If $ \frac{X_n}{1024^n} $ is an odd integer, find the smallest possible value of $n$, where $n\ge2$ is an integer. 
For each integer $k\ge2$, the decimal expansions of the numbers $1024, 1024^2, \dots, 1024^k$ are concatenated, in that order, to obtain a number $X_k$.  (For example, $X_2 = 10241048576$.)  If $ \frac{X_n}{1024^n}$ is an odd integer, find the smallest possible value of $n$, where $n\ge2$ is an integer..


*

*The answer is $\boxed{5}.$

*I tried for $X_1,X_2,X_3$ and noticed they weren't coming out as integers.

After this, the numbers were too big.

*

*I think that we can also try expanding  $X_i$ terms. But then we would like t know the number of digits in $X_i.$
For example,
we have


*$$X_2 = 1024 \cdot 10^7 + 1024^2 = 2^{17} \cdot 5^7 + 2^{20} = 2^{17} (5^7 + 2^3).$$
So $$X_2/{1024^2}$$which is not a integer.


*$$X_3= 1024 \cdot 10^{17} + 1024^2 \cdot 10^{10} + 1024^3 = 2^{27} \cdot 5^{17} + 2^{30} \cdot 5^{10} + 2^{30} = 2^{27} (5^{17} + 2^3 \cdot 5^{10} + 2^3). $$


*We have $$V_2(X_3)<V_2(1024^3).$$
Any hints?
 A: Hint: Find a nice way to write $X_n$, so that it's easy to manipulate $X_n / 1024^n$.
In fact, it is exactly what you wrote at the start for $X_2, X_3$, you just need to generalize it. (You didn't get the manipulation part though.)

 $X_n = 1024^n + \sum_{i=1}^{n-1} 1024^i \times 10^{a_i}$, where $a_i$ is to be determined.  Then,
 $$ \frac{X_n}{1024^n} = 1 + \sum_{i=1}^{n-1} \frac{2^{a_i} \times 5^{a_i} } { 2^{10 (n-i)}}.$$
 Now, study this expression.


Here's a complete (to me) solution. Fill in the details as needed.
If you're stuck, show your work when asking for help.

*

*Let $b_i = \lfloor i \times 10 \times \log_{10} 2 \rfloor + 1$. Show that the first 5 terms are $b_i = 4, 7, 10, 13, 16$.

*Show that since $2^{10} > 10^3$, hence $b_i \geq 3i+1$. (As Ross points out, $b_i = 3i+1$ for $ i \leq 97$.)

*Fix $n$. Show that $a_{n-1} = b_{n-1}, a_{i-1} = a_i + b_i,$ so $a_i$ can be determined recursively.

*Show that if $ a_i > 10 (n-i)$, then the corresponding term in the summation expression of $X_n / 1024^n$ is an even integer.

*Show that if $ a_i = 10 (n-i)$, then the corresponding term in the summation expression of $X_n / 1024^n$ is an odd integer.

*Show that if $ a_i > 10 (n-i)$,  $\forall i < n$, then $X_n / 1024^n$ is an odd integer (because of the initial term 1).

*Show that for $n=5$, $a_4 = 16, a_3 = 29, a_2 = 39, a_1 = 46$.

*Show that for $ n \geq 6$, $a_i > 10 (n-i)$ for $ i < n$. (We're adding a lot of terms that are bigger than 10 at the start, so this should be "obvious".)

*Conclude that for $n \geq 5$, $X_n$ is an odd integer.

*Show that for $ n= 4$, $a_3 = 13, a_2 = 23, a_1 = 30$.

*Conclude that for $n=4$, $X_n$ is an even integer because $a_1  = 10 (4-1)$ and $a_i > 10 (n-i)$ otherwise.

*Conclude that for $n = 3, 2$, $X_n$ is not an integer.
(It is easy to calculate directly as you did. If you want to use my method, show that it's equal to $\frac{ \text{odd}}{2^k}$, hence not an integer.)


Notes

*

*Yes, for clarity of expression, I should have used $a_{n, i }$ so that $a_n$ doesn't do double-duty. However, that would make things tedious, so I didn't do that.

*It seems like a huge coincidence that $X_4$ is odd.

A: As $1024$ is just a little larger than $10^3, 1024^k$ will be a little larger than $10^{3k}$, so will have $3k+1$ decimal digits.  According to Alpha, the number of digits does not increase above this until $k=98$.  I think you can rely on $n$ being smaller than that.
