I am told to discover the the smallest number through 4 different prime factorizations in this world.

You are watching: What is the smallest odd number with four different prime factors

I should additionally define what it means to be a element in the $\tasiilaq.netbbE$-zone. A number is element if it cannot be factored right into a product of also numbers. For all intents, weird numbers do not exist.

For example, few of the first couple of primes room 2,6,10,14,18,22... We can see that a number is element if that is congruent to 2 mod 4.

Now we also know that prime administer in this human being is not unique. Because that example, 180 has actually 3 element factorizations. Namely, (10,18), (6, 30), and (2,90).

I regulated to find a number 2940 that has 4 factorizations: (6,490), (10,294), (2,1470), and also (30,98).

However, I do not know how to go around finding the $smallest$ such number.

number-theory
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asked Feb 5 "14 in ~ 7:27

Tyler MurphyTyler Murphy
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An even integer $2n$ is element in the $\BbbE$-zone if and also only if $2n \equiv 2 \pmod4$, which wake up if and only if $n$ is odd. So, factorization right into primes amounts to writing\beginalignN &= 2n_1 \cdot 2n_2 \cdots 2n_k \\&= 2^k \cdot n_1 \cdot n_2 \cdots n_k,\endalignwhere $n_1, \ldots, n_k$ room odd integers. Due to the fact that we"re in search of small number with numerous prime factorizations, and since the product the odd numbers is odd, we may too assume the $k = 2$. (You seem to have actually done this implicitly by creating prime factorizations as pairs.)

If we want $4$ element factorizations, climate we"re in search of $4$ pairs of element factors, or one odd number through $8$ factors. (The determinants are automatically odd, too.)

In the integers, the variety of factors that the number $p_1^e_1 \cdots p_r^e_r$, whereby $p_1, \ldots, p_r$ space usual primes, is$$(1 + e_1) \cdots (1 + e_r).$$So, we have to write $8$ (the number of factors we desire) as $2 \cdot 2 \cdot 2$ or as $4 \cdot 2$. Us will think about each opportunity in turn.

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In the an initial case, us must take into consideration a product of distinct (odd!) primes $p_1 \cdot p_2 \cdot p_3$. The smallest instance of this is $3 \cdot 5 \cdot 7 = 105$, whose factors can it is in paired off as:$$\beginarrayccc1 && 3 \cdot 5 \cdot 7 \\3 && 5 \cdot 7 \\5 && 3 \cdot 7 \\7 && 3 \cdot 5\endarray$$Back in the $\BbbE$-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):$$\beginarrayccc2 && 210 \\6 && 70 \\10 && 42 \\14 && 30\endarray$$

In this case, consider a number of the form $p_1^3 \cdot p_2$. The smallest instance of this is $3^3 \cdot 5 = 135$, which is bigger than $105$, yet I"ll follow through the analysis anyway. The components in pairs are:$$\beginarrayccc1 && 3^3 \cdot 5\\3 && 3^2 \cdot 5 \\3^2 && 3 \cdot 5 \\3^3 && 5\endarray$$Back in the $\BbbE$-zone, these are prime factorizations of$$2^2 \cdot 3^3 \cdot 5= 540$$(by putting a $2$ top top each element of the pair):$$\beginarrayccc2 && 270 \\6 && 90 \\18 && 30 \\54 && 10\endarray$$

So, $420$ is the smallest positive integer through $4$ prime factorizations in the $\BbbE$-zone.