Number of Divisors

We can easily verify that with $N=60$ , we get $A=12,B=6,C=4$ so it fits the conditions of the question. Now we try to minimize $N$ , keeping in mind that $N=60$ is achievable.

For any $n>2$ , $n-1$ is not a divisor of $n$ , so $3<C<B<A<N.$ So $B≥5$ . But if $B=5$ then $C=2\neq 3$ which is a contradiction. So $B≥6$ . Since 7,8,9,10,11 all have less than 6 factors, then $A≥12$ .

Let $N=p_1^{q_1}p_2^{q_2}\ldots p_k^{q_k}$ , where the $p_i$ 's are the distinct prime factors of $N$ . So $A=(q_1+1)(q_2+1)...(q_k+1)\geq 12$ . We want to minimize $N$ , so we shall let $p_i$ be the ith smallest prime with $q_1≥q_2≥...≥q_k$ .

When $k=1$ , then $q_1≥11$ and $N≥2^{11}>60$ .
When $k=2$ , then $q_1≥\sqrt{12}-1⇒q_1≥3$ .
$\quad$ If $q_2=1$ , then $q_1≥5⇒N≥2^5⋅3>60$ .
$\quad$ If $q_2≥2$ , then $N≥2^3⋅3^2>60$ .
When $k=3$ , then $q_1≥1\sqrt[3]{12}-1⇒q_1≥2$ . Then $N≥22⋅3⋅5=60$ .
When $k=4$ , $N≥2⋅3⋅5⋅7>60.$

So we have proven that $N \geq 60$ , so the smallest $N$ is indeed 60.

Now 3 is the number of divisors of some number x. $3 = 3*1 = (2+1)*(0+1)$ . So, $x = (a^2)(b^0)$ . where $a$ and $b$ are distinct prime numbers. So since we need the smallest positive number we take the prime number that gives the smallest value for $x$ . So, here $x = (2^2)(1)=4$ .

Now 4 is the number of divisors of some number $y$ . using the similar method. $4 = (1+1)(1+1)$ . therefore, $y=(2^1)(3^1)=6$ .

Now 6 is the number of divisors of some number $z$ . $6=2*3$ or $6=6*1$ . but since we need the smallest value for z, we choose the former part. $6=(1+1)*(2+1)$ . so, $z = (3^1)*(2^2)=3*4=12$ . [ reason: we need the smallest number]. From this we get the number of divisors for the smallest positive number that we need to find is 12. Thus, using the above process $12=(2+1)(1+1)(1+1)$ . $A= (a^2)(b^1)(c^1)$ . here we have to choose the first 3 prime numbers for as the values of a, b and c such that A gets the smallest values. $A=(2^2)(3)(5)=4*3*5=60$ . So 60 is the smallest positive number, such that the number of divisors of the number of divisors of the number of divisors of its number of divisors is 3.

[Edits for clarity - Calvin]

If $n=p_1^{k_1}\cdot ...\cdot p_m^{k_m}$ , with $p_i$ being distinct primes, then the number of divisors of $n$ is $(k_1+1)\cdot...\cdot (k_m+1)$ .

Suppose $a$ is the number we are looking for, $b$ is the number of its divisors, $c$ is the number of divisors of $b$ , and $d$ is the number of divisors of $c$ .

We will first find the answer, and then justify it. Since the number of divisors of $d$ is three, $d$ is a square of a prime. The smallest such choice is $4$ . If $d=4$ , then $c$ is either a product of two primes or a cube of a prime. Two smallest such choices are $6$ and $8$ . If $c=6$ , then the smallest choice for $b$ is $2^{2}\cdot 3=12.$ The smallest choice for $a$ now is $2^{2}\cdot 3\cdot 5= 60$ . It is easy to see that all other choices of $d$ , $c$ , $b$ lead to bigger $a$ . Alternatively, one can check that if $a<60,$ then $b\leq 10$ , so $c\leq 4$ and $d\leq 3.$ Thus, the answer is $60.$

Should you put non-negative because there are a lot of negative numbers with 12 factors?

Smallest integer with number of divisor 3=4 Smallest integer to with numer of factors 4=6 same way with 4 factors =6 and with 6 factors =12 and with 12 factors=60

You can find the number of divisors of a number through prime factorization.

To calculate the number of divisors of a certain number, you can use the formula, d= a^x * b^y * c^z ... , where a, b, and, c are prime factors and x, y and z, are powers.

For example:

the number 12 has a prime factorization of 2^2 * 3. the number 2 can be chosen 0 times, 1 time, or two times = 3 the number can be used 0 times, or 1 time = 2 The number of divisors of 12 will be 3 2= 6 (1 2 3 4 6 12)

Based on the solution, we can equate d=(x+1)(y+1)(z+1) ... where x, y and z are again the exponents of the prime factors.

The Problem above tells us that the number of divisors of the number of divisors of the number of divisors of its number of divisors is 3.

We'll start with d=3

In order to have 3 divisors, we must have (x+1)(y+1)=3. Possible prime factors are: 2^2 = 4 (Observe that adding 1 to the exponent equals 3)

So 4 is the number of divisors of the number of divisors of its number of divisor. Then we now start with d=4 Possible prime factors are 2^3 =8 (Observe again that adding 1 to the exponent is equal to 4) 2*3 =6

Since we are after the smallest integer, we use d=6. Possible prime factors: 2^2 * 3 =12 2* 3^2 =18

the smallest is d=12, so for the last time we find a prime factor that has 12 factors: 2^3 * 3^2 =72 2^2 *3 *5 =60

Final answer is 60 ...

you can also refer to this site : http://mathforum.org/library/drmath/view/56169.html

Suppose that, the smallest positive integer is X the number of divisors of X is Y the number of divisors of Y is Z the number of divisors of Z is M Given that, the number of divisors of M is 3

X is the smallest positive integer thought all of those numbers( M, Z, Y, X) must be smallest. As this properties we have only one value for M which divisors number is 3 & that is 4( divisors for 4 are 1, 2, 4)

So, M = 4 Z = 6 ( 1, 2, 3, 6) Y = 12 ( 1, 2, 3, 4, 6, 12) X = 60 ( 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) Ans : 60.

Smallest integer with number of divisor 3=4 Smallest integer to with numer of factors 4=6 same way with 4 factors =6 and with 6 factors =12 and with 12 factors=60

The smallest number with 3 divisors is 4. The smallest number with 4 divisors is 6. The smallest number with 6 divisors is 12. The smallest number with 12 divisors is 60. So the answer is 60. P.S. Tau function was used to determine the number of divisors.

we can solve question in reverse direction

let us take third step divisor as 2^n 3^m 5^n..... from here no of divisors = (n+1)(m+1).... =3 (given) Only n= 2 is possible which gives 2^2 = 4

now repeating same for 2nd step divisors. (n+1)(m+1) = 4 = 2x2 so n = m = 1

So 2nd step divisor = 2^1 3^1 = 6

now for 1st step divisor (m+1)(n+1)... = 2x3

gives m=1, n=2

therefore, divisors of required number = 2^2 3^1 = 12

Now let number be (n+1)(m+1)(l+1)...= 2x2x3

gives n=1 m=1 l=2

So required number = 2^2 x 3 x 5 = 60

as smallest number is required higher power is giver to smaller number.

To yield the smallest number fitting the description, one must work backwards and find all the smallest integers along the steps. The smallest integer with said number of divisors can be determined by assuming the smallest positive integers as those divisors, or 1,2,3,etc.

The last number of divisors must have 3 divisors itself, and is 4 (with divisors 1, 2, or 4). The second to last number of divisors will be the smallest integer with 4 divisors, or 6 (1,2,3,6). The final number of divisors will be the third to smallest integer with 6 divisors, or 12 (1,2,3,4,6,12). This final number of divisors will yield an answer of 60 (1,2,3,4,5,6,10,12,15,20,30,60).

60 has no. of divisors = 12, 12 has no. of divisors = 6, 6 has no. of divisors = 4, 4 has no. of divisors = 3 so we need the answer "3". hence N=60