How to simplify $42^{25\sqrt{25^{25}}}$?

Question

Am a student preparing for GRE, I have no clue to solve this am attaching the screenshot of question:

I need you give me a short cut or tip to deal such problems...

Answer 1

The exam is wrong.

A number $n$ has $\lfloor \log_{10} n\rfloor + 1$ decimal digits. So $21^{25}$ has $34$ digits.

For the more complicated expression,

$$1+\lfloor \log_{10} 42^{25\sqrt{25^{25}}} \rfloor = 1+\lfloor \log_{10} 42^{\sqrt{25^{27}}} \rfloor = 1+\lfloor \log_{10} 42^{5^{27}} \rfloor = 1+\lfloor 5^{27} \log_{10} 42 \rfloor$$ which according to my calculator is over $10^{19}$, greater than $34+25.$

Answer 2

Let us first simply this a bit:

$$42^{25 \sqrt{25^{25}}} = 42^{25 \cdot 5^{25}} = (21^{25})^{5^{25}} 2^{5^{27}}.$$

The number of digits is obtained by the logarithm, so we calculate

$$\log_{10} 42^{25 \sqrt{25^{25}}} = 5^{25} \log_{10} 21^{25} + 5^{27} \log_{10} 2.$$

Since $n = \lfloor \log_{10} 21^{25} \rfloor + 1$, I'd say that the answer is "none of the above".

The used identities can be found here. The check with WolframAlpha was not as simple, but here it is: the number of digits of $21^{25}$ is $34$, and the number of digits of $42^{1000}$ is $1624$. I think we can agree that $25 \sqrt{25^{25}} > 1000$, and $1624 > 850 = 25 \cdot 34 = 25n$ (the other offered answers are even smaller), which I believe confirms my answer.

Answer 3

The proposed answer is off by such a large magnitude that it is easy to show its falsity without a calculator.

$21^{25}\lt100^{25}=10^{50}$ which has $51$ digits, and so if the answer is correct, $\large42^{25\sqrt{25^{25}}}$ can't have more than $51+25=76$ digits.

But, $\large42^{25\sqrt{25^{25}}}=42^{25^{13.5}}\gt 10^{25^{13.5}}\gt 10^{25^2}=10^{625}$ which has $626$ digits.

So, the given answer is wrong unless the expression is $42^{25}\cdot\sqrt{25^{25}}$ as pointed out in the comments.