Let's factor!

$12 = 2 \times 6 = 4 \times 3 = 2 \times 2 \times 3$ are all the various ways to obtain 12 through multiplication. Hence, the integer must have the form $p_1 ^{11}$ or $p_1^1 p_2 ^ 5$ or $p_1^3 p_2 ^2$ or $p_1^1 p_2^1 p_3 ^2$ .

Case 1: $p_1 ^{11}$ . The smallest value is $2^{11} = 2048$ .

Case 2: $p_1 ^1 p_2 ^5$ . The smallest value is $3 \times 2^5 = 96$ .

Case 3: $p_1 ^3 p_2 ^2$ . The smallest value is $2^3 \times 3^2 = 72$ .

Case 4: $p_1 ^1 p_2 ^1 p_3 ^2$ . The smallest value is $5 \times 3 \times 2^2 = 60$ .

Hence, the answer is 60.

as we need 12 divisors and 12 = 1 12 = 2 6=4 3=3 2 2 therefore :12= (0+1)(11+1) = (1+1)(5+1) =(3+1)(2+1)= (2+1)(1+1)(1+1) so the pair of power possible for 12 divisors are (0,11) , (1,5) , (3,2) , (2,1,1) the smallest prime number is 2,3,5,7,.... therefore for the number with exactly 12 divisors are (1^1) (2^11) = 2048 (2^5) (3^1) = 96 (2^3) (3 2) = 72 (2^2) (3^1)*(5^1) = 60

therefore the smallest integer with exactly 12 divisors is 60

Any number can be expressed as $a^m_1 * b^m_2 * c^m_3 * ...$ , where a, b and c are distinct prime divisors and $m_1, m_2, m_3, \ldots$ are positive integers. The number of positive divisors can be calculated as $(m_1 + 1) * (m_2 + 1) * (m_3 + 1) * ...$ . Since $12 = 2 * 2 * 3 = 3 * 4 = 2 * 6 = 1 * 12$ . For possibility $2 * 2 * 3$ , smallest positive integer is $2^2 * 3^1 * 5^1 = 60$ . For possibility $3 * 4$ , smallest positive integer is $2^3 * 3^2 = 72$ . For possibility $2*6$ , smallest positive integer is $2^5 * 3^1 = 96$ . For possibility $1*12$ , smallest positive integer is $2^11 = 2048$ . Note that we give the smallest prime divisor the largest exponent number to make the overall integer as small as possible, and the smallest positive integer is therefore 60

[Slight edits for clarity, can be improved on more, but I didn't want to change his (horrendous) notation $a^m_1 * b^m_2 * c^m_3 * ...$ that doesn't specify what's actually happening. Is this an infinite product? He also used $m_i$ as non-negative integers, which isn't great since we could just throw in $p_i ^0 =1$ . - Calvin]

every natural number $n$ can be factored as product and this factorization is unique. see here .

we can write : $\displaystyle n=\prod_{i=1}^{m} {p_i}^{\alpha_i}$ where $p_i$ is always a prime smaller than $n$ and $\alpha_i$ is a natural number smaller than $n$ . and $r\neq s \Rightarrow p_r \neq p_s$ for all natural $r,s$

then the number of divisors $d_n$ is given by this formula : $d_n=(\alpha_1+1)(\alpha_2+1) \cdots (\alpha_m +1)$

in this case : $12=(\alpha_1+1)(\alpha_2+1) \cdots (\alpha_m +1)$

$3*2*2=(\alpha_1+1)(\alpha_2+1) (\alpha_3+1)$

casework give us triplets for $(\alpha_1,\alpha_2,\alpha_3)$ with their permutations :

$(11,0,0)$ $(5,1,0)$ $(3,2,0)$ $(2,1,1)$

We may break up 12 as 2 * 2 3. Also , the no. of divisors is given by (a+1) * (b + 1) (c + 1) * ... where the prime factorisation of the no. is p 1^{a} * p 2^{b} * ... .So , we will assume that the no. is 2^{3-1} 3^ {2-1} 5^{2-1} or 2^{2} 3^{1}5^{1}. This yields the smallest such number. Any other way to express 12 as , say 4 3 or 6 2 will lead to only bigger numbers as can be verified. So 60 is the smallest such number.

Since 12 = 6x2 = 3x2x2, an integer has exactly 12 divisors if it has the form a^11, a^5xb or a^2xbxc. The smallest number in the first case,second case and third case are 2^11 = 2048, 2^5x3 = 96 or 2^2x3x5 = 60 respectively. Hence 60 is the smallest integer that has exactly 60 divisors.

12 can be expessed as: $1 \times 12, 2 \times 6$ , or $2 \times 3 \times 3.$

Based on the definition of divisors, all following three numbers have 12 divisors: $2^{11} = 2048, 2^5 \times 3 = 96, 2^2 \times 3 \times 5 = 60.$

Comparing these three numbers, we can get the correct answer 60.

Think of an even number less than 100 which you think has a great number of divisors. Then start to factor it to check if it's the smallest number that has exactly 12 divisors.

60=(1 and 60); (2 and 30); (3 and 20); (4 and 15); (5 and 12); (6 and 10)

If n = $a^p \times b^q \times c^r \times$ ...., where a,b,c... are prime numbers and p,q,r are natural numbers. Then the number of factors for 'n' will be $(p + 1) \times(q + 1) \times (r + 1) \times$ ....

So, here the best possible way to write 12 is $3 \times 2 \times 2$ . So, the values of p,q,r are 2,1,1 respectively. The least number that satisfies this condition is $2^2 \times 3^1 \times 5^1$ , which is equal to 60.

The easiest way to solve this problem is to figure out what the first 6 divisors are. The most likely case is that they are 1, 2, 3, 4, 5, and 6, because those are the 6 lowest factors possible. Now we can find the least common multiple of those 6 numbers, which is 60. The other six factors need to equal 60 when multiplied by those 6 factors, which is true for 60.

$60=1×60=2×30=3×20=4×15=5×12=6×10$

Thus $\boxed{60}$ is the smallest number to have 12 positive factors.

First see that the number can't be a square.

Now for twelve factors there must be 6 pairs.

6 smallest divisors are 1,2,3,4,5,6 .

Thus the number must be divisible by these.

Taking LCM of these numbers, we get the answer as 60 .

if a number can be expressed as product of primes such that the powers of the primes are p,q,r,s....,then the number of divisors of that number are (p+1)(q+1)(r+1)................. the no 60 can be expressed as (3)(2*2)(5). therefore no.of divisors=(1+1)(2+1)(1+1)=(2)(3)(2)=12.

In order for the desired integer to have exactly 12 factors, each power in its prime factorization plus one must multiply to 12. Since we are looking to minimize the desired integer, we want to minimize the sum of the exponents each prime is raised to, while still satisfying the condition that the product of each exponent + 1 is 12. The minimum sum of factors of 12 is found in its prime factorization, 2 2 3. We thus take each factor - 1 to get the exponents in the desired integer, whose prime factorization becomes p 1*p 2 p^2_3. Minimization occurs when the smallest possible prime is raised to the third power, so p_3=2. We then assign the next smallest primes to p_1 and p_2 (3 and 5). Thus the final integer equals 2^2 3*5 or 60.

The number of positive divisors of a positive integer with a prime factorization 2^e 1*3^e 2 5^e_3... is (e_1+1)(e_2+1)(e_3+1)... Therefore, (e_1+1)(e_2+1)(e_3+1)...=12. We have two cases: Case 1: (e_1+1)(e_2+1)=12. In this case, e_1=3 and e_2=2. This corresponds to the integer 2^3 3^2=72. Case 2:(e 1+1)(e 2+1)(e+3+1)=12. In this case, e 1=2, e 2=1, and e_3=1. This corresponds to the integer 2^2 3^1 5^1=60. Thus, 60 is the answer.

We can conclude from the idea in Divisors of an Integer that the exponents of the divisors will be smaller if there are more divisors. Therefore, 12 = (2)(2)(3) = (1+1)(1+1)(2+1) will be smaller than 12 = (6)(2) = (5+1)(1+1). Hence, the answer will be (2^2)(3^1)(5^1) = 60

12=2x2x3

Number with 12 positive divisors= $x^1y^1z^2$

Divisors are terms in expansion of $(1+x)(1+y)(1+z^2)$

Now smallest such number is for z= smallest prime =2 And x=3,y=5 choosing next available primes.

Let $a=\prod_{i=1}^n {p_i}^{q_i}$ be our desired positive integer.

Then the number of positive factors is $\prod_{i=1}^n (q_i+1) =12=2\cdot2\cdot3$

In order to minimize our number we must distribute the powers the most evenly and give larger powers to smaller primes, so we obtain the number $2^{3-1}\cdot3^{2-1}\cdot5^{2-1}=60$ .

the number of partitions is equal to the product of the exponents +1 of each prime factor: if n = p1 + p2 s1 ^ ^ s2 + ..... + pn ^ sn are then (s1 +1) (s2 +1) .... (sn +1) 12 = 2 * 2 * 3 then the number is n = a * b * c ² and is smollest if c=2 a=3 and b=5