Find all positive integer bases b≥9 so that the number n−1 1′s⏞11⋯10n−1 7′s⏞77⋯78n 1′s⏞11⋯1b3 is a perfect cube in base 10 for all sufficiently large positive integers n. Proposed by Yang Liu
Problem
Source: ELMO 2014 Shortlist N10, by Yang Liu
Tags: geometry, 3D geometry, modular arithmetic, number theory proposed, number theory
26.07.2014 19:06
The number in question when evaluated turns out to beb3n−b2n+1+7b2n+bn+1−7bn−13(b−1)=(bn−1)(b2n+bn(8−b)+1)3(b−1)Let the sequence of these numbers be x3n. We shall show that b−1 is a power of 3. Suppose that this is not the case. Then, ∃p∈P such that p≠3 and p|b−1. Note that b^{2n}+b^n(8-b)+1 \equiv 9 \ncong 0 \pmod{p}. Suppose p \neq 2. Then, equating the powers of p on either side using LTE, we get 3v_p(x_n) = v_p(n) for sufficiently large n. Choosing large n with v_p(n)=1 does the job. Otherwise, if p=2, again equate powers of 2 on either side using LTE. We get v_2(b+1)+v_2(n)-1=3v_3(x_n), for all sufficiently large n. But, we can easily choose large n such that the LHS is not a multiple of 3, contradiction. Therefore, b-1=3^a. Since b \ge 9, we have 3^a \ge 8 \implies a \ge 2. Suppose that a\ge 3. Then, note that v_3(b^{2n}+b^n(8-b)+1) = v_3((b^{2n}-b^{n+1})+8(b^n-1)+9) = 2because 27|b-1 \implies 27|b^n-1 and 27|b^{n+1}(b^{n-1}-1)=b^{2n}-b^{n+1}. Again, looking at the powers of 3 on either side, we have 3v_3(x_n)=2+v_3(n). Choosing large n not divisible by 3 gives a contradiction. Therefore, a=2, which means b=10. And in this case we do have x_n^3 = \frac{(10^n-1)(10^{2n}-2.10^n+1)}{27}= \left( \frac{10^n-1}{3} \right)^3
26.10.2019 06:07
Observe that the quantity in question can be written as: \frac{b^{3n} - b^{2n+1} + 7 \cdot b^{2n} + b^{n+1} - 7 \cdot b^n - 1}{3(b-1)}. Observe that as n is large, the cube root of this quantity is close to: \frac{b^n - \frac{b-7}{3}}{(3(b-1))^{\frac13}}. \qquad (1) It's easy to see from here that 3(b-1) must be a cube, and so b = 9k^3 + 1. Plugging this in, we know that for large enough n, the quantity in question must exactly be the cube of (1). This easily implies that b = 10, we're done. \square