The number of integers $a$ ($1\leq a\leq 200$) such that $a^a$ is a perfect square are is:
$A)\;105$
$B)\;103$
$C)\;107$
$D)\;109$
My work
$a^a$ is a perfect square when $a$ is even.$2^2 = 4. 4^4 = 256$ and so on.
$a = 2,4,6.....198$
$198 = 2 + (n-1)2$
$196/2 + 1 = 99$
Is my work correct? However there is no option matching this.
$\endgroup$ 54 Answers
$\begingroup$Your reasoning needs a little correction.
$(1)$ $a^a$ is a perfect square when $a$ is even. Perfectly true. The integers satisfying this condition belong to the set $A=\{2,4,6,\cdots 198,200\}$. So, the number of possibilities are :$100$.
$(2)$ You have to include those $a$'s that are perfect squares as well. In, this case, $a= k^2$. So, $a^a = (k^2)^{k^2}$ which is a perfect square. Counting all odd perfect squares, to prevent double counting of even ones from $(1)$. The integers belong to the set $B =\{1,9,25,\cdots 169\}$. So, the number of possibilities are: $7$.
We thus have a total of $107$ integers satisfying the condition. Hope it helps.
$\endgroup$ 0 $\begingroup$you are missing all of the odd perfect squares, they are:
$1,3^2,5^2,7^ 2,9^2,11^2,13^2$. You also left out $200$, so the answer is $107$
$\endgroup$ 15 $\begingroup$All the answers are ok, but I think we can be more rigorous. First, for $a=1$, $1^1=1$ is perfect square. Now, let's suppose that $a>1$, by the fundamental theorem of the arithmetic we can write $a=p_1^{r_1}\cdot p_2^{r_2}\ldots p_{k}^{r_k}$, where $p_1,p_2\ldots, p_k$ are prime numbers and $r_1,r_2,\ldots,r_k$ are positive integers. Then $$a^a=(p_1^{r_1}\cdot p_2^{r_2}\ldots p_{k}^{r_k})^a=p_1^{ar_1}\cdot p_2^{ar_2}\ldots p_{k}^{ar_k}.$$
Now, it's known that a positive integer is a perfect square if only if every one of their prime factors has an even exponent. Then if we want $a^a$ to be a perfect square it must hold that $ar_1,ar_2,\ldots, ar_k$ are even numbers. So if $a$ is even we have that condition satisfied, thus since we have $200/2=100$ even numbers, we have at the moment $100$ numbers which satisfy the required condition.
Now, let's assume that $a$ is odd. If $a$ is not a perfect square, then at least one of $r_1, r_2,\ldots, r_k$ is odd, let's say wlog that is $r_1$, then $ar_1$ is odd too, and therefore it's impossible for $a^a$ to be a perfect square. But, what happen if $a$ is an odd perfect square? in this case we can write $a=b^2$, thus $$a^a=(b^2)^{b^2}=b^{2b^2}=(b^{b^2})^2.$$
So $a^a$ is perfect square. Now the odd perfect squares less than $200$ are $9,25,49,81,121,169$ (we already counted $1$), and then we have $6$ more numbers to add. Hence, in total we have $1+100+6=107$ numbers in the interval $[1,200]$ satisfying the condition.
$\endgroup$ 2 $\begingroup$First of all, $a^a$ being a perfect square number does not necessarily indicate that $a$ is an even number. If required conditions meet, $a$ might take on odd values as well. Below you may see the required conditions.
$$a=2n+1, \space \space \space a^a=(a^n)^2 a \space \space \land \space \space a=b^2, \space \space b\in \Bbb Z $$
The number of even numbers that makes $a^a$ a perfect square number is $\frac{200-2}{2}+1=100$. On the other hand, there are odd perfect square numbers such that $$1^2, 3^2, 5^2,... 13^2$$The number of odd numbers that makes $a^a$ a perfect square number is $\frac{13-1}{2}+1=7$. The correct answer is $107$.
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