Question

The number of integers a such that \[1 \leqslant a \leqslant 100\] and \[{a^a}\] is a perfect square is
A.50
B.53
C.55
D.56

Answer 1

Hint: At first we will determine the nature of the ‘a’ and after concluding the nature of ‘a’ we will substitute the value of ‘a’ with its general form.
After the substitution, it will be easy to find the number of solutions of the inequality.

Complete step-by-step answer:
Given data: \[1 \leqslant a \leqslant 100\] and \[{a^a}\]
We know that perfect square always even in the exponent term because perfect square is of the form \[{x^{2n}}\] so that they can be rewritten as \[{\left( {{x^n}} \right)^2}\]
Now since we have given that \[{a^a}\] is also a perfect square, we can say that ‘a’ is an even number such that \[{a^a}\] can be written in the form of \[{b^2}\] where \[b = {a^{\dfrac{a}{2}}}\] .
Since ‘a’ is an even number let us assume that $a = 2n$
Now, substituting the value of ‘a’ in the given inequality,
 \[ \Rightarrow 1 \leqslant 2n \leqslant 100\]
Dividing all the sides by 2
 \[ \Rightarrow \dfrac{1}{2} \leqslant n \leqslant 50\]
Now, from the above inequality, we can say that n can take all the integers value from $\dfrac{1}{2}$ to 50
And it is clear that there are 50 integers from $\dfrac{1}{2}$ to 50 i.e. 1, 2, 3, 4,……….,50.
Option(A) is correct.

Note: Since we concluded that ‘a’ is an even number we can also say that in the inequality \[1 \leqslant a \leqslant 100\], 100 integers are satisfying the inequality and we know that even and odd number occur alternatively out of those 100, 50 are even number and 50 are odd numbers. So the required number of values of ‘a’ is 50.