What is the smallest number that has 4 factors?

Question 2 Exercise 3.5

Answer:

SOLUTION:

The smallest four prime numbers are 2,3,5 and 7.

hence, the required number is $2\times3\times5\times7=210$

Video transcript

"Hello guys, welcome to Lido homework today we'll do question number 12. Now. This is when I questioned or five immigration. So the question is I am the smallest prime number having four different prime factors. Can you find me? So as the numbers for different prime factors? So take the for smallest conjugate of prime numbers. So the for smallest conjugate of prime numbers are 2 3 5 and 7. So these are the prime factors of that of this number that we've defined. So as these are prime factors when you multiply all of these As you're going to get that number, so your answer is 2 into 3. Into 5 into 7 Therefore, your answer is 210. Thank you guys for watching. This was very easy question. If you have any doubts, please let me know in the comments below do not hesitate. Also, please like the video and subscribe the channel. Thank you so much. "

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$f(x)$ denotes the number of different prime factors of the integer $x$, excluding $1$ and $x$ itself. For example, since $12=2\times2\times3$, then $f(12)=2$.

$121=11\times 11$. So it has only one repeated prime factor, so $f(121)=1$.

Since $17$ is prime, that is, $17=1\times 17$, and we are meant to exclude $1$ and $17$, we must have that $f(17)=0$.

Write down the smallest $x$ for which $f(x)=4$.

The smallest $x$ for which $f(x)=4$ will be the product of the four smallest prime numbers. So $x=2\times3\times5\times7=210$.

Answer

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Hint:- Find four smallest prime numbers because we had to find the smallest number. And the product of the smallest factors will be the smallest number.First, we will find the first four prime numbers.As we know that, Prime numbers are those numbers which had only two factors and that were,Number itself and one.So, any number x will be a prime factor if and only if x has only two factors, 1 and x.As we know that set of some smallest numbers are {1, 2, 3, 4, 5, 6………}So, out of them the first four prime numbers will be 2, 3, 5 and 7.Because it is defined that 1 is not a prime number.And 2, 3, 5 and 7 are only divisible by 1 and the number itself.So, the product of these four numbers will give us the required smallest number.Because the product of factors of numbers will give the number itself.So, required number will be (2) * (3) * (5) * (7) = 210.Hence, the smallest number having four different prime factors will be 210.Note:- Whenever we came up with this type of problem then to calculate the required number efficiently. First, we had to find four smallest prime factors from the set of all numbers and then we had to multiply these four numbers to get the smallest number having four different prime factors.

The smallest prime factors are 2,3,5,7 thus the smallest number would be the product of these will be 210

An even integer $2n$ is prime in the $\Bbb{E}$-zone if and only if $2n \equiv 2 \pmod{4}$, which occurs if and only if $n$ is odd. So, factorization into primes amounts to writing $$ \begin{align} N &= 2n_1 \cdot 2n_2 \cdots 2n_k \\ &= 2^k \cdot n_1 \cdot n_2 \cdots n_k, \end{align} $$ where $n_1, \ldots, n_k$ are odd integers. Since we're looking for small numbers with many prime factorizations, and since the product of odd numbers is odd, we may as well assume that $k = 2$. (You seem to have done this implicitly by writing prime factorizations as pairs.)

If we want $4$ prime factorizations, then we're looking for $4$ pairs of prime factors, or an odd number with $8$ factors. (The factors are automatically odd, too.)

In the integers, the number of factors of the number $p_1^{e_1} \cdots p_r^{e_r}$, where $p_1, \ldots, p_r$ are usual primes, is $$ (1 + e_1) \cdots (1 + e_r). $$ So, we need to write $8$ (the number of factors we desire) as $2 \cdot 2 \cdot 2$ or as $4 \cdot 2$. We will consider each possibility in turn.

In the first case, we must consider a product of distinct (odd!) primes $p_1 \cdot p_2 \cdot p_3$. The smallest example of this is $3 \cdot 5 \cdot 7 = 105$, whose factors can be paired off as: $$ \begin{array}{ccc} 1 && 3 \cdot 5 \cdot 7 \\ 3 && 5 \cdot 7 \\ 5 && 3 \cdot 7 \\ 7 && 3 \cdot 5 \end{array} $$ Back in the $\Bbb{E}$-zone, these are prime factorizations of $$ 2^2 \cdot 3 \cdot 5 \cdot 7 = 420 $$ (by putting a $2$ on each factor of the pair): $$ \begin{array}{ccc} 2 && 210 \\ 6 && 70 \\ 10 && 42 \\ 14 && 30 \end{array} $$

In this case, consider a number of the form $p_1^3 \cdot p_2$. The smallest example of this is $3^3 \cdot 5 = 135$, which is larger than $105$, but I'll follow through the analysis anyway. The factors in pairs are: $$ \begin{array}{ccc} 1 && 3^3 \cdot 5\\ 3 && 3^2 \cdot 5 \\ 3^2 && 3 \cdot 5 \\ 3^3 && 5 \end{array} $$ Back in the $\Bbb{E}$-zone, these are prime factorizations of $$ 2^2 \cdot 3^3 \cdot 5= 540 $$ (by putting a $2$ on each factor of the pair): $$ \begin{array}{ccc} 2 && 270 \\ 6 && 90 \\ 18 && 30 \\ 54 && 10 \end{array} $$

So, $420$ is the smallest positive integer with $4$ prime factorizations in the $\Bbb{E}$-zone.