A = (26a^2 + 13a + 9)/a; where A belongs to integer. How many integral solutions of A are possible?

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17 Jul 2020 Number

A = (26a^2 + 13a + 9)/a; where A belongs to integer. How many integral solutions of A are possible?

OPtion

1) 2
2) 3
3) 6
4) 9
5) 8
6) 7
7) 10
8) 12
9) 24
10) None of these

Solution

(26a^2 + 13a + 9)/a = 26a + 13 + 9/a

Now we can see that whatever the value of ‘a’, 26a + 13 will always give an integral value.
So, it now depends upon 8/a only

=> a can have any integral value, which is a factor of 9.
The integers, which will satisfy this condition are ±1, ±3, ±9. Thus, in total,a can take 6 values.

Correct Option: 3 Best Solution (2)

Karan   4 years AGO

A= (26a^2+13a+9)/a= 26 a+13+9/a
A will have integer ( positive or negative) values when a will have values of -1,-3,-9,1,3,9.
Hence A can have 6 integer ( including negative integers) values.

Like? Yes (8) | No (1)  

Varun   4 years AGO

Given: A = (26.a.a/a+13.a+9)/a
A = 26a+13+9/a
26a + 13 will always be integer if we pass integer to a
Remaining is 9/a
And this will be integer if a=--9 -3 -1 1 3 or 9
=> A will be integer if we have any of this 6 values for a
= Option (3)

Like? Yes (2) | No (4)