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1 hour ago, Wind30 said:

This solution is what my colleague gave me.

My solution is to mirror image the quadrilateral along the line AD to form a Hexagon with 6 equal sides. You should be able to prove that the angle X is half of angle 110 as that Hexagon is made up of three angles 110 and three angles 130.

I thought the question is a bit cheating as there is no solution if the angles don't add up to 240... my kid have not solve this yet after many days...

Can you show your solution in greater detail and rigour? Just to be sure I'm not missing something obvious.

Because mirroring is the first thing I thought about (practically instantly). I also dismissed it very quickly because it's incorrect.

Here's the counterexample:

1822088172_geometrycounterexample.thumb.png.0124c1e4051a63017c6e530b69e5b4ae.png

Sorry about the changing labels, etc. But note that the angles have been changed to 100 and 110 (which don't add up to 240). In this case, collinearity of the three points (C', A, B' as per the labels here) doesn't hold like in the case we were originally presented.

And that makes a huge difference. You can still reflect the quadrilateral in line C'B' to get its mirror image and form a hexagon with six equal sides. But now the remaining two angles become (2*78.3) = 156.6 deg and (2*71.7) = 143.4 deg respectively. You have a hexagon with angles (going counterclockwise) of 100, 110, 143.4, 110, 100 and 156.6 degrees. Which add up to 720 deg as they should, but you can no longer assert that you have 2 sets of triplets (110, 130, 110, 130, 110, 130 degrees) like you could in the original case.

The "add to 240 degrees" (which is equivalent to collinearity of those three points) is a necessary criterion to be able to solve the problem this way. Unless you somehow showed it in your "easy reflection" approach, I'm afraid it's wrong.

 

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41 minutes ago, Turboflat4 said:

Can you show your solution in greater detail and rigour? Just to be sure I'm not missing something obvious.

Because mirroring is the first thing I thought about (practically instantly). I also dismissed it very quickly because it's incorrect.

Here's the counterexample:

1822088172_geometrycounterexample.thumb.png.0124c1e4051a63017c6e530b69e5b4ae.png

Sorry about the changing labels, etc. But note that the angles have been changed to 100 and 110 (which don't add up to 240). In this case, collinearity of the three points (C', A, B' as per the labels here) doesn't hold like in the case we were originally presented.

And that makes a huge difference. You can still reflect the quadrilateral in line C'B' to get its mirror image and form a hexagon with six equal sides. But now the remaining two angles become (2*78.3) = 156.6 deg and (2*71.7) = 143.4 deg respectively. You have a hexagon with angles (going counterclockwise) of 100, 110, 143.4, 110, 100 and 156.6 degrees. Which add up to 720 deg as they should, but you can no longer assert that you have 2 sets of triplets (110, 130, 110, 130, 110, 130 degrees) like you could in the original case.

The "add to 240 degrees" (which is equivalent to collinearity of those three points) is a necessary criterion to be able to solve the problem this way. Unless you somehow showed it in your "easy reflection" approach, I'm afraid it's wrong.

 


after you mirror the quad, prove the triangle inside is equilateral , which is trivial to prove. It is only equilateral when the angles add up to 240. I actually drew the second line of the triangle with 60 deg angle, and found the remaining angle is 35deg... Then I completed the figure and proved it’s a mirror. 

DECFCE40-F893-450C-9F1D-DDAA98DD1396.png

Edited by Wind30
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1 hour ago, Wind30 said:


after you mirror the quad, prove the triangle inside is equilateral , which is trivial to prove. It is only equilateral when the angles add up to 240. I actually drew the second line of the triangle with 60 deg angle, and found the remaining angle is 35deg... Then I completed the figure and proved it’s a mirror. 

DECFCE40-F893-450C-9F1D-DDAA98DD1396.png

I am rushing out soon, but I'll have to consider if your argument is sound. It would help if you gave more details to make it rigorous in the meantime. Thanks.

Edited by Turboflat4
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18 hours ago, Wind30 said:


after you mirror the quad, prove the triangle inside is equilateral , which is trivial to prove. It is only equilateral when the angles add up to 240. I actually drew the second line of the triangle with 60 deg angle, and found the remaining angle is 35deg... Then I completed the figure and proved it’s a mirror. 

DECFCE40-F893-450C-9F1D-DDAA98DD1396.png

I've thought about this approach quite carefully. I don't think you've got a valid proof.

It's possible to consider the situation from a few angles (no pun intended). I'm sure you used one or more of these. But all of them seem to end up hitting a wall, at least when I try them.

If (case 1) you start by mirroring the quadrilateral, then try to construct a triangle with vertices coinciding with alternate vertices of the hexagon, it's not trivial to show that it is equilateral in the case when the angle sum is 240. (It is definitely not equilateral (only isosceles) in any other case - but even in the case when angle sum is 240, I cannot see a geometric way to show the triangle is equilateral).

If (case 2) you start by drawing the first leg of the triangle within the given quadrilateral, then construct another leg 60 deg from that side and of the same length (which would make two legs of the equilateral triangle), it is not trivial to show that the point you end up with is coincident with the appropriate vertex of the mirror image if and only if the angle sum is 240.

And finally (case 3), if you start by mirroring and draw the first leg of the triangle across the line of symmetry, then construct an equal leg at a 60 deg angle, you will again find it difficult to show that the point you end up with is coincident with the appropriate vertex of the original quadrilateral.

The only time when the geometry works out perfectly to give you the equilateral triangle with vertices corresponding to alternate vertices of the hexagon formed by mirroring the quadrilateral is when that angle sum equals the magic figure of 240 deg (final two figures, with different specific angles). But as far as I can see, there is no trivial way to prove this geometrically. A trigonometric or coordinate geometry approach might work, but I cannot show it using basic geometry like taught in Primary School.

So, unless you can actually show your method, including all working (rather than saying it's "trivial"), I remain unconvinced of the validity of your approach.

My (previously presented) solution (which you stated was similar to your colleague's solution) remains the only elementary approach that I know to be correct so far.

I am happy to be proven wrong - I like learning new things.

 

 

geometry.jpg

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3 hours ago, Turboflat4 said:

I've thought about this approach quite carefully. I don't think you've got a valid proof.

It's possible to consider the situation from a few angles (no pun intended). I'm sure you used one or more of these. But all of them seem to end up hitting a wall, at least when I try them.

If (case 1) you start by mirroring the quadrilateral, then try to construct a triangle with vertices coinciding with alternate vertices of the hexagon, it's not trivial to show that it is equilateral in the case when the angle sum is 240. (It is definitely not equilateral (only isosceles) in any other case - but even in the case when angle sum is 240, I cannot see a geometric way to show the triangle is equilateral).

If (case 2) you start by drawing the first leg of the triangle within the given quadrilateral, then construct another leg 60 deg from that side and of the same length (which would make two legs of the equilateral triangle), it is not trivial to show that the point you end up with is coincident with the appropriate vertex of the mirror image if and only if the angle sum is 240.

And finally (case 3), if you start by mirroring and draw the first leg of the triangle across the line of symmetry, then construct an equal leg at a 60 deg angle, you will again find it difficult to show that the point you end up with is coincident with the appropriate vertex of the original quadrilateral.

The only time when the geometry works out perfectly to give you the equilateral triangle with vertices corresponding to alternate vertices of the hexagon formed by mirroring the quadrilateral is when that angle sum equals the magic figure of 240 deg (final two figures, with different specific angles). But as far as I can see, there is no trivial way to prove this geometrically. A trigonometric or coordinate geometry approach might work, but I cannot show it using basic geometry like taught in Primary School.

So, unless you can actually show your method, including all working (rather than saying it's "trivial"), I remain unconvinced of the validity of your approach.

My (previously presented) solution (which you stated was similar to your colleague's solution) remains the only elementary approach that I know to be correct so far.

I am happy to be proven wrong - I like learning new things.

 

 

geometry.jpg


I use case 2. Below are the steps of the construction. The second pic just construct an isosceles triangle with angle 35. Once u have that triangle out, u can see the two sides are equal because of similar triangles. Since the middle angle is 60 the final side must also be equal. Once u prove that, it is easy to construct the hexagon with the mirror. I did not assume the location of any vertices during construction. U can actually see the kite at the third picture so u can get the answer there already as the angle is 110 divided 2

 

F1B19DBE-CF9F-416A-B73C-516967C067EC.png

06748FAF-7040-4453-B8E3-3E0D7F51E756.png

E312E86E-0975-4DA1-95CC-C0A5D494D600.png

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6 hours ago, Wind30 said:


I use case 2. Below are the steps of the construction. The second pic just construct an isosceles triangle with angle 35. Once u have that triangle out, u can see the two sides are equal because of similar triangles. Since the middle angle is 60 the final side must also be equal. Once u prove that, it is easy to construct the hexagon with the mirror. I did not assume the location of any vertices during construction. U can actually see the kite at the third picture so u can get the answer there already as the angle is 110 divided 2

 

F1B19DBE-CF9F-416A-B73C-516967C067EC.png

06748FAF-7040-4453-B8E3-3E0D7F51E756.png

E312E86E-0975-4DA1-95CC-C0A5D494D600.png

Thanks for clarifying.

You're using a variant of case 2, where you're drawing the line segment using the base angle from the first "small" isosceles triangle.

In the given problem, the base angle is 0.5*(180-110) = 35 deg.

So you're drawing a 35 angle at the 130 deg vertex.

You can, of course, immediately say that the middle angle is 60 degrees. No issue there.

The problem is that you have not shown that your new line segment terminates on a vertex of the mirror quadrilateral (in the bottom half of the hexagon).

In fact, in general, it does not.

An alternative way to prove your argument is to show that your line segment is perpendicular to the "base" of the quadrilateral (the base is also the line of mirror symmetry). The only time you'll have a right angle here is with the magic sum of 240. But there is no way of showing this geometrically here either, too few angles are known.

You can't "just draw" an extension of your line segment, and claim it terminates at the vertex of the mirror quadrilateral. You have to show it, either directly, or by showing perpendicularity between that line segment and the base of the quadrilateral. If you do not do either of these, you have no basis to assert you're even forming a triangle (let alone isosceles).

I've attached a figure to show you what I'm talking about.The middle figure is what you're given in the question (same angles as original question, magic sum 240). Here everything works out "swee-swee", but there's no way to show either coincidence with the mirror vertex or perpendicularity with the base with basic geometry (and without relying on accurate scale construction, like I've done here). Everything works, but you can't show it trivially.

The right-most case exemplifies another variant with magic sum 240 but with different angles. Note that here, you calculate the base angle as 41 deg, but everything still works out - coincidence with the mirror vertex and perpendicularity with the base. The magic sum being 240 is necessary and sufficient.

The left-most case is the counter-example. Here, the magic sum is not 240, the base angle is 40 and you can see that the line doesn't come close to a vertex on the mirror side. Neither does perpendicularity hold. Obviously, the middle angle is not 60 degrees (and you can't construct your equilateral triangle).

I guess you might argue that the construction only works if the middle angle is 60 degrees, and you can certainly construct the same length for the other leg, joining the last side to give your equilateral triangle. But you still have to show that the bottom vertex of that triangle is exactly coincident with the mirror vertex of the quadrilateral. Alternatively, show that the leg of that triangle forms a right angle with the base of the quadrilateral (which is the line of symmetry).

If you've not rigorously shown that either of the above hold, I'm afraid that's not valid.

geometry2.thumb.jpg.67e2b4fbfb7ab4a03e8a31d9885a1cf2.jpg

 

 

Edited by Turboflat4
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3 hours ago, Turboflat4 said:

Thanks for clarifying.

You're using a variant of case 2, where you're drawing the line segment using the base angle from the first "small" isosceles triangle.

In the given problem, the base angle is 0.5*(180-110) = 35 deg.

So you're drawing a 35 angle at the 130 deg vertex.

You can, of course, immediately say that the middle angle is 60 degrees. No issue there.

The problem is that you have not shown that your new line segment terminates on a vertex of the mirror quadrilateral (in the bottom half of the hexagon).

In fact, in general, it does not.

 

 

I did not claim it terminates at the vertex of the mirror quadrilateral. If you look at my figure 3, the mirror quadrilateral is not complete but the answer is already there. There is no need for the mirror quad at all. 

Don't you agree I can draw another isosceles triangle at my figure 2, with angles 35, 35, 110 without any assumptions?  

Once you have the second isoceles triangle you have two similar isoceles triangles (35,35,110 angles) you can prove that the last line I draw in my figure three forms an equilateral triangle since you have a triangle with two equal sides and angle 60. With that you have a kite quadrilateral (2 pairs of adj equal sides), x is just half of 110.

The MAIN thing is proving the equilateral triangle in figure three. 

BTW the kite figure proves what you requested.... "An alternative way to prove your argument is to show that your line segment is perpendicular to the "base" of the quadrilateral "

I am not sure why you keep saying I make assumptions when I did not in my last post. Don't read my previous posts as there are assumptions made when I was just writing without analyzing deeply. 

I am not sure how to explain because it is pretty clear... 

 

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37 minutes ago, Wind30 said:

 

I did not claim it terminates at the vertex of the mirror quadrilateral. If you look at my figure 3, the mirror quadrilateral is not complete but the answer is already there. There is no need for the mirror quad at all. 

Don't you agree I can draw another isosceles triangle at my figure 2, with angles 35, 35, 110 without any assumptions?  

Once you have the second isoceles triangle you have two similar isoceles triangles (35,35,110 angles) you can prove that the last line I draw in my figure three forms an equilateral triangle since you have a triangle with two equal sides and angle 60. With that you have a kite quadrilateral (2 pairs of adj equal sides), x is just half of 110.

The MAIN thing is proving the equilateral triangle in figure three. 

BTW the kite figure proves what you requested.... "An alternative way to prove your argument is to show that your line segment is perpendicular to the "base" of the quadrilateral "

I am not sure why you keep saying I make assumptions when I did not in my last post. Don't read my previous posts as there are assumptions made when I was just writing without analyzing deeply. 

I am not sure how to explain because it is pretty clear... 

 

OK, it is clear that your previous posts were misleading as there were implied assumptions. That is why I couldn't follow what you were doing.

You had mentioned the "mirror quad" before. That's why I thought your argument hinged on that.

Also, I believe you meant "congruent" rather than "similar", which makes a big difference. Congruence means figures are identical in size and shape and may be superimposed with only translation, rotation and reflection. No scaling needed.

I now believe you have a perfectly valid proof, except you are not stating it very rigorously (which is what led to so much misunderstanding). No matter, I can help with this.

Let's first go through your argument.

It is completely possible to construct the isosceles triangle with base angle 35 deg (and apex 110 deg).

By itself, this is not helpful in the "general (no magic sum = 240)" case: 

282978315_geometrycounterexample.thumb.png.b290101846930823a5f9921eef58d48d.png

Here, with base angles of 40 deg (and apex 100), you still get your congruent isosceles triangle. But the crucial thing is the absence of the 60 degree middle angle.

Now, let's express your argument rigorously. It depends upon these steps for a tight chain of reasoning:

1) first draw the isosceles as you've done. Show that it is congruent with the other isosceles (you can use the SAS side-angle-side property).

2) next compute the middle angle, which turns out to be 60 degrees. 

3) deduce based on the previously established congruence that the third sides are also equal.

4) deduce that you now have two sides of an isosceles triangle with apex angle 60 degrees, making it equilateral.

5) deduce that the juxtaposed equilateral and isosceles triangles make a kite quadrilateral.

6) use the property of the kite quadrilateral to deduce that its diagonals are perpendicular.

7) use that to calculate the unknown angle.

If that was your chain of reasoning, I agree completely that is perfectly valid. In fact, it is more elegant than my original method.

I've written up in hand on my LCD tablet. Space constraints mean that it's a bit untidy, but every logical step is fleshed-out adequately (as should be for a proper proof).

IMG_20200112_162909.thumb.jpg.4e8a3056f4f6604ad8687a77d740bcea.jpg

I think we can both put this behind us now.😀

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1 hour ago, Turboflat4 said:

OK, it is clear that your previous posts were misleading as there were implied assumptions. That is why I couldn't follow what you were doing.

You had mentioned the "mirror quad" before. That's why I thought your argument hinged on that.

Also, I believe you meant "congruent" rather than "similar", which makes a big difference. Congruence means figures are identical in size and shape and may be superimposed with only translation, rotation and reflection. No scaling needed.

I now believe you have a perfectly valid proof, except you are not stating it very rigorously (which is what led to so much misunderstanding). No matter, I can help with this.

Let's first go through your argument.

It is completely possible to construct the isosceles triangle with base angle 35 deg (and apex 110 deg).

By itself, this is not helpful in the "general (no magic sum = 240)" case: 

282978315_geometrycounterexample.thumb.png.b290101846930823a5f9921eef58d48d.png

Here, with base angles of 40 deg (and apex 100), you still get your congruent isosceles triangle. But the crucial thing is the absence of the 60 degree middle angle.

Now, let's express your argument rigorously. It depends upon these steps for a tight chain of reasoning:

1) first draw the isosceles as you've done. Show that it is congruent with the other isosceles (you can use the SAS side-angle-side property).

2) next compute the middle angle, which turns out to be 60 degrees. 

3) deduce based on the previously established congruence that the third sides are also equal.

4) deduce that you now have two sides of an isosceles triangle with apex angle 60 degrees, making it equilateral.

5) deduce that the juxtaposed equilateral and isosceles triangles make a kite quadrilateral.

6) use the property of the kite quadrilateral to deduce that its diagonals are perpendicular.

7) use that to calculate the unknown angle.

If that was your chain of reasoning, I agree completely that is perfectly valid. In fact, it is more elegant than my original method.

I've written up in hand on my LCD tablet. Space constraints mean that it's a bit untidy, but every logical step is fleshed-out adequately (as should be for a proper proof).

IMG_20200112_162909.thumb.jpg.4e8a3056f4f6604ad8687a77d740bcea.jpg

I think we can both put this behind us now.😀

Ya , my earlier posts were a bit misleading. But once u prove the equilateral triangle, u can prove the mirror quad. That was my first solution. The kite was thought of after I tried to explain here...

this question took me quite a lot of thinking though... probably more than an hour on and off....

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1 hour ago, Wind30 said:

Ya , my earlier posts were a bit misleading. But once u prove the equilateral triangle, u can prove the mirror quad. That was my first solution. The kite was thought of after I tried to explain here...

this question took me quite a lot of thinking though... probably more than an hour on and off....

The kite is crucial. The fact that the long diagonal is the perpendicular bisector of the short one is that which permits mirroring to the hexagon. 

Good question.

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Hypersonic

Alright smart people, time to use your brain. Here is a question -

Prove that for any 6 consecutive integers there is a prime number that divides exactly one of them.

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Hypersonic
4 hours ago, Vid said:

Alright smart people, time to use your brain. Here is a question -

Prove that for any 6 consecutive integers there is a prime number that divides exactly one of them.

@Turboflat4 bro, need your help with ☝️ question 😁

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57 minutes ago, Vid said:

@Turboflat4 bro, need your help with ☝️ question 😁

Not trivial bro. Have to think about it.

As it stands, the statement is trivially false (0 is an integer but every prime divides it). But if you amend it slightly to 6 consecutive naturals, it might be true. Let me see if I can make progress later.

Where did this problem come from? 

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Hypersonic
28 minutes ago, Turboflat4 said:

Not trivial bro. Have to think about it.

As it stands, the statement is trivially false (0 is an integer but every prime divides it). But if you amend it slightly to 6 consecutive naturals, it might be true. Let me see if I can make progress later.

Where did this problem come from? 

My son gave it to me. He has been stuck on this qns for a month. He spent hours and hours on it but can’t solve. He got so frustrated with it today so I’m trying to get divine help 🙏😁

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LOL "divine" 😂

OK, @Vid, let's give it a shot, yah?

First restate the problem so it's correct. Among any 6 consecutive natural numbers (integers greater than or equal to one), prove that there exists a prime that divides exactly one of them.

Denote the natural number sequence: k, k+1, k+2, k+3, k+4, k+5

Case 1: k is even.

In this case, only need to study the odds k+1, k+3, k+5.

Case 1a: Let's say 3 divides k+1.

You're guaranteed that 3 will divide neither k+3 nor k+5 (since the difference between each of these integers and (k+1) is not a multiple of 3 -> the fancy way of saying this is "modulo 3").

Then k+3 has a minimum prime factor of 5.

If 5 is indeed a factor of k+3, then 5 divides no other number in the sequence (by reasoning modulo 5), so 5 is a prime factor unique to k+3 (done).

If 5 is not a factor of k+3, then k+3 must have a minimum prime factor of 7 or greater. There can only be a single multiple of 7 or more in a sequence of six consecutive naturals, so this prime factor will be unique to k+3, and we're done (done).

(Case 1a proven)

Case 1b: Let's say 3 does not divide k+1. Hence k+1 has a minimum prime factor of 5.

If 5 is indeed a factor of k+1, then none of the other numbers in the sequence can be a multiple of 5 (reasoning modulo 5). Hence 5 is a unique prime factor of k+1 (done).

If 5 is not a factor of k+1, then k+1 has a minimum prime factor of 7 or greater. Note that no other number in a sequence of six consecutive naturals can be a multiple of this prime, so the prime is unique to k+1 (done).

(Case 1b proven)

Case 2: k is odd.

In this case, only need to study the odds k, k+2, k+4

Case 2a: Let's say 3 divides k. Then 3 divides neither k+2 nor k+4 (reasoning modulo 3).

Then k+2 has a minimum prime factor of 5.

If 5 is indeed a factor of k+2, it is not a factor of any other number in the sequence, so it is a unique prime factor of k+2 (done).

If 5 is not a factor of k+2, then k+2 has a minimum prime factor of 7 or greater. In this case, no other number in the sequence can have this prime as a factor, so it will be a unique prime factor to k+2 (done).

(Case 2a proven)

Case 2b: Let's say 3 does not divide k. Then k has a minimum prime factor of 5.

Let's say 5 is a factor of k.

Note that in this sub-case, you *cannot* conclude that 5 is a factor unique to k, since k+5 will also be a multiple of 5. A different argument is needed.

First observe that neither k+2 nor k+4 will be a multiple of 5 (reasoning modulo 5).

Therefore, k+2 must have a prime factor of 3 or a prime factor greater than or equal to 7.

If the latter case holds, then the prime factor will be unique to k+2 (done).

If the former case holds, then k+4 cannot be a multiple of 3 (reasoning modulo 3). It must have a prime factor greater than or equal to 7, and this will be unique to k+4 (done).

(Case 2b proven).

All cases proven. (QED).

Not that tough after all, eh? 🤣

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Sorry @Vid, I just noticed a small gap in my proof above. Please use the following instead (I added a case zero just to handle the k=1 case, which otherwise interferes with the reasoning).

First restate the problem so it's correct. Among any 6 consecutive natural numbers (integers greater than or equal to one), prove that there exists a prime that divides exactly one of them.

Denote the natural number sequence: k, k+1, k+2, k+3, k+4, k+5

Case 0: k = 1.

In this case, the sequence is simple 1, 2, 3, 4, 5, 6 and it is trivially obvious that 5 is a prime that satisfies the requirement. (Case 0 proven).

Case 1: k is an even number greater than 1.

In this case, only need to study the odds k+1, k+3, k+5.

Case 1a: Let's say 3 divides k+1.

You're guaranteed that 3 will divide neither k+3 nor k+5 (since the difference between each of these integers and (k+1) is not a multiple of 3 -> the fancy way of saying this is "modulo 3").

Then k+3 has a minimum prime factor of 5.

If 5 is indeed a factor of k+3, then 5 divides no other number in the sequence (by reasoning modulo 5), so 5 is a prime factor unique to k+3 (done).

If 5 is not a factor of k+3, then k+3 must have a minimum prime factor of 7 or greater. There can only be a single multiple of 7 or more in a sequence of six consecutive naturals, so this prime factor will be unique to k+3 (done).

(Case 1a proven)

Case 1b: Let's say 3 does not divide k+1. Hence k+1 has a minimum prime factor of 5.

If 5 is indeed a factor of k+1, then none of the other numbers in the sequence can be a multiple of 5 (reasoning modulo 5). Hence 5 is a unique prime factor of k+1 (done).

If 5 is not a factor of k+1, then k+1 has a minimum prime factor of 7 or greater. Note that no other number in a sequence of six consecutive naturals can be a multiple of this prime, so the prime is unique to k+1 (done).

(Case 1b proven)

Case 2: k is an odd number greater than 1.

In this case, only need to study the odds k, k+2, k+4

Case 2a: Let's say 3 divides k. Then 3 divides neither k+2 nor k+4 (reasoning modulo 3).

Then k+2 has a minimum prime factor of 5.

If 5 is indeed a factor of k+2, it is not a factor of any other number in the sequence, so it is a unique prime factor of k+2 (done).

If 5 is not a factor of k+2, then k+2 has a minimum prime factor of 7 or greater. In this case, no other number in the sequence can have this prime as a factor, so it will be a unique prime factor to k+2 (done).

(Case 2a proven)

Case 2b: Let's say 3 does not divide k. Then k has a minimum prime factor of 5.

Let's say 5 is a factor of k.

Note that in this sub-case, you *cannot* conclude that 5 is a factor unique to k, since k+5 will also be a multiple of 5. A different argument is needed.

First observe that neither k+2 nor k+4 will be a multiple of 5 (reasoning modulo 5).

Therefore, k+2 must have a prime factor of 3 or a prime factor greater than or equal to 7.

If the latter case holds, then the prime factor will be unique to k+2 (done).

If the former case holds, then k+4 cannot be a multiple of 3 (reasoning modulo 3). It must have a prime factor greater than or equal to 7, and this will be unique to k+4 (done).

(Case 2b proven).

All cases proven. (QED).

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6 hours ago, Turboflat4 said:

 Case 2b: Let's say 3 does not divide k. Then k has a minimum prime factor of 5.

Let's say 5 is a factor of k.

Note that in this sub-case, you *cannot* conclude that 5 is a factor unique to k, since k+5 will also be a multiple of 5. A different argument is needed.

First observe that neither k+2 nor k+4 will be a multiple of 5 (reasoning modulo 5).

Therefore, k+2 must have a prime factor of 3 or a prime factor greater than or equal to 7.

If the latter case holds, then the prime factor will be unique to k+2 (done).

If the former case holds, then k+4 cannot be a multiple of 3 (reasoning modulo 3). It must have a prime factor greater than or equal to 7, and this will be unique to k+4 (done).

(Case 2b proven).

All cases proven. (QED).

Amend the above to:

Case 2b: Let's say 3 does not divide k. Then k has a minimum prime factor of 5.

Either 5 is a factor of k or 5 is not a factor of k.

If 5 is not a factor of k then k has to have a minimum prime factor of 7 or greater, which will be unique within the sequence (done). 

Let's say 5 is a factor of k.

Note that in this sub-case,... 

(rest is the same). 

Sorry @Vidplease add the bolded portion in to complete the proof. If your son can find any gaps, let me know (also means he's sibei smart). 😁

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Hypersonic
25 minutes ago, Turboflat4 said:

Amend the above to:

Case 2b: Let's say 3 does not divide k. Then k has a minimum prime factor of 5.

Either 5 is a factor of k or 5 is not a factor of k.

If 5 is not a factor of k then k has to have a minimum prime factor of 7 or greater, which will be unique within the sequence (done). 

Let's say 5 is a factor of k.

Note that in this sub-case,... 

(rest is the same). 

Sorry @Vidplease add the bolded portion in to complete the proof. If your son can find any gaps, let me know (also means he's sibei smart). 😁

You sibei smart... although I have no idea what everything is 🤣🤣

Let me try to digest some of them too. 😁

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