How many factors does the number 19 x 29 x 59 x 79 have?

[justinepark]

Having trouble with this problem, appreciate help.

Excluding 1 and itself, how many factors does the number 19 x 29 x 59 x 79 have?

 

[mathmari]

Having trouble with this problem, appreciate help.

Excluding 1 and itself, how many factors does the number 19 x 29 x 59 x 79 have?

$$n=p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$$
where $p_i$ are primes ($p_i \neq p_j$) and $a_i >0$.

The number of factors of $n$ is equal to $$\tau(n)=(a_1+1)(a_2+1) \dots (a_k+1)$$

 

[kaliprasad]

Having trouble with this problem, appreciate help.

Excluding 1 and itself, how many factors does the number 19 x 29 x 59 x 79 have?

the number $19 *29 *59 * 79$ is product of 4 primes and for each factor there is choice of 2 for each prime so 2^4 ot 16 factors as per How to Find How Many Factors Are in a Number: 4 Steps

to illustrate the factors are

1 $19^0*29^0*59^0*79^0$ or 1
2 $19^1*29^0*59^0*79^0$ or 19
3 $19^0*29^1*59^0*79^0$ or 29
4 $19^1*29^1*59^0*79^0$ or 19 * 29
5 $19^0*29^0*59^1*79^0$ or 59
6 $19^1*29^0*59^1*79^0$ or 19 * 59
7 $19^0*29^1*59^1*79^0$ or 29 * 59
8 $19^1*29^1*59^1*79^0$ or 19 * 29 * 79

and 8 product of the above combinations and 79

if one has to exclude 1and itself then 14 factors

 

[mathmari]

the number $19 *29 *59 * 79$ is product of 4 primes and for each factor there is choice of 2 for each prime so 2^4 ot 16 factors as per How to Find How Many Factors Are in a Number: 4 Steps

to illustrate the factors are

1 $19^0*29^0*59^0*79^0$ or 1
2 $19^1*29^0*59^0*79^0$ or 19
3 $19^0*29^1*59^0*79^0$ or 29
4 $19^1*29^1*59^0*79^0$ or 19 * 29
5 $19^0*29^0*59^1*79^0$ or 59
6 $19^1*29^0*59^1*79^0$ or 19 * 59
7 $19^0*29^1*59^1*79^0$ or 29 * 59
8 $19^1*29^1*59^1*79^0$ or 19 * 29 * 79

and 8 product of the above combinations and 79

if one has to exclude 1and itself then 14 factors

That's right!

Using the formula I wrote in my previous post, we have the following:

$$n=19^1 \cdot 29^1 \cdot 59^1 \cdot 79^1$$

$$\tau{(19 \cdot 29 \cdot 59 \cdot 79)}=2 \cdot 2 \cdot 2 \cdot 2=2^4=16$$

At the $16$ factors, $1$ and itself are included.

Therefore, the number of factors excluding $1$ and itself is equal to $14$.

 

[mathbalarka]

Well, it's more or less enough to note that $\tau$ is multiplicative instead of using that formula : elts of $\{19, 29, 59, 79\}$ are pairwise coprime. Indeed, they are all primes.

$$\tau(19 \cdot 29 \cdot 59 \cdot 79) = \tau(19) \cdot \tau(29) \cdot \tau(59) \cdot \tau(79) = 2 \cdot 2 \cdot 2 \cdot 2 = 2^4$$

In an "english" translation, if $p$ is a prime then there are only two factors of $p$ : $\{1, p\}$. For $p_1 \cdot p_2 \cdot p_3 \cdot p_4$, $p_i$s being prime, looking for number of factors is essentially equivalent to finding how many ways one can pick a single ball from each box, given 4 boxes with 2 balls inside marked as $1$ and $p$. It's an easy exercise in combinatorics to show that there are indeed $2^4$ ways to do it.