PUMaC 2009 · 数论（B 组） · 第 7 题

PUMaC 2009 — Number Theory (Division B) — Problem 7

Suppose that for some positive integer n , the first two digits of 5 and 2 are identical. Suppose the first two digits are a and b in this order. Find the two-digit number ab .

解析

Suppose that for some positive integer n , the first two digits of 5 and 2 are identical. Find the number formed by these two digits. Solution. 31. Suppose a is the number formed by the two digits. From the condition, we k n k l n l must have 10 a < 2 < 10 ( a + 1) and 10 a < 5 < 10 ( a + 1) for some positive integers k and l . Then, we can multiply these two equations together to obtain k + l 2 n k + l 2 10 a < 10 < 10 ( a + 1) . 2 2 Now, note that a is a two-digit number, and so 10 ≤ a ≤ 99. so, 10 ≤ a , and in fact, 2 2 4 ( a + 1) ≤ 100 = 10 . So, our equation now has two additional bounds, and becomes k + l +2 k + l 2 n k + l 2 k + l +4 10 ≤ 10 a < 10 < 10 ( a + 1) ≤ 10 2 3 and hence it follows that n = k + l + 3. Once we have this, it becomes easy to see a < 10 < √ 2 ( a + 1) , i.e. a < 1000 < a + 1, whence a = 31.