Re: Number Properties: Shortcut for no. of Primes in a Factorial
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12 Oct 2020, 11:01
I assume you're asking about a question like this: What is the largest integer k for which 5^k is a divisor of 400!
and I assume you're asking about the method one could use, where we divide 400 by small powers of 5, so we can count how many 5's are in 400!. Using that method:
400/5 = 80, so there are 80 multiples of 5 in 400!
400/25 = 16, so there are 16 multiples of 5^2 in 400!, each of which gives us one extra '5'
400/125 = 3 with a remainder, so there are 3 multiples of 5^3 in 400!, each of which gives us one extra '5'
Now we can stop. If we can only take 5^3 out of 400 three times, naturally we can't take 5^3 out of 400 five times. So it will certainly be true that the quotient is zero when we divide 400 by (5)(5^3) = 5^4. So the answer to the question above is 80+16+3= 99.
So assuming you're using this method for powers of a prime p, you can stop once your quotient q is less than p. You can certainly stop when your quotient is 1, but you can also stop in other situations as well.
I'd add though that learning special time-saving tricks with this method is almost certainly pointless for GMAT purposes. The numbers in actual GMAT questions testing these concepts are always so small that you don't even need to use this method at all, and GMAT questions about this are rare anyway. It's very unlikely that it will matter how fast you are at solving them. For some unrealistic prep company questions, however, this method can be useful, or even indispensable, when the numbers are large.