According to the definition, $u(2023) = 3202$ or $u(2023) = 3022$?

zhaoli is right, I failed at translation. I'm definitely going to need more practice in both Spanish and English It is always tempting to simplify the statement of the problem into an elegant and concise sentence, especially when translating between languages, and I can swear it backfires every time I've done that. In this question, I was tempted to avoid a longer explanation by using the word ``exchange'' in the place of ``move,'' but I made a bad mistake in defining $u(N)$. The question defines $u(N)$ as the number obtained by making the ones digit the left-most digit of $N$, that is, taking the last, right-most digit (the ones digit) of $N$ and moving it leftwards through the digits of $N$ until it becomes the first (left-most) digit of $N$; for example, $u(2023) = 3202$

Part (1) Let d = number of digits of N Let N=10A+B, where A has (d-1) digits and B only has one digit. Then, $u(N)=10^{d-1}B+A$ $\frac{u(N)}{N}=\frac{10^{d-1}B+A}{10A+B}=\frac{23}{35}$ $35(10^{d-1}B+A)=230A+23B$ $(35\cdot 10^{d-1}-23)B=230A-35A$ $(35\cdot 10^{d-1}-23)B=195A$ $(35\cdot 10^{d-1}-23)B=3\cdot 5 \cdot 13 \cdot A$ $\frac{(35\cdot 10^{d-1}-23)}{3\cdot 13}B=5A$ $\frac{2(35\cdot 10^{d-1}-23)}{3\cdot 13}B=10A$ $\frac{(70\cdot 10^{d-1}-46)}{3\cdot 13}B+B=10A+B=N$ for d=6, we have: $\frac{(70\cdot 10^5-46)}{3\cdot 13}B+B=10A+B=N$ $179486B+B=10A+B=N$ $179487B=10A+B=N$ The last digit of 179487B has to be B, so, the last digit of 7B has to be B. The only time this happens is when B=5 and we have: $N=179487 \cdot 5 = 897435$ Part (2) For N to exist we need $\frac{(70\cdot 10^{d-1}-46)}{3\cdot 13}$ to be integer Since $(70\cdot 10^{d-1}) \equiv 1 \; mod \; 3$, and $-46 \equiv -1 \; mod \; 3$, then $(70\cdot 10^{d-1}-46) \equiv 0 \; mod \; 3$, so it is divisible by 3 for all values of d Now we need to check if it is divisible by 13. Since $-46 \equiv -7 \; mod \; 13$, we need to check for which d is $(70\cdot 10^{d-1}) \equiv 7 \; mod \; 13$ or check for which d is $(7\cdot 10^{d}) \equiv 7 \; mod \; 13$. At d=5, $700000 \equiv 2 \; mod \; 13 \not \equiv 7 \; mod \; 13$ At d=4, $70000 \equiv 8 \; mod \; 13 \not \equiv 7 \; mod \; 13$ At d=3, $7000 \equiv 6 \; mod \; 13 \not \equiv 7 \; mod \; 13$ At d=2, $700 \equiv 11 \; mod \; 13 \not \equiv 7 \; mod \; 13$ At d=1, $70 \equiv 5 \; mod \; 13 \not \equiv 7 \; mod \; 13$ This means that for any d<6, $(70\cdot 10^{d-1}-46) \not \equiv 0 \; mod \; 13$ and thus, there is no N with less than 6 digits that satisfies $\frac{u(N)}{N}=\frac{23}{35}$