This question was found last Saturday when I helped my brother find the logic question.
This question comes from the puzzles of mr. s and Mr. P, a professor at Stanford University in the United States.
Questions: questions about Mr. S and Mr. P
There are two natural numbers x, y, 2 <= x <= Y <= 99. Mr. S knows the sum of the two numbers, and Mr. P knows the product P of the two numbers, the two of them had the following conversation:
S
: I'm sure you don't know what these two numbers are, but I don't know either.
P:
When you say this, I will know what these two numbers are.
S:
Me too. Now I know.
Can you deduce the two numbers through their sessions? (Of course, Mr. S and Mr. P are both very smart)After seeing it, I immediately fell in love with it, but it was not hard to feel ashamed until today. It is not difficult to think logically. In fact, the first sentence of S was half understood in the evening. by the afternoon of the next day, the first sentence of S and the sentence of P were thoroughly understood, I should have done it at that time, but I ignored the second sentence of S. So I thought over the past few days that I was missing a condition, but I was not sure, if I can't think of it today, I read the question again and found that I used less than one condition. So I entered the second sentence of S, and then I thought and verified it. Next let's talk about my ideas (P, which is written in inferences, indicates the prime number, and P in uppercase is not the prime number ). 1. If S is the following number, according to 2 ≤ x ≤ y ≤ 99, Mr. S can directly know X and Y
- S = 4, x = 2, y = 2
- S = 5, x = 2, y = 3
- When S is 197, x = 98, y = 99
- When S is 198, x = 99, y = 99
According to the second half of Mr. S's First sentence, s cannot be any of the four numbers above. 2. If both X and Y are prime numbers, then Mr. P can easily calculate X and Y, for example, P = 481 = 13 × 37, when learning p. Therefore, we know from the first half of Mr. S's first sentence that s cannot be expressed as the sum of two prime numbers, that is, X and Y cannot be both prime numbers. For example, if S = 84, there may be X = 5, y = 79, then P may be P = 5 × 79 = 395, mr P can easily obtain X and Y. Therefore, if S is 84, Mr. s cannot be sure that Mr. P does not know X and Y, so
S
Is the number of sums that cannot be expressed as two prime numbers.
(A) You can reduce the value range of S. 3. If there is only one prime number in X or Y and the number is greater than 50, then Mr. P can still solve X and Y only when he knows p. Let p be decomposed into the product of the following prime numbers: P = p0p1p2... in Pn, PN is a prime number greater than 50. Because PN> 50 is used for any prime number P, there are PNP> 100. Therefore, there must be a number equal to PN in X and Y, the other number is equal to p0p1p2... pn-1. The minimum prime number greater than 50 is 53, so if 53 + 2 ≤ S ≤ 53 + 99, when 55 ≤ S ≤ 152, mr. S is not certain that Mr. P cannot calculate X and Y. The maximum prime number smaller than 99 is 97, so when 97 + 2 ≤ S ≤ 97 + 99, that is, when 99 ≤ S ≤ 196, mr. S is not certain that Mr. P cannot calculate X and Y. Therefore, according to Mr. S's First sentence, the value range of S is: 6 ≤ S ≤ 54 (B) 4. according to the verification of godebach's conjecture, we can easily determine that all the even numbers in the range [6, 54] can be expressed as the sum of two prime numbers (if you are not afraid of trouble, you can also verify it yourself ), S is an odd number in the range [7, 53. Because 2 is a prime number, there are some odd numbers that can also be expressed as the sum of two prime numbers, finding out the prime numbers in [3, 53], adding them to 2, and then removing the range [6, 54] an odd number expressed as the sum of two prime numbers: 7 = 2 + 59 = 2 + 713 = 2 + 1115 = 2 + 1319 = 2 + 1721 = 2 + 1925 = 2 + 2331 = 2 + 29; 33 = 2 + 3139 = 2 + 3743 = 2 + 4145 = 2 + 4349 = 2 + 47. Therefore, according to Mr. S's First sentence, it can be determined that the value of S is only one of the following: 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53
Set set a = {11,17, 35, 51, 53}, then
S
ε(C) 5. according to Mr. P's sentence, Mr. P obtained the conclusion c Based on Mr. S's First sentence, and calculated x and y based on known p.
Let the prime number of P be decomposed:
P = p0p1p2... pi... Pn (n
> 2)
There is a unique
X = p0p1p2... Pi
Y = PI + 1... PN
Yes
X + y = s
εThe conclusion (condition) is named D, and the P meeting the condition forms the Set B. For example, if p = 130 is set, the prime number decomposition for P is as follows: P = 2 × 5 × 13. At this time, the solutions of X and Y may be) or (130), but only () this group of solutions satisfied: 10 + 13 = 23, a so when P =, according to the Conclusion C, Mr. P can solve X and Y. But when P =, 28 ,...... Mr. P can also find X and Y. 6. According to Mr. P's sentence, Mr. S has also solved X and Y, that is, according to condition d, S has solved X and Y. That is, in S, the sum expression s = x0 + y0s = X1 + y1s = x2 + y2 ...... S = xn + YN only has a group of (XI, Yi) product P meet the condition d, that is, meet p = xi x Yi P in B so that Mr. S can solve X and Y. Verify the elements in set a one by one: When S = 11, S = 2 + 9, P = 2 × 9 = 18, B S = 3 + 8, P = 3 × 8 = 24 ε B s = 4 + 7, P = 4 × 7 = 28 ε B when S = 17, only S = 4 + 13, P = 4 × 13 = 52, B when S = 23, S = 4 + 19, P = 4 × 19 = 76, B S = 7 + 16, P = 7 × 16 = 112 records B s = 10 + 13, P = 10 × 13 = 130 records B ...... In short, only when S = 17, there is a unique group (4, 13) that satisfies S = 4 + 13 = 17, a P = 4 × 13 = 52, B.
X = 4, y = 13I wrote a program to verify this verification process.
