Finding Numbers with a Certain Number of FactorsDate: 3/12/96 at 16:18:46 From: Patrick Krause Subject: Factoring Question Dr. Math: I am stuck on the following question: Given that twelve is the least postive integer with six different positive factors (1,2,3,4,6,12). What is the least positive integer with exactly twenty-four positive factors? I know by writing a factoring program that the answer is 360 (factors: 1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180,360) but I cannot make the connection between the factors of 12 and the way to find a number with 24 positive factors. I have tried prime factorization, I have tried everything that I can to make this connection, but I have cannot make the connection. I do not see what knowing the factors of 12 have to do with the factors of 360. Can you give me some help with this? Thank you. Date: 5/30/96 at 11:38:32 From: Doctor Alain Subject: Re: Factoring Question Let P1, P2, ... , Pn be the prime factors of a given number X, and let E1, E2, ..., En be the greatest exponents of the respective prime factors such that their powers are still factors of X. That is to say that X = P1^E1 * P2^E2 * ... * Pn^En. Then all the positive factors F of X are in the form F = P1^e1 * P2^e2 * ... * Pn^en, where e1, e2, ... , en are all integer numbers greater or equal to zero and less than or equal to their respective Ei (e1 less than or equal to E1, e2 less than or equal to E2, ... , en less than or equal to En). Any set of integers e1, e2, ... , en satisfying these conditions will produce a factor of X. The set of numbers satisfying these conditions is one-to-one onto with the set of positive factors of X. So the number N of positive factors of X is given by N = (E1+1) * (E2+1) * ... * (En+1). [There are E1+1 ways to chose e1, E2+1 ways to chose e2, ..., En+1 ways to chose en] Now let's go back to your problem. You want a number with 24 positive factors, 24 is 3*8 = 3*4*2 = 3*2*2*2 so the numbers E1, E2, ... , En will be either (#1) E1=2 (E1+1=3), E2=7 (E2+1=8) [24 = 3*8] or (#2) E1=2, E2=3, E3=1 [24 = 3*4*2] or (#3) E1=2, E2=1, E3=1, E4=1. [24 = 3*2*2*2] These are the only ways you can get N = (E1+1) * (E2+1) * ... * (En+1)=24. So all integers X with 24 positive factors can be written in one of the following forms (#1) X = P1^2 * P2^7 (#2) X = P1^2 * P2^3 * P3 (#3) X = P1^2 * P2 * P3 * P4 The smallest number of form #1 is obviously 3^2 * 2^7 = 1152. The smallest number of form #2 is obviously 3^2 * 2^3 * 5 = 360. The smallest number of form #3 is obviously 2^2 * 3 * 5 * 7 = 420. -Doctor Alain, The Math Forum |
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