A = (225a^2 + 45a + 10)/a; where A belongs to integer. How many integral solutions of A are possible

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17 Sep 2020 Number

A = (225a^2 + 45a + 10)/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

(225a^2 + 45a +10)/a = 225a + 45 + 10/a

Now we can see that whatever the value of ‘a’, 225a + 45 will always give an integral value.
So, it now depends upon 10/a only

=> a can have any integral value, which is a factor of 10.
The integers, which will satisfy this condition are ±1, ±2, ±5, ±10. Thus, in total,a can take 8 values.

Correct Option: 5 Best Solution (4)

Bhargav   4 years AGO

solution set ={1,-1,2,-2,5,-5,10,-10}
total=8

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jaswanth Reddy Ganapathireddy   4 years AGO

A=(225*a^2+45*a+10)/a
A=225*a+45+(10/a)
Given A be integer
then A~(10/a) to be integer
This can be possible by 8 numbers
-10,-5,-2,1,1,2,5,10
By substituting 0(zero) the value will become infinite.
But infinite is not an integer,it may be decimal also.

Answer - 8

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Rahul Bathla   4 years AGO

Given statement can be written as 225a+45+10/a
Now for this to be integer
10/a should be integer
Possible value are =1,2,5,10,-1,-2,-5,-10
Correct ans is option 5

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G.S.Kantharao   4 years AGO

A=(225a^2+45a+10)/a
=225a+45+10/a
So factors of 10={-10,-5,-2,-1,1,2,5,10}
No of integral solutions =8

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