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PUMaC 2011 · 数论（B 组） · 第 10 题

PUMaC 2011 — Number Theory (Division B) — Problem 10

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

Number Theory B

  1. [ 3 ] What is the largest prime factor of 7999488?
  2. [ 3 ] Two robots are programmed to communicate numbers using different bases. The first robot states: “I communicate in base 10, which interestingly is a perfect square. You communicate in base 16, which is not a perfect square.” The second robot states: “I find it more interesting that the sum of our bases is the factorial of an integer.” The second robot is referring to the factorial of which integer?
  3. [ 4 ] The only prime factors of an integer n are 2 and 3. If the sum of the divisors of n (including itself) is 1815, find n .
  4. [ 4 ] What is the largest positive integer n < 1000 for which there is a positive integer m satisfying lcm( m, n ) = 3 m × gcd( m, n )? p p
  5. [ 5 ] Find the sum of all primes p such that 7 − 6 + 2 is divisible by 43.
  6. [ 6 ] For how many ordered triplets of three positive integers is it true that their product is four more than twice their sum?
  7. [ 7 ] Find the sum of all positive integers k with k ≤ 1000 such that there exists an integer n > k that satisfies ( ) 2 n k ∑ ∑ 3 i = 3 i . i = k +1 i =1
  8. [ 8 ] Let d ( n ) denote the number of divisors of n (including itself). You are given that ∞ 2 ∑ 1 π = . 2 n 6 n =1 Find p (6), where p ( x ) is the unique polynomial with rational coefficients satisfying ∞ ∑ d ( n ) p ( π ) = . 2 n n =1 1
解析
  1. Thus, b − 2 and c − 2 are a pair of (positive) factors of 10, and so must be equal to 1 , 10 or 2 , 5. Thus, we get the solutions { a, b, c } = { 1 , 3 , 12 } and { 1 , 4 , 7 } , which can be permuted in 12 ways (6 for each solution). If a = 2, we get that 2 bc − 2 b − 2 c = 8, so bc − b − c + 1 = ( b − 1)( c − 1) = 5. Thus, b − 1 and c − 1 are a pair of factors of 5, and so must be equal to 1 , 5. This gives the solution { a, b, c } = { 2 , 2 , 6 } , which can be permuted in 3 ways. If a = 3, we get that 3 bc − 2 b − 2 c = 10, so 9 bc − 6 b − 6 c + 4 = (3 b − 2)(3 c − 2) = 34. Then 3 b − 2 , 3 c − 2 are a pair of factors of 34 and (3 b − 2 , 3 c − 2) = (1 , 34) , (2 , 17). In the first pair of solutions, we have b = 1 ≤ a = 3, and in the second pair of solutions, we do not get integer values for b, c . In total, we have 15 ordered triplets which satisfy the given conditions. n n k ∑ ∑ ∑ 2 2 2 2 2 2 n ( n +1) k ( k +1) k ( k +1) 3 3 3