Numbers with 12 Factors....
Date: 11/15/96 at 02:59:25 From: Shane Conrad Subject: Numbers with 12 Factors.... I have to find two numbers which have exactly 12 factors, including 1, the number itself, 5 and 7. I found one such number, 140 (140 has factors of 1,2,4,5,7,10,14,20,28,35,70,140) but I had to do a lot of guessing. Is there another way to figure problems like this out without trying a lot of numbers?
Date: 04/01/97 at 19:39:52 From: Doctor Yiu Subject: Re: Numbers with 12 Factors.... Dear Shane, Suppose you factor a number into a product of prime powers: the first prime appearing A times, the second prime appearing B times, .... the last prime appearing Z times. Then that number has exactly (1 + A) times (1 + B) times ... all the way up to (1+Z) divisors, including 1 and itself. This is a theorem in a subject called number theory. If you want to know why it holds, write back to us! Since your number is divisible by 5 and 7, you can take the first prime to be 5 and the second to be 7, and ask whether it can have other primes or not. Note that 1+A and 1+B cannot be smaller than 2. Since there are altogether 12 divisors, these (1+A), (1+B), and others if any, should multiply to 12. Now the only ways to write 12 as a product of numbers with at least two of them not smaller than 2 are (i) 2 times 6 (ii) 3 times 4 (iii) 2 times 2 times 3 In (i) and (ii), the number has only 5 and 7 for its prime divisors. It must be 5^1 times 7^5 = 84035, or 5^5 times 7^1 = 21875, or 5^2 times 7^3 = 8575, or 5^3 times 7^2 = 6125. In (iii), the number has one more PRIME divisor. Let it be p. This number is either 5 times 7 times p^2 = 35p^2, or 5 times p times 7^2 = 245p, or p times 7 times 5^2 = 175p. Your example corresponds to the first of these, with p = 2. I hope this helps. If you are confused about anything, please do write back! Good luck. -Doctors Yiu and Sydney, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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