N =(18n² 9n 8)/n, where n belongs to integer. How many integral solutions of N are possible?.please explain it.?

Answers

Vasudha Verma
Feb 18, 2021
hey this d=sol is given by edurev study material however I also don't understand the solution
The given expression can be broken as: 18n2/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n. 
  • Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n only ⇒ n can have any integral value, which is a factor of 8. 
  • The integers, which will satisfy this condition are �1, �2, �8, �4. Thus, in total, n can take 8 values.

hey this d=sol is given by edurev study material however I also don't understand the solutionThe given expression can be broken as: 18n2/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n.Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n only ⇒ n can have any integral value, which is a factor of 8.The integers, which will satisfy this condition are �1, �2, �8, �4. Thus, in total, n can take 8 values.
hey this d=sol is given by edurev study material however I also don't understand the solutionThe given expression can be broken as: 18n2/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n.Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n only ⇒ n can have any integral value, which is a factor of 8.The integers, which will satisfy this condition are �1, �2, �8, �4. Thus, in total, n can take 8 values.