0
$\begingroup$

Define the function $F(N) = (9[N])^2$, meaning: the number made of $N$ consecutive $9$'s digits, raised to the power $2$. For example, $F(4) = 9999^2$.

Looking at some example values, it seems like there are only $4$ possible digits in the resulting number: $0,1,8,9$.


Can you give me an expression for the multiplicities of each of these digits, as a function of N? For example, $F(4) = 999^2 = 998001$, which has two $0$'s, one $1$, one $8$, and two $9$'s.

Also, it seems like (again, only testing on some small values of $N$), a given digit will have all of its occurrences consecutively, e.g. for $F(7)=9999999^2$, the result is $99999980000001$, which has all of the $9$'s consecutively, and all of the $0$'s consecutively. Is this true in general for any non-negative integer $N$ that the digits will have this consecutive property?

$\endgroup$
3
$\begingroup$

$$(10^N-1)^2 = 10^{2N} - 2 \cdot 10^N + 1 = \sum_{j=N+1}^{2N-1} 9 \cdot 10^j + 8 \cdot 10^N + 1$$ so there are $N-1$ 9's, followed by $8$, $N-1$ 0's and finally $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.