which odd number has 12 factors?

Answer 1

This is a wild guess.

3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 = the number.
The answer will be the product of the first 12 primes that aren't 2. (If 2 were a factor, the answer would be even, not odd.)

For the rest, use a calculator. And check to make sure all those numbers are prime. Sorry this is such a whack answer.

Answer 2

Since you didn't mention whether prime factors are of concern, I assume 'all' factors should be considered. Therefore, in order to be an odd number, it requires all its prime factors to be odd. Also if we assume the number as A=a^n*b^m*c^p*... (where a,b,c are prime odd factors) then the total number of factors would be (n+1)*(m+1)*(p+1)*...
for example, if we think of 3^1*5^1*7^1 then the total factors would be 2*2*2=8
so if we find three numbers greater than 1 whose product be 12 (which is obviously 2,2,3) then it is necessary and sufficient to put any arbitrary 3 odd prime numbers (i.e. prime numbers except than 2) powered to the found 3 values minus one:

Example 1: a=3, b=5, c=7
A=3^1*5^1*7^2=735
Factors: 1,3,5,7,49,15,21,35,
147,245,105,735

Example 2: a=5, b=11, c=3
A=5^1*11^1*3^2=495
Factors: 1,3,5,11,9,15,33,55,
165,45,99,495

You'll find the rest...

Answer 3

There are infinite, but one is 315:

The factors are {1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315}

They are given by x=p^2•q•r where p,q,r are all odd primes.

They are also given by x=p^3•q^2, where p and q are odd primes.

Example of second is 775=3^3•5^2

The factors are {1, 3, 5, 9, 15, 25, 27, 45 ,75, 135, 225, 775}

Answer 4

But its factors need not all be prime.

So, 819 = 1 * 819 = 3 *273 = 7 * 117 = 9 * 91 = 13 * 63 = 21 * 39

Here also 819 has 12 factors, 1,3,7,9,13,21,39,63,91,117,273 & 819.

Answer 5

any odd number can have any number of factors, it just depends on which set of numbers you want to consider

Answer 6

1*2*3*4*5*6*7*8*9*10*11*12=479001600

Answer 7

the number is 3710369067405

Answer 8

you know, I was going to answer this but then I gave up trying.