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It is known that "if three prime numbers a, B and C greater than 3 satisfy the relation 2A + 5B = C, then a + B + C is a multiple of integer n". What is the maximum possible value of integer n in this theorem? Please prove your conclusion
It is proved that: ∵ a + B + C = a + B + 2A + 5B = 3 (a + 2b). Obviously, 3 | a + B + C, if the remainder of a and B divided by 3 is RA and Rb respectively, then RA ≠ 0, Rb ≠ 0. If ra ≠ Rb, then RA = 2, Rb = 1 or RA = 1, Rb = 2, then 2A + 5B = 2 (3m + 2) + 5 (3N + 1) = 3 (2m + 5N + 3), or 2A + 5B = 2 (3P + 1)
If any three prime numbers a, B, C greater than 3 satisfy the relation 2A + 5B = C, then a + B + C must be a multiple of some integer (constant) n, and the maximum value of n is
Analysis: according to the meaning of the question, we take two groups of values for observation and analysis
(1) If a = 11, B = 5, then C = 22 + 25 = 47, a + B + C = 63
(2) If a = 13, B = 7, then C = 26 + 35 = 61, a + B + C = 81
∵ (63,81) = 9 ∵ N and the maximum possible value is 9
It is proved that: ∵ 2A + 5B = C ∵ a + B + C = a + B + 2A + 5B = 3A + 6B = 3 (a + 2b) ∵ 3 | a + B + C
Let a and B be divided by 3 and the remainder be RA and Rb. Since a and B are prime numbers, the values of RA and Rb must be 1 or 2
(1) If RA ≠ Rb, then one of them must be 1 and the other 2
∵1+2=3 ∴ c=2a+5b=2(a+b)+3b ∴3|c
This is contrary to the fact that C is prime, so this situation does not exist
(2) If RA = Rb, then 3 | a-b. ∵ a + 2B = 3B + (a-b) ∵ 3 | a + 2B ∵ 9 | a + B + C
The proposition holds, that is, n = 9
Three prime numbers a, B, C greater than 3 satisfy the relation 2A + 5B = C, then a + B + C is a multiple of integer n, what is the maximum possible value of integer n? And prove the conclusion
Analysis: according to the meaning of the question, we take two groups of values for observation and analysis: (1) a = 11, B = 5, then C = 22 + 25 = 47, a + B + C = 63 (2) a = 13, B = 7, then C = 26 + 35 = 61, a + B + C = 81 ∫ (63,81) = 9. The maximum possible value is 9
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