Find the number of positive integer n for which
n<=1991 and 6 is factor of n^2+3n+2.
Can someone pls help with it as quickly as possible
Answers
Given : n≤1991 and 6 is factor of n²+3n+2.
To Find : Numbers of positive integer n
Solution:
n² + 3n+ 2 = 6p
n = 6k
=> (6k)² + 3(6k) + 2 = 6*3k(2k + 1) + 2 = 6p + 2 not a factor of 6
n = 6k+1
=> (6k+1)² + 3(6k+1) + 2 = 36k² + 12k + 1 + 18k + 3 + 2 = 36k² + 30k + 6
= 6(6k² + 5k + 1) = 6p Factor of 6
n = 6k+2
=> (6k+2)² + 3(6k+2) + 2 = 36k² + 24k + 4 + 18k + 6 + 2 = 36k² + 42k + 12
= 6(6k² + 7k + 2) = 6p Factor of 6
n = 6k+3
=> (6k+3)² + 3(6k+3) + 2 = 36k² + 36k + 9 + 18k + 9 + 2 = 36k² + 54k + 20
= 6(6k² + 9k + 3) +2 = 6p +2 not a factor of 6
n = 6k+4
=> (6k+4)² + 3(6k+4) + 2 = 36k² + 48k + 16 + 18k + 12 + 2 = 36k² + 66k + 30 = 6(6k² + 11k + 5) = 6p a factor of 6
n = 6k+5
=> (6k+5)² + 3(6k+5) + 2 = 36k² + 60k + 25 + 18k + 15 + 2 = 36k² + 78k + 42 = 6(6k² + 13k + 7) = 6p a factor of 6
n = 6k & 6k + 3 are not resulting in factor 6 of n² + 3n+ 2
=> 3k
1 to 1991 are 1991 numbers
3 , 6 , 9........................1989
= 3 *( 1 , 2 , 3 , ...................663 )
663 numbers
1991 - 663 = 1328
1328 number of positive integer n for which n≤1991 and 6 is factor of n²+3n+2.
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mate I can't understand what you have written