<s>[INST] 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?
 [/INST] 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.
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