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Topic: Can you guys solve this difficult college math problem? Help, I am having trouble.?**Question:**
What is the smallest positive integer n that every possible coloring of the integers from 1 to n, with each integer red or blue, has at least one arithmetic progression of 3 different integers of the same color.

July 20, 2019 / By Breann

Let's try to create the longest possible string which does NOT have an arithmetic progression of 3 integers of the same color. We fail when such a progression occurs. Use R for red and B for blue, X for unknown. Without loss of generality*, assume 1 is red. RRRX fail 1-2-3 RRBRRR fail 4-5-6 RRBRRBR fail 1-4-7 RRBRRBBR fail 2-5-8 RRBRRBBB fail 6-7-8 RRBRBR fail 2-4-6 RRBRBBR fail 1-4-7 RRBRBBB fail 5-6-7 RRBBRRR fail 5-6-7 RRBBRRBR fail 2-5-8 RRBBRRBBR fail 1-5-9 RRBBRRBBB fail 7-8-9 RRBBRBRR fail 2-5-8 RRBBRBRB fail 4-6-8 RRBBRBBR fail 2-5-8 RRBBRBBB fail 6-7-8 RRBBB fail 3-4-5 RBRRR fail 3-4-5 RBRRBRR fail 1-4-7 RBRRBRBR fail 4-6-8 RBRRBRBB fail 2-5-8 RBRRBBR fail 1-4-7 RBRRBBB fail 5-6-7 RBRBR fail 1-3-5 RBRBBRRR fail 6-7-8 RBRBBRRB fail 2-5-8 RBRBBRBRR fail 3-6-9 RBRBBRBRB fail 5-7-9 RBRBBRBB fail 2-5-8 RBRBBB fail 4-5-6 RBBRRR fail 4-5-6 RBBRRBR fail 1-4-7 RBBRRBBRR fail 1-5-9 RBBRRBBRB fail 3-6-9 RBBRRBBB fail 6-7-8 RBBRBRR fail 1-4-7 RBBRBRB fail 3-5-7 RBBRBBR fail 1-4-7 RBBRBBB fail 5-6-7 RBBB fail 2-3-4 That's every single possibility. Six times, we managed to get a sequence 8 numbers long which didn't have a progression in it, so clearly the answer to the puzzle is NOT 8. It must be more than 8. We were unable to build a sequence of length 9, so the answer is 9. Every possible coloring of the numbers 1-9 will result in a sequence of three numbers which form an arithmetic progression, because there are no colorings of length 9 which don't. *If 1 is actually blue, we can just switch our labels and let R=Royal blue and B= Berry red. Either way, we can label 1 "R".

👍 176 | 👎 4

Did you like the answer? (∞,5)(5,∞) ARN except for x=5 The domain of a function is the values of x the function can take. The only asymptote or break in the function you will have is when x=5 because it is impossible to divide by zero.

All Real Numbers except x = 5 because that would give you y= 1/(5-5) + 7 y= 1/0 + 7 1/0 is not possible

progression >= 3 objects in the series. the worst case scenario is RRBBR... or BBRRB... , when both blue and red cannot complete the 3 consecutive. but even like this, there's a pattern behind it. it repeats every other 4 in a cycle. but after 2 cycles, the next object (the 1st in 3rd cycle) shall complete some series, the extreme being the 1st, 5th and 9th series. so answer is 9. below is 2 of the possible distribution of red and blue. 1247 and 35689 (369) 12569 and 3478 (159)

👍 70 | 👎 -3

you want the smallest interger, positive or negative? im in 9th grade so I wouldnt know this all I know is that the smallest number is negative infity to the infinitieth power, Im not even sure that exist... so forget what I said nvm...

👍 67 | 👎 -10

Let's number the rectangles 1 to 15. You started with 15 rectangles. Combine 1 and 2; 2 and 3; 3 and 4; etc, that's 14 total. Combine 1, 2 and 3; 2, 3 and 4; etc; that's 13 total Combine 1, 2, 3 and 4; 2, 3, 4, and 5; etc, that's 12 total. Continue to the end and you get: 15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 120

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