What is the smallest number with 12 factors?

Answer 1

(ITS 60. Factors: 1,2,3,4,5,6,10,12,15,20,30,60) The smallest positive integer with 12 divisors, including 1 and itself, is 60. They are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Notice that six of these divisors are smaller than the square-root of 60, and six of them are larger. (Only a number that is a perfect square can have an odd number of divisors, incidentally.)How can you find this result? To find quickly the number of divisors in any positive integer, first factorise it into prime factors, using exponents to count repetitions. Let's use 72 as an example: 72 = 8 X 9 = 2³ X 3². Thus, 3 occurs twice, and 2 occurs thrice. Every divisor of 72 contains 3 as a factor 0, 1, or 2 times; and it contains 2 as a factor either 0, 1, 2, or 3 times; and it can contain no other prime factor besides 2 and 3. Because there are 3 choices of how many times the divisor of 72 may contain 3 as a factor and 4 choices of how many times it may contain 4 as a factor, there are twelve possible divisors of 72. You can check this by asking how many divisors 72 has that are less than 9, the smallest positive integer whose square is greater than 72. It turns out that there are six, namely, 1, 2, 3, 4, 6, and 8. Its other six divisors must be at least 9, and they are found, by division, to be 9, 12, 18, 24, 36, and 72.Applying this logic, we now seek the smallest positive integer with twelve divisors: We can readily see that the number we seek will be a product of powers of the smallest primes. Moreover, it is plausible, even before multiplying it out, that it will be either the product of 2² X 3 X 5 or the product of 2³ X 3². And why is this? If we use 2² X 3 X 5, each of its divisors must present us 3 X 2 X 2 = 12 choices, as to the number of times each of its prime factors may be repeated in its prime factorization (obtained by adding 1 to each of the exponents and taking their product - noting that 1 is the exponent of 3 and 5, whilst 2 is the exponent we have shown for 2). Therefore, it will have 12 divisors. If we use 2³ X 3², each of its divisors must present us 4 X 3 = 12 choices, obtained by the same method.We can easily see that 2^11 (giving us 12 choices and, therefore, 12 divisors) is going to be much too large. On the other hand, four or more distinct prime factors (such as 2, 3, 5, and 7) will present us with with at least 2^4 choices and 2^4 = 16 divisors, which is more than we need. Now, since we can readily see that 2² X 3 X 5 = 60 is smaller than 2³ X 3² = 72, it is clear that 60 is the number we seek.