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

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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.

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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$$\beginalignN &= 2n_1 \cdot 2n_2 \cdots 2n_k \\&= 2^k \cdot n_1 \cdot n_2 \cdots n_k,\endalign$$where $n_1, \ldots, n_k$ 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.