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PUMaC 2012 · 数论（A 组） · 第 11 题

PUMaC 2012 — Number Theory (Division A) — Problem 11

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

Number Theory A

  1. [ 3 ] Albert has a very large bag of candies and he wants to share all of it with his friends. At first, he splits the candies evenly amongst his 20 friends and himself and he finds that there are five left over. Ante arrives, and they redistribute the candies evenly again. This time, there are three left over. If the bag contains over 500 candies, what is the fewest number of candies the bag can contain? 2012
  2. [ 3 ] How many ways can 2 be expressed as the sum of four (not necessarily distinct) positive squares? x x k k
  3. [ 4 ] Let the sequence { x } be defined by x ∈ { 5 , 7 } and, for k ≥ 1 , x ∈ { 5 , 7 } . For n 1 k +1 5 7 5 7 5 7 5 7 5 5 7 7 5 5 7 7 example, the possible values of x are 5 , 5 , 5 , 5 , 7 , 7 , 7 , and 7 . Determine the 3 sum of all possible values for the last two digits of x . 2012
  4. [ 4 ] Find the sum of all possible sums a + b where a and b are nonnegative integers such that a b 4 + 2 + 5 is a perfect square.
  5. [ 5 ] Call a positive integer x a leader if there exists a positive integer n such that the decimal n 3 representation of x starts (not ends) with 2012. For example, 586 is a leader since 586 =
  6. How many leaders are there in the set { 1 , 2 , 3 , ..., 2012 } ? p n − 1
  7. [ 6 ] Let p = 2012 and p = 2012 for n > 1 . Find the largest integer k such that p − p 1 n 2012 2011 k is divisible by 2011 .
  8. [ 7 ] Let a , b , and c be positive integers satisfying 4 2 2 4 a + a b + b = 9633 2 2 2 2 5 2 a + a b + 2 b + c = 3605 . What is the sum of all distinct values of a + b + c ?
  9. [ 8 ] Find the largest possible sum m + n for positive integers m, n ≤ 100 such that m + 1 ≡ n 2 − 1 m − 1 n a 3 (mod 4) and there exists a prime number p and nonnegative integer a such = m + p . m − 1 1
解析
  1. Thus, the only possible residues for c are 0 and 1, so the possible residues for c are 0 , ± 1. The only possible residues for squares modulo 11 are 0, 1, 3, 4, 5, and 9. Considering the right-hand side modulo 11, we find that the residue is 6, so ( ) 2 5 2 2 c ≡ 1 (mod 11) , a + b + 1 ≡ 5 (mod 11) . Going back to the second given equation, we see that we only have to check for c = 1 , 2 , 3 , 4 , 5. 5 But 2 ≡ − 1 (mod 11), so we only consider c = 1 , 3 , 4 , 5. Considering our summed equation now, we see that c cannot be 1 because the last digit of a square cannot be 8. We can check 2 2 that c = 3 is the only possible solution, with a + b + 1 = 114. The only solutions are a = 7, b = 8 and a = 8, b = 7, so a + b + c = 18 .