dunglamtoan
24/12/2017 at 21:14-
With x = 0
=> A = 0 + 0 + 6 = 6 (Removed) Because \(6⋮\left(1;2;3;6\right)\)
With x = 1
=> A = 1 + 1 + 6 = 8 (Removed) Because \(8⋮\left(1;2;4;8\right)\)
Similar , with x = 2 then A still not a prime number
With x \(\ge\) 3 => x has three forms
\(\left\{{}\begin{matrix}x=3k\\x=3k+1\\x=3k+2\end{matrix}\right.\)
+) With x = 3k => A = 9k2 + 3k + 6 = 3.(3k2 + k + 2) (Removed)
+) With x = 3k + 1 => A = (3k + 1)2 + (3k + 1) + 6 = 9k2 + 6k + 1 + 3k + 1 + 6 = 9k2 + 9k + 8
A = 9k(k + 1) + 8
We can see : k(k + 1) \(⋮\) 2 because it is a multiplication of 2 continuous numbers
8 \(⋮\) 2 too
So A \(⋮\) 2 => x = 3k + 1 (removed)
+) With x = 3k + 2 => A = (3k + 2)2 + (3k + 2) + 6 = 9k2 + 12k + 4 + 3k + 2 + 6 = 9k2 + 15k +12
A = 3.(3k2 + 5k + 4) \(⋮\) 3 (Removed)
So , without any integer value of x that satisfy A is a prime number
Selected by MathYouLike -
We have:
\(x^2+x=x\left(x+1\right)⋮2\) is the product of two consecutive integers, so divisible by 2
\(\Rightarrow A=x^2+x+6⋮2\)
But A is a prime number and we only have an even prime number is 2
So A = 2
=> \(x^2+x+6=2\)
\(\Rightarrow x\in\varnothing\)
So there isn't any integer x satisfy the above condition
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Faded 22/01/2018 at 12:42With x = 0
=> A = 0 + 0 + 6 = 6 (Removed) Because 6⋮(1;2;3;6)
With x = 1
=> A = 1 + 1 + 6 = 8 (Removed) Because 8⋮(1;2;4;8)
Similar , with x = 2 then A still not a prime number
With x ≥
3 => x has three forms
⎧⎪⎨⎪⎩x=3kx=3k+1x=3k+2
+) With x = 3k => A = 9k2 + 3k + 6 = 3.(3k2 + k + 2) (Removed)
+) With x = 3k + 1 => A = (3k + 1)2 + (3k + 1) + 6 = 9k2 + 6k + 1 + 3k + 1 + 6 = 9k2 + 9k + 8
A = 9k(k + 1) + 8
We can see : k(k + 1) ⋮
2 because it is a multiplication of 2 continuous numbers
8 ⋮
2 too
So A ⋮
2 => x = 3k + 1 (removed)
+) With x = 3k + 2 => A = (3k + 2)2 + (3k + 2) + 6 = 9k2 + 12k + 4 + 3k + 2 + 6 = 9k2 + 15k +12
A = 3.(3k2 + 5k + 4) ⋮
3 (Removed)
So , without any integer value of x that satisfy A is a prime number