What is the remainder when we divide 125! By 10^31?
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When we divide 125! By 10^31, then the Remainder is 0.
Step-by-step Explanation :
125! = 1*2*3*4.............125
10^31 = 10*10*10*........(31 zeroes)
Total Number of zeroes in 125! :
=> 125/5 = 25
=> 125/25 = 5
=> 125/125 = 1
=> by adding we get 25 + 5 + 1 = 31
10^31 has 31 zeroes. It's perfectly divisible.
Hence, remainder is 0.
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