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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.

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  • $\begingroup$ I think the problem is you're giving a sufficient condition for those numbers to be perfect squares but non a necessary condition, so you're not catching them all with your reasoning. $\endgroup$ – Harnak Dec 25 '16 at 16:39
  • $\begingroup$ Harnak, what do you mean? Elaborate $\endgroup$ – pi-π Dec 25 '16 at 16:40
  • $\begingroup$ I mean that all $a^a$ when $a$ is even are perfect squares, but this is not necessary, i.e. there may be some even when $a$ is odd. Of course, $a$ cannot be prime. $\endgroup$ – Harnak Dec 25 '16 at 16:42
  • $\begingroup$ But how do I calculate that? $\endgroup$ – pi-π Dec 25 '16 at 16:44
  • $\begingroup$ I have no idea for now :p I was just pointing out that you probably missed some of the odd numbers, that's why you get $99$ instead of one of the possible answers. $\endgroup$ – Harnak Dec 25 '16 at 16:45
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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.

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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.

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    $\begingroup$ very nice, this is the exact same process I went through when solving, although I didn't treat $1$ seperately. $\endgroup$ – Jorge Fernández Hidalgo Dec 25 '16 at 17:13
  • $\begingroup$ @JorgeFernándezHidalgo thanks for your comment. And yes, it's not necessary to separate $1$, but I did for pedagogical reasons. $\endgroup$ – Xam Dec 25 '16 at 17:17
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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 $107$

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    $\begingroup$ How is 3^3=27 a perfect square?? $\endgroup$ – pi-π Dec 25 '16 at 16:43
  • $\begingroup$ For which integer value of $a$ does the equation $a^a=3^2$ hold??? $\endgroup$ – barak manos Dec 25 '16 at 16:45
  • $\begingroup$ $(3^2)^{3^2}=(3^9)^2$ $\endgroup$ – Jorge Fernández Hidalgo Dec 25 '16 at 16:46
  • $\begingroup$ So why not write $9^9$? And where are all the others ($15^{15},17^{17},\dots$)? And how does that get you to $107$? $\endgroup$ – barak manos Dec 25 '16 at 16:46
  • $\begingroup$ Please show a proper calculation, I cannot understand $\endgroup$ – pi-π Dec 25 '16 at 16:47

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