Numbers with Large Exponents
Date: 08/22/2001 at 21:11:30 From: T Subject: Large exponential numbers I have not been able to figure this out. The problem is: How many zeros end the number 2^300 * 5^600 * 4^400 ? I was thinking of the end patterns of the numbers but I could not put anything together. Much appreciated, T
Date: 08/23/2001 at 10:48:45 From: Doctor Jubal Subject: Re: Large exponential numbers Hi T - thank you for writing Dr. Math. You need to figure out how many times your number is divisible by 10. Any number divisible by 10 ends in a zero (40, 90, 120, etc.). Any number divisible by 100 ends in two zeros (500, 14300, 6600, etc.). And so forth: the number of zeros at the end of a number is the number of powers of ten it is divisible by. Now, the prime factorization of ten is 10 = 2*5, so 10^n = 2^n * 5^n. That is to say, in order for a number to be divisible by 10^n (the number ends in n zeros), that number has to be divisible by 2^n and 5^n - it has to contain at least n 2's and n 5's in its prime factorization. Let's look at 40. Its prime factorization is 40 = 2^3 * 5. It is divisible by three powers of 2, but only one 5, so it is only divisible by one power of ten, and ends in one zero. For another example, let's look at 2^3 * 5^9 * 4^2. It contains nine powers of five. Each power of four is actually two powers of 2 (4 = 2^2), so the two powers of 4 are actually four powers of 2, for a total of seven powers of two (2^3 * 4^2 = 2^3 * 2^(2*2) = 2^3 * 2^4 = 2^(3+4) = 2^7). So the prime factorization of this number is 2^7 * 5^9. Each power of 10 needs one 2 and one 5, so we have seven powers of 10 with two 5's left over. We expect this number to end in seven zeros, and if we actually calculate it out, we get 250,000,000 which does end in seven zeros. I'll leave you to apply these ideas to your actual problem. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/
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