Find the number of positive integers n for which n ≤ 2020 and 6 is a factor of n^2+3n+2
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Answers
Given : n ≤ 2020 and 6 is a factor of n^2+3n+2
To find : number of positive integers Satisfying given condition
Solution:
6 is a factor of n² + 3n + 2
n can be of form 3k , 3k + 1 , 3k + 1
n² + 3n + 2 = (3k)² + 3(3k) + 2
= 9k² + 9k + 2
= 9k(k+1) + 2
k(k+1) is always Divisible by 2 hence 9k(k + 1) is Divisible by 6
but + 2 is there which is not divisible by 6
hence n = 3k not Divisible by 6
n² + 3n + 2 = (3k+1)² + 3(3k+1) + 2
= 9k² + 6k + 1 + 9k + 3 + 2
= 9k(k+1) + 6(k + 1)
Divisible by 6
n² + 3n + 2 = (3k+2)² + 3(3k+2) + 2
= 9k² + 12k + 4 + 9k + 6 + 2
= 9k(k+1) + 12(k + 1)
Divisible by 6
Hence number Divisible by 6
for n = 1 , 2 , 4 , 5 , 7 , 8 ................................ 2017 , 2018 , 2020
2019/3 = 673 ( 2 numbers fro every 3 ) & 2020
2 * 673 + 1
= 1347
= 1347 integers fro which n² + 3n + 2 is Divisible by 6
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