2001 AIME I 第 8 题

If $N$ has base-$7$ digits $d_2 d_1 d_0,$ the condition says $100 d_2 + 10 d_1 + d_0$ $= 2(49 d_2 + 7 d_1 + d_0)$

That simplifies to $2 d_2 = 4 d_1 + d_0.$ A fourth digit is impossible because $10^3 - 2 \cdot 7^3 = 314$ is bigger than the other terms could balance.

Suppose $N$ has base-$7$ digits $d_k \ldots d_1 d_0.$ The condition is $\sum d_i \, 10^i = 2 \sum d_i \, 7^i,$ that is $\sum d_i \,(10^i - 2 \cdot 7^i) = 0.$ The coefficients $10^i - 2 \cdot 7^i$ for $i = 0, 1, 2, 3$ are $-1,$ $-4,$ $2,$ $314.$ Every coefficient for $i\ge3$ is positive and at least $314.$ Thus if any digit in a place $i\ge3$ were nonzero, its positive contribution could not be canceled by the two negative terms, whose total magnitude is at most $4\cdot6+6=30.$ So $N$ has at most three base-$7$ digits.

For three digits the condition reads $2 d_2 = 4 d_1 + d_0.$ To maximize $N = 49 d_2 + 7 d_1 + d_0,$ take $d_2 = 6,$ so $4 d_1 + d_0 = 12;$ the largest value of $7 d_1 + d_0$ comes from $d_1 = 3,$ $d_0 = 0.$

Thus $N = 49 \cdot 6 + 7 \cdot 3 = 315,$ whose base-$7$ representation is $630 = 2 \cdot 315.$