We have $n^3+3 = n(n^2+7) - (7n-3)$
Because $n^2+7| n^3 + 3$ so $n^2+ 7 | 7n - 3$
As 7n - 3 is positive so we have
$n^2 + 7 \le 7n- 3$
Or $n^2 - 7n + 4 \le 0$ giving $ n < 7$
By trying the values of n from 1 to 6 we get 1
Solution set $\{2,5\}$
We have $n^3+3 = n(n^2+7) - (7n-3)$
Because $n^2+7| n^3 + 3$ so $n^2+ 7 | 7n - 3$
As 7n - 3 is positive so we have
$n^2 + 7 \le 7n- 3$
Or $n^2 - 7n + 4 \le 0$ giving $ n < 7$
By trying the values of n from 1 to 6 we get 1
Solution set $\{2,5\}$
No comments:
Post a Comment