Discussion
The last non-zero digit of 20! is:
Answer: B
To find the last non-zero digit, we divide the factorial by its powers of 10, which means removing pairs of 2 and 5. There are more 2s than 5s. Let's count them. 5s in 20! = \(\lfloor 20/5 \rfloor = 4\). 2s in 20! = \(10+5+2+1 = 18\). We remove four 5s and four 2s. We need the unit digit of \(20! / 10^4\). This is a complex calculation. A known formula is that the last non-zero digit of n! is \((2^{a_2-a_5} \times (\text{unit digits of other primes})) \pmod{10}\), where \(a_2\) and \(a_5\) are powers of 2 and 5. A simpler method involves finding the unit digit of the product of numbers after removing all 5s and an equal number of 2s. The calculation is complex but the result for 20! is known to be 4.
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