Question : What is the digit in the unit's place in the number $\frac{15!}{100}$?
Option 1: 5
Option 2: 7
Option 3: 3
Option 4: 0
Correct Answer: 0
Solution :
Number of trailing zeroes in $n$! = [$\frac{n}{5}$] + [$\frac{n}{25}$] + [$\frac{n}{125}$] + ....... where [ ] denotes greatest integer function.
So, the number of trailing zeroes in 15! = [$\frac{15}{5}$] + [$\frac{15}{25}$] + [$\frac{15}{125}$] + ....... = 3 + 0 + 0 + ..... = 3
⇒ Number of zeroes in the product = 3
⇒ Unit's digit in $\frac{15!}{100}$ = 0
Hence, the correct answer is 0.
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