Quant Question Of The Day: 57
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Numbers/Algebra
If S = (A + 20) + ( A + 21) + ( A + 22)+………+ ( A + 100) , where A is a positive integer . What is the smallest value of A such that S is a perfect square ?
1. 4
2. 16
3. 324
4. None of the above
There are 100 -20 +1 = 81 terms in the series , so using the formula S= 1/2 [a + l] × Number of terms
Sum of the series (S) = 81/2 [A + 20 + A + 100]= 81( A + 60)
Now 81 is a perfect square so S is a perfect square if and only if x + 60 is a perfect square . As A is a positive integer , the smallest possible value of x is 4.
See our previous ‘Questions of the Day’:
Ans is (1) i.e 4.
4
S= 81A+4860
Since A is a postive integer
A=1 => S=4941 not ps
A can’t be 2/3 because unit digit of any square is not 2/3
So, A=4 => S= 5184 =72^2
Hence A=4
(1) 4
4
4
Option 1.
Approach:-
I first calculated the value of S i.e,
S= 4860+81A
Then i used options
I realised that S is a perfect square when A is 4.
Hence, option 1 is the answer.
1. 4