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is it m that the qs was referring to? be careful with your typing, it freaked me out trying to find P
m = 1x2x3x4x...x30
there are 10 multiples of 3 (3x1, 3x2, 3x3,...3x10)
3 mutilples of 3^2 (9x1, 9x2, 9x3)
1 multiple of 3^3 (27)
therefore the number of times factor 3 appears in the product m is 10+3+1 = 14. _________________
Carson, this is what I tell my students about any problem that has factors, multiples, products, divisors, or any other reference to perfect integers:
The number in the denominator must disappear completely to make an integer.
If I know that x/y is an integer, than I know that somehow, y will cancel with x.
This problem is broken down essentially to 30!/3^n, and we know it's an integer, because 3^n is a factor of 30!. Since we're just talking about a string of 3's on the bottom, we need to know how many 3's are on top to figure out the maximum number on the bottom. And since the 3's are coming only from the multiples of 3 in 30!, we need only to consider, as jpv and the others said, 3,6,9,13,15,18,21,24,27,30. Prime factor each of these numbers, and you'll find that there are 14 3's among them.
So if n was 15, let's say, then there would be 15 3's on the bottom and only 14 on top, and the bottom wouldn't canel out completely, and you wouldn't have an integer. So it has to be 14.