M4maths Previous Puzzles

20 Aug 2020 Number

A = (126a^2 + 113a + 11)/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) 4
10) None of these

Solution

(126a^2 + 133a +11)/a = 126a + 133 + 11/a

Now we can see that whatever the value of ‘a’, 126a + 133 will always give an integral value.
So, it now depends upon 11/a only

=> a can have any integral value, which is a factor of 11.
The integers, which will satisfy this condition are ±1, ±11. Thus, in total,a can take 4 values.

Correct Option: 9 Best Solution (0)