a number is divisible by 10 if it has factors 5 and 2
10!=10*....5*...2 has \(10^n\) with the n=2
\((5!)^2\) has (5*...2) raised to the power of 2 results in n=2
\(10^2(9*8*7*6*4*3-2*16*9)=32*9*10^2(7*6*3-1)=32*9*10^2*125\) ---> contains \(2^5\) and \(5^3\) besides \(10^2\). The number n=5. Answer is C.

this question resembles me one of the very hard ones from magoosh and would take me 3+ min. of the solution time

Question says 'is divisible by'. Technically, it is divisible by 10^6.

I would say it will take under a minute if your concepts of exponents are sound

\(10! - 2(5!)^2\)

\((10)(9)(8)(7)(6)(5!) - 2(5!)(5!)\)
\(5![(10)(9)(8)(7)(6) - 2(5!)]\)
\(5![30240 - 2(120)]\)
\(120[30240 - 240]\)
\(120(30000)\)
\(3600000 = 36(10^5)\)

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