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Determine the form of the prime factorization of all positive integers n with d(n) = 6....

Question

Determine the form of the prime factorization of all positive integers n with d(n) = 6.

Determine the form of the prime factorization of all positive integers n with d(n) = 6.



Answers

Show that $n^{3}-n$ is divisible by 6 for all positive integers $n$

They want us to prove that 60 factor of in Q plus three and squared plus two in for all values off in. Okay, so let's go ahead and see what we can do with this. So we're gonna do this by induction, So let's go ahead and call this just our statement yet. Now we need to check our base case, which is going to be in. Is he gonna one Since we're doing this for all natural numbers in and that's going to be so, it will just plug it one someone cute plus three times one squared, plus two times one. So that all adds up to six. And being a factor six or having a factor of six is just a fancy way of saying gets weaken divided by six. So six is divisible by six. So it's one so checks up all right now Or are induction hypothesis we will need to do the all the link. Well, we want to say Piquet is a statement. So six actor Oh que cute plus three k squared plus two k And then we want to show that PK plus one It is also true, which is the statement. Six factor of K plus one cute plus three K plus one squared plus, uh, two K plus one. All right, now, let's go ahead and start the proof. So remember, we always want tell the reader what kind of proof were doing because there's multiple ways to prove things. So by way, uh, induction, assume p K is true. Now we want to start with this, Okay? Plus one cute times, three K plus one squared, plus two times K plus one. So we somehow want to get this k cubed lost three K squared, plus two k to come out from this and just looking at it, I really don't see a good way to do that. And if we were to factor out a K plus one it doesn't look like would help either, because we just have, like, a K squared term in the middle. So, uh, that doesn't look like it's Ah, good way to try to go about this. So why don't we just expand everything? So let's see. So if we were to expand this, we should get que que plus three k squared plus three K plus one. If we were to expand this here That should give us plus three k squared plus six K plus three and then that last term is just gonna be two k plus two. So let's add everything. Go. Um, we only have one case square term, so that would just be K squared. We have three k squared three k squared. So that's going to give us six k squared. We have three K plus six K plus two case, that's 11 K And unless we have one plus two plus three, which is six. Okay, so we somehow want to get this cake plus three k squared, plus two K to come up. So I'm gonna write that down right here. Que cute plus three k squared plus two K. Well, that shoes pulls some of these terms out, and right over here, we're gonna have K cubed plus three K squared plus two K. So if we do that, then so that case squared or kick, it becomes just zero. This K squared should become plus three case. Where and then over here, this 11 K is going to become, um, nine K. And in the six just days there All right, so we have this now, so we know for sure This portion in blue is visible by six by our induction hypothesis by we don't know. Sorry. Let me just say this is PK appear, but we don't know if this term here is divisible by six. So one of the tricks we tried in some of the other problems is to just keep pulling out mawr terms of it until we get this to be something that's all divisible by six. But you might know. So we try to do that again. So we would have negative k cubed here, and then we would have, um, no k squared. And then we'd have that nine k, but it would be seven k. Then here would be plus six still. And then this would be two times que que plus three k squared plus two K like this. And then we would say, Okay, well, this still isn't anything that's divisible by six, and we just keep repeating. And since this nine here isn't an even number will never be able to get anything that's divisible by six. If we keep subtracting because I'll always be odd. So we might need to try something else or this one. So what do we do? Some scratch work over here on the side or this So scratch? Well, we can factor three outs. Let's try that. So we have three times k squared plus three K plus two. And then this quadratic is also factor. What would be K plus one? Okay, plus two. Okay, well, if something is divisible by six, that means it's divisible by two and three. So we already know this is gonna be divisible by three since we're multiplying by three. So what we need to ask ourselves is this expression visible by two. And if you think about it for a little, you should be able to agree with me that this will be divisible by two since K plus one in K plus two are consecutive integers. And since it's consecutive integers, one of them has to be even is if you think about it. So if we start counting so 01234567 and so on. It alternates between even and odd. So one of these has to be even and thus the definition of even is that it's divisible by two. So that means this is divisible by six. So let's just go ahead and rewrite this real fast looks k cubed plus three casework plus two K And then over here, this is going to be plus three times K plus one times K plus two. Yeah. So this is the statement we're working with now. Well, by induction hypothesis, So by our induction hypothesis, ih que que plus three k squared plus two K is divisible I six and now to show that three K plus one times K plus two is divisible, we're pretty much is going to explain the same thing we did over here about this being three times some even number so so and three times K plus one times K plus two is divisible by six cents. It is too visible by three and it must be divisible by two since K plus one and K plus two or consecutive integers and one must be either. All right, so what we just show is now or I should before we do that. So since both of these are true so since both terms are divisible by six, so must be there some and so we pretty much shown that PK plus one is also a true statement of it being divisible by six. So let's just go ahead and wrap this upper class by saying so since P K plus one is true, assuming P K is true, then peon is true for all in elements of the natural numbers. And then once you do this, you could put your proof box and a smiley face because you're glad you're done with it.

How can we find that the prime factory ization of 26? Well, to do this, we can use a factor tree where we brought draw two branches. And then we think, what two numbers could multiply together to give us 26 while we know 26 is even so, it's divisible by two and two times 13 is equal to 26. Then we would keep going and drawing branches if we have composite numbers on each branch. But in this case she was prime. So we circle it and so is 13. We can write this as the sentence of prime factory ization, as 26 is equal to two times 13.

Computer that that's all given Borchin multiplied a office here and talk who in part off X? It means six might be blinded. Constant be gone. Well, but it this number in Toto factors. Yeah. Yeah. Did this number into two factors side that the same isn't off? Each pack does Must be minus seven. Basic us. Um, also right here. You and more of its multi planet city. And Barbara picks mine. Us too. And you're who and power of X today in about Abed's? Yes. I mean, who inbound up experiment hair from? That's the ban. Then the men in time did he? That's come out, Get in and blowing up ands mine us do minus one woman here T and about of eggs minus. Then did you, in about of its mileage to common hair, did he? And power of X minus tool woman. They're the remaining time toe end part of eggs. Minus one is fact us our solution of giving fortune. Thank you

Were asked to find the smallest positive integer with exactly end different positive factors. In part, a were given that N is equal to three. So defying the smallest such positive imager. Let's consider the factors of the positive managers until we find the energy or with exactly three factors. So on the left I'm going to write the factors, the integer and and on the right, all right. It's factors. So we have vinegar is one. This has factor one managers to says The two factors. One and two finishers. Three. This has the two factors one in three and if the integer is four four has yet factors one, two and four. So for is the smallest positive imagery, with exactly three different factors. So it follows that for part A. Our answer is four. So I'll circuit with us in Red Part B were given that N is four. So I want to find the smallest positive injure with exactly four different positive factors. So, looking back through our list, we saw that none of the Pasi majors, less than equal to four, have four factors. Even so, we have to keep going. Five has factors one in five. Six has the factors. 12 three and six. This is four factors, so it follows that six is the smallest positive integer with exactly for different factors. So this is our answer for part B in part C. Forgiven. The end is five. So you want to find the smallest positivity with exactly five different positive factors. So far, all the answers we considered have at most four positive factors. So we need to keep working. Seven has positive factors. 17 eight has positive factors. One to four and eight nine Has positive factors. One. Three in nine 10 as positive factors. One to five and 10 11 Has positive factors. One and 11 12. His positive factors. 123 four, six 12 So that's six factors right there. However, this is mawr than five factors. You want to find the smallest number with exactly five factors. So moving on we have 13 is positive factors. One in 13 14 has positive factors. 127 and 14 15 is positive factors. One, three, five and 15 16 has positive factors. One to four, eight and 16 These air five positive factors, so it follows that 16 is the smallest positive imager, with exactly five different positive factors. So this is our answer for part C for Part D. Were given that and is equal to six. So we're trying to find the smallest positive manager with exactly six different positive factors. We actually already encountered this one so recall that 12 had exactly six positive factors. So it follows that the answer to Parc de is 12 and finally in part E were given. That end is 10. So we're skipping along here rest to find the smallest positive injure with exactly 10 different positive factors. So this might take a bit longer. So consider 17 this positive factors 1 17 18 is positive factors. 12 three, six, nine and 18 19 is positive factors. One of 19 2 factors, 20 has factors. 12 four, five, 10 20 So six positive factors 21 has thief actors 13 seven 21 4 Positive factors 22 Has factors 12 11 22 four Factors 23 has factors 1 23 two Factors 24 has factors one to three four six eight 12 24 So eight factors We're getting close, but we're not there yet. 25 as one five 25. This factors three factors. 26 has the factors one to 13 26 four factors 27 Has factors one, three nine. 27 So four factors 28 as factors one to four seven 14 28 So six Factors 29 has factors one and 29 two Factors 30 Has factors one to three, five, six 10. 15 30 Again eight factors So we're getting close. 31 Has factors one in 31 two factors 32 as factors one to four eight 16 32 six Factors. 33 as factors. One, three, 11. 33 So four factors. 34 as factors. One to 17 and 34 So four factors 35 has factors one five seven. 35 So four factors 36 has factors one to three, four six nine. 12 18 36 So nine factors or even closer now need one more factor. 37 as factors. One in 37 So two factors 38 has factors one to 19 38 So four factors 39 has factors. One 3 13 39. So four factors 40 These factors one to four, five, eight, 10 2040 So eight factors Not quite there. 41 has factors one and 41 So two factors 42 has factors. One to three, 67 14 21 42 Eight Factors again close but we're two away. 43 Has factors 1. 43 two Factors 44 Has factors 12 four, 11 22 4 Before So six factors 45 as factors. One, three five nine. 15 in 45. So six factors 46 as factors. One to 33. That was a mistake. Not 33 23 and 46. So four Factors 47 has factors. One in 47 two factors. Mhm 48 has factors one to three, four six eight 12 16 24 48 for a total of 10 factors. So it follows that the smallest positive integer with exactly 10 different positive factors is 48


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