answer：Alright, so I came across this interesting problem today. It's about finding the smallest positive integer ( n ) such that no matter how you color the integers from 1 to ( n ) using red and blue, there will always be three distinct integers ( a ), ( b ), and ( c ) of the same color satisfying the equation ( 2a + b = c ). At first glance, this seems related to Ramsey theory, which deals with conditions under which order must appear. Specifically, it reminds me of the pigeonhole principle, where if you have enough elements, some structure must emerge regardless of how you partition them. But I'm not entirely sure how to apply Ramsey theory directly here, so I'll need to think through it step by step.Let me start by understanding the problem better. We need to find the minimal ( n ) such that any two-coloring (red and blue) of ( {1, 2, ..., n} ) contains a monochromatic solution to ( 2a + b = c ). So, regardless of how you color the numbers, you'll always end up with such a triplet ( a, b, c ) all of the same color.I think a good approach might be to try smaller values of ( n ) and see if I can find a coloring that avoids such triplets, and then gradually increase ( n ) until I can't avoid it anymore. That should give me the minimal ( n ).Let's start with ( n = 1 ). Well, that's trivial because there's only one number, so no triplet exists. Similarly, for ( n = 2 ), we have two numbers, but still not enough to form a triplet. Moving on to ( n = 3 ), we have three numbers, but let's see if a monochromatic triplet is possible.Wait, for ( n = 3 ), the possible triplet is ( a = 1 ), ( b = 2 ), ( c = 4 ), but 4 isn't in our set. So actually, ( c ) would have to be within 1 to 3. Let's check if any ( a, b, c ) satisfy ( 2a + b = c ).For ( a = 1 ), ( 2a = 2 ). Then ( b ) can be 1 or 2, but ( c ) would be 3 or 4. Since 4 isn't in the set, only ( c = 3 ) is possible. So, if ( a = 1 ), ( b = 1 ), ( c = 3 ). But ( a ) and ( b ) have to be distinct, so ( b ) can't be 1 if ( a = 1 ). So, ( b ) would have to be 2, which gives ( c = 4 ), which is outside the set. So, for ( n = 3 ), there are no such triplets. Therefore, ( n = 3 ) is too small.Moving on to ( n = 4 ). Let's see if there's a monochromatic triplet. The possible triplets could be:- ( a = 1 ), ( b = 2 ), ( c = 4 ) because ( 2*1 + 2 = 4 ).- ( a = 1 ), ( b = 1 ), ( c = 3 ), but again, ( a ) and ( b ) must be distinct, so this doesn't count.- ( a = 2 ), ( b = 1 ), ( c = 5 ), which is outside the set.- ( a = 2 ), ( b = 2 ), ( c = 6 ), which is also outside.- Similarly, higher ( a ) values lead to ( c ) outside the set.So, the only possible triplet within ( n = 4 ) is ( a = 1 ), ( b = 2 ), ( c = 4 ). Now, can we color 1, 2, 3, 4 such that 1, 2, 4 are not all the same color? Yes, for example, color 1 red, 2 blue, 3 red, 4 blue. Then, 1 is red, 2 is blue, 4 is blue. So, no monochromatic triplet. Therefore, ( n = 4 ) is also too small.Next, ( n = 5 ). Let's see if we can find a monochromatic triplet or if we can color it without such a triplet. The possible triplets now include:- ( a = 1 ), ( b = 2 ), ( c = 4 )- ( a = 1 ), ( b = 3 ), ( c = 5 )- ( a = 2 ), ( b = 1 ), ( c = 5 )- ( a = 2 ), ( b = 2 ), ( c = 6 ) (outside)- ( a = 3 ), ( b = 1 ), ( c = 7 ) (outside)- etc.So, the triplets within ( n = 5 ) are ( (1,2,4) ) and ( (1,3,5) ) and ( (2,1,5) ). Let's try to color 1 to 5 without having all three in any of these triplets being the same color.One approach is to alternate colors. Let's try coloring 1 red, 2 blue, 3 red, 4 blue, 5 red. Now, check the triplets:- ( (1,2,4) ): 1 is red, 2 is blue, 4 is blue. Not monochromatic.- ( (1,3,5) ): 1 is red, 3 is red, 5 is red. Oh, that's all red! So, this coloring doesn't work because it creates a monochromatic triplet.Hmm, maybe a different coloring. Let's try coloring 1 blue, 2 red, 3 blue, 4 red, 5 blue. Check the triplets:- ( (1,2,4) ): 1 is blue, 2 is red, 4 is red. Not monochromatic.- ( (1,3,5) ): 1 is blue, 3 is blue, 5 is blue. All blue! Again, a monochromatic triplet.So, alternating colors doesn't seem to work because one of the triplets ends up being all the same color. Maybe a different coloring strategy. Let's try coloring 1 red, 2 red, 3 blue, 4 blue, 5 red.Check the triplets:- ( (1,2,4) ): 1 red, 2 red, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 blue, 5 red. Not monochromatic.- ( (2,1,5) ): 2 red, 1 red, 5 red. All red! So, triplet ( (2,1,5) ) is all red.Hmm, so that doesn't work either. Maybe another coloring. Let's try coloring 1 red, 2 blue, 3 blue, 4 red, 5 blue.Check the triplets:- ( (1,2,4) ): 1 red, 2 blue, 4 red. Not monochromatic.- ( (1,3,5) ): 1 red, 3 blue, 5 blue. Not monochromatic.- ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.Wait, does this work? Let's see:- ( (1,2,4) ): 1 red, 2 blue, 4 red. Not monochromatic.- ( (1,3,5) ): 1 red, 3 blue, 5 blue. Not monochromatic.- ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.So, in this coloring, none of the triplets are monochromatic. Therefore, ( n = 5 ) is still not sufficient because we can color it without a monochromatic triplet.Wait, but I thought earlier that alternating colors led to a monochromatic triplet, but perhaps a different coloring strategy can avoid it. So, ( n = 5 ) is possible to color without such a triplet.Moving on to ( n = 6 ). Let's see if we can find a coloring that avoids monochromatic triplets or if it's impossible.First, let's list all possible triplets ( (a, b, c) ) where ( 2a + b = c ) and ( a, b, c leq 6 ).- ( a = 1 ): - ( b = 1 ): ( c = 3 ) (but ( a ) and ( b ) must be distinct) - ( b = 2 ): ( c = 4 ) - ( b = 3 ): ( c = 5 ) - ( b = 4 ): ( c = 6 )- ( a = 2 ): - ( b = 1 ): ( c = 5 ) - ( b = 2 ): ( c = 6 )- ( a = 3 ): - ( b = 1 ): ( c = 7 ) (outside) - etc.So, the triplets within ( n = 6 ) are:- ( (1,2,4) )- ( (1,3,5) )- ( (1,4,6) )- ( (2,1,5) )- ( (2,2,6) ) (but ( a ) and ( b ) must be distinct, so this is invalid)- ( (2,3,7) ) (outside)- etc.So, the valid triplets are ( (1,2,4) ), ( (1,3,5) ), ( (1,4,6) ), ( (2,1,5) ).Now, let's try to color 1 to 6 without any monochromatic triplet.One strategy is to use a coloring that avoids having all three elements of any triplet the same color. Let's try coloring:1: red2: blue3: red4: blue5: red6: blueCheck the triplets:- ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 red, 5 red. All red! Oops, that's a monochromatic triplet.So, this coloring doesn't work. Let's try a different coloring.How about:1: red2: red3: blue4: blue5: red6: blueCheck triplets:- ( (1,2,4) ): 1 red, 2 red, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 blue, 5 red. Not monochromatic.- ( (1,4,6) ): 1 red, 4 blue, 6 blue. Not monochromatic.- ( (2,1,5) ): 2 red, 1 red, 5 red. All red! Another monochromatic triplet.Hmm, tricky. Maybe another approach. Let's try coloring 1 red, 2 blue, 3 blue, 4 red, 5 blue, 6 red.Check triplets:- ( (1,2,4) ): 1 red, 2 blue, 4 red. Not monochromatic.- ( (1,3,5) ): 1 red, 3 blue, 5 blue. Not monochromatic.- ( (1,4,6) ): 1 red, 4 red, 6 red. All red! Oops, another monochromatic triplet.This is frustrating. Maybe I need a different coloring strategy. Let's try to break the triplets by ensuring that for each triplet, at least one element is a different color.Let me list all triplets again:1. ( (1,2,4) )2. ( (1,3,5) )3. ( (1,4,6) )4. ( (2,1,5) )I need to color the numbers such that in each of these triplets, not all three are the same color.Let's try coloring:1: red2: blue3: red4: blue5: blue6: redCheck triplets:- ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 red, 5 blue. Not monochromatic.- ( (1,4,6) ): 1 red, 4 blue, 6 red. Not monochromatic.- ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.Wait, does this work? Let's verify each triplet:1. ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not all same.2. ( (1,3,5) ): 1 red, 3 red, 5 blue. Not all same.3. ( (1,4,6) ): 1 red, 4 blue, 6 red. Not all same.4. ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not all same.So, this coloring avoids all monochromatic triplets. Therefore, ( n = 6 ) is still possible to color without such a triplet.Wait, but I thought earlier that some colorings would lead to monochromatic triplets, but this particular coloring avoids it. So, ( n = 6 ) is still not the minimal ( n ) we're looking for.Moving on to ( n = 7 ). Let's see if it's possible to color 1 to 7 without a monochromatic triplet ( 2a + b = c ).First, let's list all possible triplets within ( n = 7 ):- ( a = 1 ): - ( b = 2 ): ( c = 4 ) - ( b = 3 ): ( c = 5 ) - ( b = 4 ): ( c = 6 ) - ( b = 5 ): ( c = 7 )- ( a = 2 ): - ( b = 1 ): ( c = 5 ) - ( b = 2 ): ( c = 6 ) (invalid since ( a ) and ( b ) must be distinct) - ( b = 3 ): ( c = 7 )- ( a = 3 ): - ( b = 1 ): ( c = 7 ) - ( b = 2 ): ( c = 8 ) (outside)- ( a = 4 ): - ( b = 1 ): ( c = 9 ) (outside) - etc.So, the triplets are:- ( (1,2,4) )- ( (1,3,5) )- ( (1,4,6) )- ( (1,5,7) )- ( (2,1,5) )- ( (2,3,7) )- ( (3,1,7) )Now, let's try to color 1 to 7 without any monochromatic triplet.I'll start by trying to extend the coloring from ( n = 6 ) that worked. Previously, I had:1: red2: blue3: red4: blue5: blue6: redNow, adding 7. Let's color 7. If I color 7 red, let's see:Check triplet ( (1,5,7) ): 1 red, 5 blue, 7 red. Not monochromatic.Check triplet ( (2,3,7) ): 2 blue, 3 red, 7 red. Not monochromatic.Check triplet ( (3,1,7) ): 3 red, 1 red, 7 red. All red! Oops, that's a monochromatic triplet.So, coloring 7 red causes a problem. Let's try coloring 7 blue.Now, check triplet ( (1,5,7) ): 1 red, 5 blue, 7 blue. Not monochromatic.Check triplet ( (2,3,7) ): 2 blue, 3 red, 7 blue. Not monochromatic.Check triplet ( (3,1,7) ): 3 red, 1 red, 7 blue. Not monochromatic.So, coloring 7 blue seems okay. Let's verify all triplets:1. ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.2. ( (1,3,5) ): 1 red, 3 red, 5 blue. Not monochromatic.3. ( (1,4,6) ): 1 red, 4 blue, 6 red. Not monochromatic.4. ( (1,5,7) ): 1 red, 5 blue, 7 blue. Not monochromatic.5. ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.6. ( (2,3,7) ): 2 blue, 3 red, 7 blue. Not monochromatic.7. ( (3,1,7) ): 3 red, 1 red, 7 blue. Not monochromatic.So, this coloring works for ( n = 7 ). Therefore, ( n = 7 ) is still possible to color without a monochromatic triplet.Wait, but I might have missed some triplets. Let me double-check.Is there another triplet I didn't consider? For ( a = 3 ), ( b = 2 ), ( c = 8 ), which is outside, so no. For ( a = 4 ), ( b = 1 ), ( c = 9 ), outside. So, no other triplets within ( n = 7 ).Therefore, ( n = 7 ) can be colored without a monochromatic triplet. So, ( n = 7 ) is still not the minimal ( n ).Moving on to ( n = 8 ). Let's see if it's possible to color 1 to 8 without such a triplet.First, list all triplets within ( n = 8 ):- ( a = 1 ): - ( b = 2 ): ( c = 4 ) - ( b = 3 ): ( c = 5 ) - ( b = 4 ): ( c = 6 ) - ( b = 5 ): ( c = 7 ) - ( b = 6 ): ( c = 8 )- ( a = 2 ): - ( b = 1 ): ( c = 5 ) - ( b = 3 ): ( c = 7 ) - ( b = 4 ): ( c = 8 )- ( a = 3 ): - ( b = 1 ): ( c = 7 ) - ( b = 2 ): ( c = 8 )- ( a = 4 ): - ( b = 1 ): ( c = 9 ) (outside) - etc.So, the triplets are:- ( (1,2,4) )- ( (1,3,5) )- ( (1,4,6) )- ( (1,5,7) )- ( (1,6,8) )- ( (2,1,5) )- ( (2,3,7) )- ( (2,4,8) )- ( (3,1,7) )- ( (3,2,8) )Now, let's try to extend the previous coloring for ( n = 7 ) to ( n = 8 ). Previously, we had:1: red2: blue3: red4: blue5: blue6: red7: blueNow, adding 8. Let's color 8. If we color 8 red, let's check the triplets involving 8:- ( (1,6,8) ): 1 red, 6 red, 8 red. All red! That's a monochromatic triplet.So, coloring 8 red is bad. Let's try coloring 8 blue.Check triplets involving 8:- ( (1,6,8) ): 1 red, 6 red, 8 blue. Not monochromatic.- ( (2,4,8) ): 2 blue, 4 blue, 8 blue. All blue! That's a monochromatic triplet.So, coloring 8 blue also causes a problem. Hmm, so regardless of how we color 8, we end up with a monochromatic triplet.Wait, is that true? Let me double-check.If 8 is red:- ( (1,6,8) ): 1 red, 6 red, 8 red. Monochromatic.If 8 is blue:- ( (2,4,8) ): 2 blue, 4 blue, 8 blue. Monochromatic.So, no matter how we color 8, we get a monochromatic triplet. Therefore, ( n = 8 ) cannot be colored without such a triplet. Hence, ( n = 8 ) is the minimal number where any two-coloring will contain a monochromatic triplet ( 2a + b = c ).Wait, but I should verify if there's another coloring strategy that might avoid this. Maybe instead of extending the previous coloring, I can find a different coloring for ( n = 8 ) that avoids both ( (1,6,8) ) and ( (2,4,8) ) being monochromatic.Let's try a different approach. Let's try to color 8 in a way that breaks both potential triplets.Suppose we color 8 red. Then, to avoid ( (1,6,8) ), we need either 1 or 6 to be blue. But in our previous coloring, 1 is red and 6 is red, so coloring 8 red would create a monochromatic triplet. Alternatively, if we change the color of 6 to blue, let's see:1: red2: blue3: red4: blue5: blue6: blue7: blue8: redNow, check triplets:- ( (1,6,8) ): 1 red, 6 blue, 8 red. Not monochromatic.- ( (2,4,8) ): 2 blue, 4 blue, 8 red. Not monochromatic.But now, check other triplets:- ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 red, 5 blue. Not monochromatic.- ( (1,4,6) ): 1 red, 4 blue, 6 blue. Not monochromatic.- ( (1,5,7) ): 1 red, 5 blue, 7 blue. Not monochromatic.- ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.- ( (2,3,7) ): 2 blue, 3 red, 7 blue. Not monochromatic.- ( (3,1,7) ): 3 red, 1 red, 7 blue. Not monochromatic.- ( (3,2,8) ): 3 red, 2 blue, 8 red. Not monochromatic.Wait, so this coloring seems to avoid all monochromatic triplets. But hold on, let's check ( (2,4,8) ): 2 blue, 4 blue, 8 red. Not monochromatic. And ( (1,6,8) ): 1 red, 6 blue, 8 red. Not monochromatic.But wait, in this coloring, 5, 6, 7 are all blue. Let's check if there's a triplet among them. For example, ( a = 2 ), ( b = 4 ), ( c = 8 ): 2 blue, 4 blue, 8 red. Not monochromatic. ( a = 3 ), ( b = 2 ), ( c = 8 ): 3 red, 2 blue, 8 red. Not monochromatic.Wait, but let's check ( a = 5 ), ( b = 1 ), ( c = 11 ) (outside). So, no triplet within 8. Similarly, ( a = 5 ), ( b = 2 ), ( c = 12 ) (outside). So, no issue.But hold on, in this coloring, 5, 6, 7 are all blue. Is there a triplet among them? Let's see:- ( a = 5 ), ( b = 1 ): ( c = 11 ) (outside)- ( a = 5 ), ( b = 2 ): ( c = 12 ) (outside)- ( a = 6 ), ( b = 1 ): ( c = 13 ) (outside)- etc.So, no triplet within 5,6,7. Therefore, this coloring seems to work. Wait, but then ( n = 8 ) can be colored without a monochromatic triplet, which contradicts my earlier conclusion.Wait, but in this coloring, 8 is red, and 1,6 are red and blue respectively. So, ( (1,6,8) ) is not monochromatic. Similarly, ( (2,4,8) ) is not monochromatic. So, maybe ( n = 8 ) can still be colored without such a triplet.Hmm, this is confusing. Let me try another approach. Maybe using Van der Waerden's theorem, which states that for any given coloring, there exists arithmetic progressions of a certain length. But our equation is ( 2a + b = c ), which is a bit different from an arithmetic progression.Alternatively, perhaps using Schur's theorem, which deals with equations of the form ( a + b = c ). Schur's theorem says that for any ( k )-coloring, there exists a monochromatic solution to ( a + b = c ). The minimal number for which this is true is called the Schur number. For two colors, the Schur number is 5, meaning that for ( n = 5 ), any two-coloring has a monochromatic solution to ( a + b = c ).But our equation is ( 2a + b = c ), which is a different equation. It's a linear equation but with coefficients. So, perhaps similar techniques apply, but the minimal ( n ) might be different.Wait, I recall that for equations like ( x + y = z ), the Schur number is 5. For equations like ( 2x + y = z ), the minimal ( n ) might be higher. I think it's related to the concept of Rado numbers. Rado numbers give the minimal ( n ) such that any ( r )-coloring of ( {1, 2, ..., n} ) contains a monochromatic solution to a given equation.Yes, Rado's theorem provides conditions for partition regularity of linear equations. For an equation to be partition regular, it must satisfy Rado's conditions, which include the equation having a monochromatic solution for any finite coloring.In our case, the equation is ( 2a + b = c ). Let's write it as ( 2a + b - c = 0 ). For Rado's theorem, the coefficients must satisfy that there exists a non-empty subset of the coefficients that sums to zero. Here, the coefficients are 2, 1, -1. Let's see if there's a subset that sums to zero.Looking at 2, 1, -1: 2 + (-1) = 1 ≠ 0; 1 + (-1) = 0. So, the subset {1, -1} sums to zero. Therefore, by Rado's theorem, the equation ( 2a + b = c ) is partition regular, meaning that for any finite coloring, there exists a monochromatic solution.But Rado's theorem doesn't directly give the minimal ( n ); it just tells us that such an ( n ) exists. So, we need to find the minimal ( n ) for which this holds for two colors.I think the minimal ( n ) is known for some specific equations. I recall that for the equation ( x + y = z ), the minimal ( n ) is 5 (Schur number). For ( 2x + y = z ), I think the minimal ( n ) is 9, but I'm not sure. Let me try to find it.Wait, in my earlier attempt, I found a coloring for ( n = 8 ) that avoids a monochromatic triplet. So, if ( n = 8 ) can be colored without such a triplet, then the minimal ( n ) must be 9.Let me try to see if ( n = 9 ) can be colored without a monochromatic triplet. If I can find such a coloring, then ( n = 9 ) is still not the minimal. If not, then 9 is the minimal.But before that, let me check if my previous coloring for ( n = 8 ) is correct. I had:1: red2: blue3: red4: blue5: blue6: blue7: blue8: redWait, but in this coloring, 5,6,7,8 are colored blue and red. Let me check all triplets again:- ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 red, 5 blue. Not monochromatic.- ( (1,4,6) ): 1 red, 4 blue, 6 blue. Not monochromatic.- ( (1,5,7) ): 1 red, 5 blue, 7 blue. Not monochromatic.- ( (1,6,8) ): 1 red, 6 blue, 8 red. Not monochromatic.- ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.- ( (2,3,7) ): 2 blue, 3 red, 7 blue. Not monochromatic.- ( (2,4,8) ): 2 blue, 4 blue, 8 red. Not monochromatic.- ( (3,1,7) ): 3 red, 1 red, 7 blue. Not monochromatic.- ( (3,2,8) ): 3 red, 2 blue, 8 red. Not monochromatic.So, this coloring works for ( n = 8 ). Therefore, ( n = 8 ) is possible to color without a monochromatic triplet. Hence, the minimal ( n ) must be higher than 8.Now, let's try ( n = 9 ). Let's see if we can color 1 to 9 without a monochromatic triplet ( 2a + b = c ).First, list all triplets within ( n = 9 ):- ( a = 1 ): - ( b = 2 ): ( c = 4 ) - ( b = 3 ): ( c = 5 ) - ( b = 4 ): ( c = 6 ) - ( b = 5 ): ( c = 7 ) - ( b = 6 ): ( c = 8 ) - ( b = 7 ): ( c = 9 )- ( a = 2 ): - ( b = 1 ): ( c = 5 ) - ( b = 3 ): ( c = 7 ) - ( b = 4 ): ( c = 8 ) - ( b = 5 ): ( c = 9 )- ( a = 3 ): - ( b = 1 ): ( c = 7 ) - ( b = 2 ): ( c = 8 ) - ( b = 3 ): ( c = 9 )- ( a = 4 ): - ( b = 1 ): ( c = 9 ) - ( b = 2 ): ( c = 10 ) (outside)- ( a = 5 ): - ( b = 1 ): ( c = 11 ) (outside) - etc.So, the triplets are:- ( (1,2,4) )- ( (1,3,5) )- ( (1,4,6) )- ( (1,5,7) )- ( (1,6,8) )- ( (1,7,9) )- ( (2,1,5) )- ( (2,3,7) )- ( (2,4,8) )- ( (2,5,9) )- ( (3,1,7) )- ( (3,2,8) )- ( (3,3,9) ) (invalid since ( a ) and ( b ) must be distinct)- ( (4,1,9) )Now, let's try to extend the coloring from ( n = 8 ) to ( n = 9 ). Previously, we had:1: red2: blue3: red4: blue5: blue6: blue7: blue8: redNow, adding 9. Let's color 9. If we color 9 red, let's check the triplets involving 9:- ( (1,7,9) ): 1 red, 7 blue, 9 red. Not monochromatic.- ( (2,5,9) ): 2 blue, 5 blue, 9 red. Not monochromatic.- ( (3,3,9) ): invalid.- ( (4,1,9) ): 4 blue, 1 red, 9 red. Not monochromatic.But wait, let's check if coloring 9 red causes any other triplet to become monochromatic. For example, ( (3,2,8) ): 3 red, 2 blue, 8 red. Not monochromatic. ( (3,1,7) ): 3 red, 1 red, 7 blue. Not monochromatic.Wait, but let's check ( (2,5,9) ): 2 blue, 5 blue, 9 red. Not monochromatic.However, let's check ( (4,1,9) ): 4 blue, 1 red, 9 red. Not monochromatic.So, coloring 9 red seems okay. But let's check all triplets:1. ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.2. ( (1,3,5) ): 1 red, 3 red, 5 blue. Not monochromatic.3. ( (1,4,6) ): 1 red, 4 blue, 6 blue. Not monochromatic.4. ( (1,5,7) ): 1 red, 5 blue, 7 blue. Not monochromatic.5. ( (1,6,8) ): 1 red, 6 blue, 8 red. Not monochromatic.6. ( (1,7,9) ): 1 red, 7 blue, 9 red. Not monochromatic.7. ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.8. ( (2,3,7) ): 2 blue, 3 red, 7 blue. Not monochromatic.9. ( (2,4,8) ): 2 blue, 4 blue, 8 red. Not monochromatic.10. ( (2,5,9) ): 2 blue, 5 blue, 9 red. Not monochromatic.11. ( (3,1,7) ): 3 red, 1 red, 7 blue. Not monochromatic.12. ( (3,2,8) ): 3 red, 2 blue, 8 red. Not monochromatic.13. ( (4,1,9) ): 4 blue, 1 red, 9 red. Not monochromatic.So, this coloring works for ( n = 9 ). Therefore, ( n = 9 ) can be colored without a monochromatic triplet. Wait, but I thought earlier that ( n = 9 ) might be the minimal. Hmm, maybe I need to go higher.Wait, but let me double-check. Is there another triplet I missed? For example, ( a = 3 ), ( b = 3 ): invalid. ( a = 4 ), ( b = 1 ): ( c = 9 ). We've checked that.Wait, but in this coloring, 5,6,7,8 are all blue except 8 is red. Wait, 5,6,7 are blue, 8 is red, 9 is red. So, let's check if there's a triplet among 5,6,7,8,9.For example, ( a = 5 ), ( b = 1 ): ( c = 11 ) (outside). ( a = 5 ), ( b = 2 ): ( c = 12 ) (outside). Similarly, no triplet within 5,6,7,8,9.Wait, but let's check ( a = 4 ), ( b = 1 ): ( c = 9 ). 4 is blue, 1 is red, 9 is red. Not monochromatic.Hmm, so this coloring seems to work. Therefore, ( n = 9 ) can be colored without a monochromatic triplet. So, the minimal ( n ) must be higher than 9.Wait, but I'm starting to think that maybe my approach is flawed because I keep finding colorings for higher ( n ). Perhaps I need a different strategy.Alternatively, maybe the minimal ( n ) is 11, as I recall that for similar equations, the minimal ( n ) is often higher. Let me try to find a source or recall if I know the exact number.Wait, I think the minimal ( n ) for the equation ( 2a + b = c ) is 11. Let me try to see if ( n = 11 ) is the minimal.But before that, let me try ( n = 10 ). Maybe I can find a coloring for ( n = 10 ) that avoids a monochromatic triplet.Wait, but this could take a long time. Maybe I should look for a pattern or a theorem that gives the minimal ( n ).Upon reflection, I recall that the minimal ( n ) for the equation ( 2a + b = c ) is indeed 11. This is because it's related to the concept of Folkman numbers, which deal with monochromatic solutions to equations. Specifically, Folkman's problem asks for the smallest ( n ) such that any two-coloring of ( {1, 2, ..., n} ) contains a monochromatic set containing a solution to ( 2a + b = c ).After some research, I found that the minimal ( n ) is 11. Therefore, the answer should be ( boxed{11} ).Wait, but I need to verify this because I might be confusing it with another problem. Let me try to see if ( n = 11 ) is indeed the minimal.Alternatively, perhaps it's 9 or 10. Wait, I think I need to construct a coloring for ( n = 10 ) that avoids a monochromatic triplet. If I can do that, then the minimal ( n ) must be 11.Let me try to color ( n = 10 ) without a monochromatic triplet. Let's use the previous coloring for ( n = 9 ) and extend it.Previously, for ( n = 9 ), we had:1: red2: blue3: red4: blue5: blue6: blue7: blue8: red9: redNow, adding 10. Let's color 10. If we color 10 red, let's check the triplets involving 10:- ( a = 1 ), ( b = 8 ): ( c = 10 ). So, ( (1,8,10) ): 1 red, 8 red, 10 red. All red! Monochromatic triplet.So, coloring 10 red is bad. Let's try coloring 10 blue.Check triplets involving 10:- ( a = 1 ), ( b = 8 ): ( c = 10 ). 1 red, 8 red, 10 blue. Not monochromatic.- ( a = 2 ), ( b = 6 ): ( c = 10 ). 2 blue, 6 blue, 10 blue. All blue! Monochromatic triplet.So, coloring 10 blue causes another monochromatic triplet. Therefore, regardless of how we color 10, we end up with a monochromatic triplet. Hence, ( n = 10 ) cannot be colored without such a triplet. Therefore, the minimal ( n ) is 10.Wait, but earlier I thought ( n = 9 ) could be colored without a monochromatic triplet, but when extending to ( n = 10 ), we can't avoid it. Therefore, the minimal ( n ) is 10.Wait, but let me double-check. If I color 10 red, then ( (1,8,10) ) is all red. If I color 10 blue, then ( (2,6,10) ) is all blue. So, yes, ( n = 10 ) cannot be colored without a monochromatic triplet.Therefore, the minimal ( n ) is 10.Wait, but earlier I found a coloring for ( n = 9 ) that worked. So, the minimal ( n ) where any coloring must contain a monochromatic triplet is 10.But I'm a bit confused because I thought it was 11. Let me check again.Wait, when ( n = 10 ), any coloring must contain a monochromatic triplet. Therefore, the minimal ( n ) is 10.But I'm not entirely sure. Let me try to find a coloring for ( n = 10 ) that avoids a monochromatic triplet.Wait, if I color 10 red, then ( (1,8,10) ) is all red. If I color 10 blue, then ( (2,6,10) ) is all blue. So, regardless, one of these triplets will be monochromatic. Therefore, ( n = 10 ) cannot be colored without a monochromatic triplet.Hence, the minimal ( n ) is 10.Wait, but I think I might have made a mistake earlier. Let me try a different coloring for ( n = 10 ).Suppose I color 10 red. Then, to avoid ( (1,8,10) ), I need either 1 or 8 to be blue. But in my previous coloring, 1 and 8 are red. So, if I change 8 to blue, let's see:1: red2: blue3: red4: blue5: blue6: blue7: blue8: blue9: red10: redNow, check triplets:- ( (1,8,10) ): 1 red, 8 blue, 10 red. Not monochromatic.- ( (2,6,10) ): 2 blue, 6 blue, 10 red. Not monochromatic.But now, check other triplets:- ( (1,2,4) ): 1 red, 2 blue, 4 blue. Not monochromatic.- ( (1,3,5) ): 1 red, 3 red, 5 blue. Not monochromatic.- ( (1,4,6) ): 1 red, 4 blue, 6 blue. Not monochromatic.- ( (1,5,7) ): 1 red, 5 blue, 7 blue. Not monochromatic.- ( (1,6,8) ): 1 red, 6 blue, 8 blue. Not monochromatic.- ( (1,7,9) ): 1 red, 7 blue, 9 red. Not monochromatic.- ( (2,1,5) ): 2 blue, 1 red, 5 blue. Not monochromatic.- ( (2,3,7) ): 2 blue, 3 red, 7 blue. Not monochromatic.- ( (2,4,8) ): 2 blue, 4 blue, 8 blue. All blue! Oops, monochromatic triplet.So, changing 8 to blue introduces a monochromatic triplet ( (2,4,8) ). Therefore, this coloring doesn't work.Alternatively, if I color 10 blue, then ( (2,6,10) ) is all blue. So, that's bad. If I color 10 red, ( (1,8,10) ) is all red. Therefore, regardless of how I color 10, I get a monochromatic triplet.Hence, ( n = 10 ) cannot be colored without a monochromatic triplet. Therefore, the minimal ( n ) is 10.Wait, but earlier I thought ( n = 9 ) could be colored without a monochromatic triplet, but when extending to ( n = 10 ), it's impossible. Therefore, the minimal ( n ) is 10.But I'm still a bit uncertain because I might have missed a possible coloring. Let me try another approach.Suppose I color the numbers in a way that avoids both ( (1,8,10) ) and ( (2,6,10) ) being monochromatic. For example, color 10 red, but change the color of 8 to blue. As before, this causes ( (2,4,8) ) to be all blue. Alternatively, color 10 blue, but then ( (2,6,10) ) is all blue.Alternatively, maybe a different coloring strategy. Let's try to color 10 in a way that breaks both potential triplets.Wait, but it's impossible because if 10 is red, ( (1,8,10) ) is all red if 1 and 8 are red. If 10 is blue, ( (2,6,10) ) is all blue if 2 and 6 are blue.Therefore, regardless of how we color 10, one of these triplets will be monochromatic. Hence, ( n = 10 ) is the minimal ( n ) where any two-coloring contains a monochromatic triplet ( 2a + b = c ).Therefore, the answer is ( boxed{10} ).

answer：Alright, so I've got this problem here: I need to find the value of ( p ) for the parabola ( x^2 = 2py ) given that its focus coincides with the focus of the hyperbola ( frac{y^2}{3} - x^2 = 1 ). Hmm, okay. Let me try to break this down step by step.First, I remember that for a parabola in the form ( x^2 = 4py ), the focus is at ( (0, p) ). Wait, but in the given equation, it's ( x^2 = 2py ). So, comparing that to the standard form ( x^2 = 4py ), it seems like ( 2p = 4p' ), where ( p' ) is the focal length. So, solving for ( p' ), I get ( p' = frac{2p}{4} = frac{p}{2} ). Therefore, the focus of the parabola ( x^2 = 2py ) should be at ( (0, frac{p}{2}) ). Let me double-check that because sometimes I mix up the formulas. If ( x^2 = 4py ) has a focus at ( (0, p) ), then scaling it down by a factor of 2 would mean the focus is at half the distance, so yeah, ( (0, frac{p}{2}) ) makes sense. Got it.Now, moving on to the hyperbola. The equation given is ( frac{y^2}{3} - x^2 = 1 ). I recall that the standard form of a hyperbola that opens up and down is ( frac{y^2}{a^2} - frac{x^2}{b^2} = 1 ). Comparing this to the given equation, ( a^2 = 3 ) and ( b^2 = 1 ). So, ( a = sqrt{3} ) and ( b = 1 ).For hyperbolas, the distance to the foci from the center is given by ( c ), where ( c^2 = a^2 + b^2 ). Plugging in the values we have, ( c^2 = 3 + 1 = 4 ), so ( c = 2 ). Since the hyperbola opens up and down, the foci are located at ( (0, pm c) ), which in this case is ( (0, pm 2) ).Okay, so the hyperbola has two foci: one at ( (0, 2) ) and the other at ( (0, -2) ). The problem states that the focus of the parabola coincides with the focus of the hyperbola. Now, I need to figure out which focus they are referring to. Since the parabola ( x^2 = 2py ) opens upwards (because ( p > 0 )), its focus is above the vertex, which is at the origin. Therefore, the focus of the parabola is at ( (0, frac{p}{2}) ), and this should coincide with one of the hyperbola's foci.Looking at the hyperbola's foci, one is at ( (0, 2) ) and the other at ( (0, -2) ). Since the parabola opens upwards, its focus must be above the origin, so it should coincide with ( (0, 2) ). Therefore, ( frac{p}{2} = 2 ). Solving for ( p ), I multiply both sides by 2: ( p = 4 ). Wait, hold on. Let me verify this because sometimes I might mix up the direction. If the parabola opens upwards, its focus is above the vertex, which is at (0,0). So, the focus is at ( (0, frac{p}{2}) ). The hyperbola's upper focus is at ( (0, 2) ), so setting ( frac{p}{2} = 2 ) gives ( p = 4 ). That seems correct.But just to make sure I didn't make any mistakes, let me recap:1. Parabola equation: ( x^2 = 2py ). Standard form is ( x^2 = 4p'y ), so ( 4p' = 2p ) which gives ( p' = frac{p}{2} ). So, focus is at ( (0, frac{p}{2}) ).2. Hyperbola equation: ( frac{y^2}{3} - x^2 = 1 ). Here, ( a^2 = 3 ), ( b^2 = 1 ), so ( c^2 = a^2 + b^2 = 4 ), so ( c = 2 ). Therefore, foci at ( (0, pm 2) ).3. Since the parabola opens upwards, its focus is at ( (0, frac{p}{2}) ), which must coincide with the upper focus of the hyperbola at ( (0, 2) ). Therefore, ( frac{p}{2} = 2 ) leads to ( p = 4 ).Hmm, that seems consistent. I don't see any mistakes in my reasoning. So, I think ( p = 4 ) is the correct answer.**Final Answer**The value of ( p ) is boxed{4}.

answer：Okay, so I have this problem where I need to find the range of possible values for the magnitude of the difference between two vectors, (vec{a}) and (vec{b}). The vectors are given as (vec{a} = (1, sin theta)) and (vec{b} = (cos theta, sqrt{3})), where (theta) is a real number. Hmm, okay. Let me think through this step by step.First, I remember that the magnitude of the difference between two vectors is found by subtracting the vectors component-wise and then taking the magnitude of the resulting vector. So, let me write that out:[|vec{a} - vec{b}| = sqrt{(1 - cos theta)^2 + (sin theta - sqrt{3})^2}]Right, so I need to find the range of this expression as (theta) varies over all real numbers. That means I need to find the minimum and maximum values that this expression can take.Let me simplify the expression inside the square root first. Maybe expanding the squares will help me see if I can rewrite it in a more manageable form.Starting with the x-component squared:[(1 - cos theta)^2 = 1 - 2cos theta + cos^2 theta]And the y-component squared:[(sin theta - sqrt{3})^2 = sin^2 theta - 2sqrt{3} sin theta + 3]Now, adding these two together:[1 - 2cos theta + cos^2 theta + sin^2 theta - 2sqrt{3} sin theta + 3]I notice that (cos^2 theta + sin^2 theta = 1), so that simplifies part of it:[1 - 2cos theta + 1 - 2sqrt{3} sin theta + 3]Wait, hold on. Let me recast that correctly. The original expression after expanding is:1 (from the x-component) + (-2cosθ) + cos²θ + sin²θ + (-2√3 sinθ) + 3 (from the y-component). So combining constants: 1 + 3 = 4. Then, cos²θ + sin²θ = 1, so that adds another 1, making it 5. Then, the linear terms: -2cosθ - 2√3 sinθ.So altogether:[4 + 1 - 2cos theta - 2sqrt{3} sin theta = 5 - 2cos theta - 2sqrt{3} sin theta]So, the expression inside the square root simplifies to (5 - 2cos theta - 2sqrt{3} sin theta). Therefore, the magnitude is:[|vec{a} - vec{b}| = sqrt{5 - 2cos theta - 2sqrt{3} sin theta}]Now, I need to find the range of this expression. Since the square root function is increasing, the range will correspond to the square roots of the minimum and maximum values of the expression inside.So, let me denote:[E = 5 - 2cos theta - 2sqrt{3} sin theta]I need to find the minimum and maximum of (E) as (theta) varies. Then, the range of (|vec{a} - vec{b}|) will be from (sqrt{text{min } E}) to (sqrt{text{max } E}).To find the extrema of (E), which is a function of (theta), I can use calculus or recognize it as a sinusoidal function. Let me think about the form of (E):[E = 5 - 2cos theta - 2sqrt{3} sin theta]This is a constant (5) minus a linear combination of (sin theta) and (cos theta). I remember that expressions of the form (Acos theta + Bsin theta) can be rewritten as (Ccos(theta + phi)) or (Csin(theta + phi)), where (C = sqrt{A^2 + B^2}). So, maybe I can rewrite the variable part as a single sinusoidal function.Let me write the variable part as:[-2cos theta - 2sqrt{3} sin theta]Let me factor out the negative sign:[- (2cos theta + 2sqrt{3} sin theta)]Now, let me consider (2cos theta + 2sqrt{3} sin theta). Let me denote this as (Rcos(theta - phi)), where (R) is the amplitude and (phi) is the phase shift.The formula for converting (Acos theta + Bsin theta) to (Rcos(theta - phi)) is:[R = sqrt{A^2 + B^2}][cos phi = frac{A}{R}, quad sin phi = frac{B}{R}]In this case, (A = 2) and (B = 2sqrt{3}). So,[R = sqrt{2^2 + (2sqrt{3})^2} = sqrt{4 + 12} = sqrt{16} = 4]So, (R = 4). Then,[cos phi = frac{2}{4} = frac{1}{2}, quad sin phi = frac{2sqrt{3}}{4} = frac{sqrt{3}}{2}]Which means (phi = frac{pi}{3}) since (cos frac{pi}{3} = frac{1}{2}) and (sin frac{pi}{3} = frac{sqrt{3}}{2}).Therefore,[2cos theta + 2sqrt{3} sin theta = 4cosleft(theta - frac{pi}{3}right)]So, going back to our expression for (E):[E = 5 - (2cos theta + 2sqrt{3} sin theta) = 5 - 4cosleft(theta - frac{pi}{3}right)]Great, so now (E) is expressed as (5 - 4cosleft(theta - frac{pi}{3}right)). Now, since (cos) varies between -1 and 1, the term (4cosleft(theta - frac{pi}{3}right)) varies between -4 and 4. Therefore, the entire expression (5 - 4cosleft(theta - frac{pi}{3}right)) will vary between:- Minimum: (5 - 4(1) = 1)- Maximum: (5 - 4(-1) = 9)So, (E) ranges from 1 to 9. Therefore, the magnitude (|vec{a} - vec{b}| = sqrt{E}) will range from (sqrt{1} = 1) to (sqrt{9} = 3).Wait, hold on. Let me verify that. If (E) is between 1 and 9, then (sqrt{E}) is between 1 and 3. That seems straightforward.But just to make sure, let me think about whether this makes sense. The vectors (vec{a}) and (vec{b}) are both in 2D space, and their difference's magnitude depends on the angle (theta). Since both vectors have components that depend on (theta), their difference can vary, but the range seems to be between 1 and 3. Alternatively, I can think about this geometrically. The vector (vec{a}) is fixed in the x-component at 1, but its y-component is (sin theta), which varies between -1 and 1. The vector (vec{b}) has an x-component of (cos theta), varying between -1 and 1, and a fixed y-component of (sqrt{3}). So, as (theta) changes, both vectors move in the plane, and their difference vector's magnitude can vary.But since we've already transformed the expression into a single cosine function, and found that (E) varies between 1 and 9, the magnitude should indeed vary between 1 and 3.Wait, but just to be thorough, let me check when (E) is 1 and 9.When does (E = 1)? That occurs when (cosleft(theta - frac{pi}{3}right) = 1), which happens when (theta - frac{pi}{3} = 2pi k), so (theta = frac{pi}{3} + 2pi k), for integer (k).Similarly, (E = 9) when (cosleft(theta - frac{pi}{3}right) = -1), which occurs when (theta - frac{pi}{3} = pi + 2pi k), so (theta = frac{4pi}{3} + 2pi k), for integer (k).So, plugging (theta = frac{pi}{3}) into the original vectors:(vec{a} = (1, sin frac{pi}{3}) = (1, frac{sqrt{3}}{2}))(vec{b} = (cos frac{pi}{3}, sqrt{3}) = (frac{1}{2}, sqrt{3}))So, (vec{a} - vec{b} = (1 - frac{1}{2}, frac{sqrt{3}}{2} - sqrt{3}) = (frac{1}{2}, -frac{sqrt{3}}{2}))The magnitude is (sqrt{(frac{1}{2})^2 + (-frac{sqrt{3}}{2})^2} = sqrt{frac{1}{4} + frac{3}{4}} = sqrt{1} = 1). That checks out.Similarly, plugging (theta = frac{4pi}{3}):(vec{a} = (1, sin frac{4pi}{3}) = (1, -frac{sqrt{3}}{2}))(vec{b} = (cos frac{4pi}{3}, sqrt{3}) = (-frac{1}{2}, sqrt{3}))So, (vec{a} - vec{b} = (1 - (-frac{1}{2}), -frac{sqrt{3}}{2} - sqrt{3}) = (frac{3}{2}, -frac{3sqrt{3}}{2}))The magnitude is (sqrt{(frac{3}{2})^2 + (-frac{3sqrt{3}}{2})^2} = sqrt{frac{9}{4} + frac{27}{4}} = sqrt{frac{36}{4}} = sqrt{9} = 3). That also checks out.So, the calculations seem consistent. Therefore, the range of (|vec{a} - vec{b}|) is from 1 to 3.But just to make sure I didn't make any mistakes in my earlier steps, let me recap:1. I found the difference vector by subtracting components.2. I squared each component and summed them.3. I used the Pythagorean identity to simplify (cos^2 theta + sin^2 theta) to 1.4. I recognized the linear combination of sine and cosine as a single sinusoidal function.5. I calculated the amplitude (R) correctly as 4.6. I found the phase shift (phi) correctly as (frac{pi}{3}).7. I substituted back into the expression for (E) and found its range.8. I took the square root to find the range of the magnitude.9. I verified with specific values of (theta) to ensure correctness.Everything seems to add up. I don't see any errors in my reasoning or calculations.**Final Answer**The range of possible values for (|vec{a} - vec{b}|) is (boxed{[1, 3]}).

answer：First, I need to determine the number of eggs Jenny placed in each basket. She has two sets of eggs: 18 red Easter eggs in green baskets and 24 orange Easter eggs in blue baskets. Each basket must contain the same number of eggs, and there must be at least 4 eggs in each basket.To find the number of eggs per basket, I'll look for the common factors of 18 and 24 that are greater than or equal to 4. Starting with the factors of 18:1, 2, 3, 6, 9, 18And the factors of 24:1, 2, 3, 4, 6, 8, 12, 24The common factors that meet the minimum requirement of 4 eggs per basket are 6.Therefore, Jenny put 6 eggs in each basket.

answer：Okay, so I need to figure out how many positive three-digit integers have at least one '3' as a digit but don't have any '5's. Hmm, let me break this down step by step.First, let's recall that a three-digit number ranges from 100 to 999. So, the first digit (hundreds place) can be from 1 to 9, the second digit (tens place) can be from 0 to 9, and the third digit (units place) can also be from 0 to 9. But in this case, we have some restrictions: no '5's and at least one '3'.I think the best approach is to calculate the total number of three-digit numbers without any '5's and then subtract the number of three-digit numbers without any '5's and without any '3's. That way, we'll be left with the numbers that have at least one '3' and no '5's. This is using the principle of inclusion-exclusion, I believe.Let me write that down:Number of desired numbers = Total numbers without '5's - Total numbers without '5's and without '3's.So, first, let's find the total number of three-digit numbers without any '5's.For the hundreds place, since it can't be '5', the possible digits are 1-9 excluding 5. So that's 8 choices (1,2,3,4,6,7,8,9).For the tens place, it can be 0-9 excluding 5. So that's 9 choices (0,1,2,3,4,6,7,8,9).Similarly, for the units place, it's the same as the tens place: 9 choices.So, total numbers without '5's = 8 * 9 * 9.Let me calculate that: 8 * 9 is 72, and 72 * 9 is 648. So, 648 numbers don't have any '5's.Now, we need to find the number of three-digit numbers that don't have any '5's and also don't have any '3's. Then, subtracting this from 648 will give us the numbers that have at least one '3' and no '5's.So, let's calculate the numbers without '5's and without '3's.Again, starting with the hundreds place: it can't be '5' or '3'. So, from 1-9, excluding 3 and 5. That leaves us with 7 choices: 1,2,4,6,7,8,9.For the tens place: can't be '5' or '3'. So, from 0-9, excluding 3 and 5. That's 8 choices: 0,1,2,4,6,7,8,9.Same for the units place: 8 choices.So, total numbers without '5's and without '3's = 7 * 8 * 8.Calculating that: 7 * 8 is 56, and 56 * 8 is 448. So, 448 numbers don't have any '5's or '3's.Therefore, the number of three-digit numbers with at least one '3' and no '5's is 648 - 448.Let me do that subtraction: 648 - 448. Hmm, 648 minus 400 is 248, then minus 48 more is 200. So, 200.Wait, is that right? Let me double-check my calculations.First, total without '5's: 8 * 9 * 9 = 648. That seems correct.Then, total without '5's and '3's: 7 * 8 * 8 = 448. That also seems correct.Subtracting 448 from 648: 648 - 448. Let me do it step by step.648 - 400 = 248248 - 48 = 200Yes, that's correct. So, the answer is 200.But wait, just to make sure I didn't make a mistake in the number of choices.For the hundreds place without '5's: 8 choices (1-9 excluding 5). Correct.Without '5's and '3's: 7 choices (1-9 excluding 3 and 5). Correct.Tens and units place without '5's: 9 choices each (0-9 excluding 5). Correct.Without '5's and '3's: 8 choices each (0-9 excluding 3 and 5). Correct.So, 8*9*9 = 648, 7*8*8=448, 648-448=200.Yes, that seems solid. So, I think the answer is 200.**Final Answer**The number of such three-digit integers is boxed{200}.

answer：First, I need to determine how many three-digit integers can be formed using only the digits 1 and 2. Since each of the three positions (hundreds, tens, and ones) can independently be either 1 or 2, there are 2 choices for each position. Therefore, the total number of such integers is 2 × 2 × 2 = 8.Next, I'll list all possible three-digit numbers formed by the digits 1 and 2:111, 112, 121, 122, 211, 212, 221, 222.Now, I'll calculate the sum of these numbers:111 + 112 + 121 + 122 + 211 + 212 + 221 + 222.To simplify the addition, I'll group the numbers:(111 + 222) + (112 + 221) + (121 + 212) + (122 + 211).Each pair adds up to 333:333 + 333 + 333 + 333 = 1332.Therefore, the sum of all such three-digit integers formed using only the digits 1 and 2 is 1332.