a qualifying factor as \(2^a3^b5^c7^d\). Two conditions stack on each exponent: it must be even (perfect square) and \(\ge\) the exponent in 420 (multiple of 420), while staying within \(N\).
\(a:\) even, \(2\le a\le 11 \Rightarrow \{2,4,6,8,10\}\) — 5 choices.
\(b:\) even, \(b\ge 1\Rightarrow b\ge 2,\ b\le 11 \Rightarrow \{2,4,6,8,10\}\) — 5 choices.
\(c:\) even, \(c\ge 1\Rightarrow c\ge 2,\ c\le 9 \Rightarrow \{2,4,6,8\}\) — 4 choices.
\(d:\) even, \(d\ge 1\Rightarrow d\ge 2,\ d\le 10 \Rightarrow \{2,4,6,8,10\}\) — 5 choices.

Step 5: Multiply.
\[ 5\times5\times4\times5 = 500. \]

Final answer: 500 — option (A).

Answer 2

Approach: Start from the bigger picture: first count every perfect-square factor of \(N\), ignoring the 420 condition, then narrow down to the ones that are also multiples of 420. This two-stage view also doubles as a check against a common trap answer.

Step 1: Factorise \(N\). \(10! = 2^8 3^4 5^2 7\), so \(N = 2^3 3^7 5^7 7^9 \times 2^8 3^4 5^2 7 = 2^{11} 3^{11} 5^9 7^{10}\).

Step 2: Count ALL perfect-square factors of \(N\) (no 420 condition yet). For a perfect square, every exponent just needs to be even:
\[ \left(\left\lfloor\tfrac{11}{2}\right\rfloor+1\right)\left(\left\lfloor\tfrac{11}{2}\right\rfloor+1\right)\left(\left\lfloor\tfrac{9}{2}\right\rfloor+1\right)\left(\left\lfloor\tfrac{10}{2}\right\rfloor+1\right) = 6\times6\times5\times6 = 1080. \]
(This matches option (B) exactly, the count you would wrongly get if you forgot the "multiple of 420" requirement.)

Step 3: Now impose the 420 floor. \(420 = 2^2\times3\times5\times7\), so being a multiple of 420 additionally forces each even exponent to be at least 2, trimming the choices for each prime from \(\{0,2,4,\dots\}\) down to \(\{2,4,6,\dots\}\), one fewer option per prime:
\[ (6-1)(6-1)(5-1)(6-1) = 5\times5\times4\times5 = 500. \]

Final Answer: 500 – option (A). The 1080 figure from Step 2 confirms why that option is a tempting but incomplete answer.