The highest power of 9 dividing 99 factorial completely is
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This is the same as asking what is the highest multiple of 3 that divides 99!, divided by two and rounded down. So lets look at how many multiples of 3 there are.
First, every third number will be divisible by 3, starting with 3. This gives 33 3s.
Next, every third multiple of 3 is divisible twice, so we'll count them extra. Starting from 9, this gives 11 3s.
Next, every third multiple of those will be divisible 3 times, so they get another count on top. These are 27, 54, and 81, or 3 extra 3s.
Finally, 81 actually has 4 multiples of 3, so we'll add it in one last time for a grand total of 33+11+3+1=48 3s, or 24 multiples of 9. Thus 9^24 will divide 99! evenly.
First, every third number will be divisible by 3, starting with 3. This gives 33 3s.
Next, every third multiple of 3 is divisible twice, so we'll count them extra. Starting from 9, this gives 11 3s.
Next, every third multiple of those will be divisible 3 times, so they get another count on top. These are 27, 54, and 81, or 3 extra 3s.
Finally, 81 actually has 4 multiples of 3, so we'll add it in one last time for a grand total of 33+11+3+1=48 3s, or 24 multiples of 9. Thus 9^24 will divide 99! evenly.
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the common factor if dividing the
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