First 100 natural numbers in decimal system are converted to base 6 and their product if found. Find the no. of trailing 0’s in the product when converted in base 6?
Explanation:
Product of first 100 natural numbers = 100!.
When we do the same exercise in decimal system, the number of zeros is same as the highest power of 10.
Since, we are doing this exercise in base 6, the number of zeros will be same as highest power of 6.
So we have to find the highest power of 6 in 100!.
6 = 2 × 3.
∴ Highest power of 6 will be same as highest power of 3 in 100!.
Highest power of 3 is sum of all the quotients when 100 is successively divided by 2 i.e., 33 + 11 + 3 + 1 = 48.
Hence, 48.
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