18What is so special about 495? Is there any 3–digit number does not fall into this black hole?Is there an upper limit of number of steps to get to this black hole?Is there any type of numbers require more steps than other types?Is there a black hole for 2–digit numbers?
19What about 4-digit numbers. What about 5-digit numbers What about 4-digit numbers? What about 5-digit numbers? What about 6-digit numbers? Is there a pattern?
21Is “9” a Black Hole. Is there any other Black Hole for 2–digit numbers Is “9” a Black Hole? Is there any other Black Hole for 2–digit numbers? Which 2–digit number takes the longest to reach the Black Hole?
22What about 4–digit numbers What about 4–digit numbers? 2341: 4321 – 1234 = – 0378 = – 2358 = – 1467 = 6174 Is 6174 the Black Hole for 4–digit numbers?
23What about 5–digit numbers What about 5–digit numbers? 12345: – = – = – = – = – = Is the Black Hole for 5–digit numbers?
24Try : – = – = – = – = – = – = – = – = What happened? No Black Hole for 5–digit numbers?
2522220: – = – = – = – = – = – = – = – = However, is different from Could it be that 5–digit numbers have more than one Black Holes?
26What about 6–digit, 7–digit, …, n–digit numbers?
41Facts about Permutable Primes: 1 Facts about Permutable Primes: 1. Must compose from digits 1, 3, 7, and 9 if it is not 1-digit. 2. There are no permutable primes with number of digits 3 < n < 6×10175 except those that are composed all 1’s. (Not Proven) 3. There are no permutable primes with only digit 1 other than what are listed in the previous slide.(Not Proven)
42Facts about Permutable Primes: 4 Facts about Permutable Primes: 4. No permutable prime exists which contains 3 different numbers of the 4 digits 1, 3, 7, 9. (For example: 731 = 17 × 43) 5. No permutable primes that are composed of 2 or more of each of the 4 digits 1, 3, 7, 9.
43= A composite number that changes any digit and still composite. “Unprimable” Numbers= A composite number that changesany digit and still composite.What is the smallest “unprimable” number?Example: 200
44For many generations, prime numbers have always had mystical and magical appeals to mathematicians. Dating back to Euclid’s days, he had already proved that there are infinitely many prime numbers. However, from what we had seen before, it is very difficult to prove even some of the simple properties about prime numbers.
45There are infinitely many prime numbers (Proved by Euclid about 2000 years ago). There are infinitely many twin prime numbers (still not yet proved).
46There is always at least one prime number exists between n and 2n where n is a natural number larger than 1. (Proved in 1850 by Chebyshev but he used very difficult and deep mathematics in his proof. However, Erdos, the second most prolific mathematicians in history, used a very simply proof to prove this result when he was a freshman in college.)
47There are some numbers that are comparable to prime numbers There are some numbers that are comparable to prime numbers. The great late Indian mathematician Ramanujan had considered some numbers called “highly composite” numbers.
48Prime numbers have the minimum number of factors…only two…1 and itself Prime numbers have the minimum number of factors…only two…1 and itself. Highly composite numbers have the maximum number of factors.
49Prime numbers have the minimum number of factors…only two…1 and itself Prime numbers have the minimum number of factors…only two…1 and itself. Highly composite numbers have the maximum number of factors. A composite number is Highly Composite if it has more distinct factors than any composite numbers before it.
504 has 3 distinct factors. 6 has 4 distinct factors 4 has 3 distinct factors. 6 has 4 distinct factors. 8 has 4 distinct factors. 9 has 3 distinct factors. 10 has 4 distinct factors.
514 has 3 distinct factors…. Yes 6 has 4 distinct factors… 4 has 3 distinct factors…. Yes 6 has 4 distinct factors…. Yes 8 has 4 distinct factors…. No 9 has 3 distinct factors…. No 10 has 4 distinct factors…. No
5312 has 6 distinct factors …. Yes What is the next highly composite number?
5424 has 8 distinct factors …. Yes What is the next highly composite number?
5536 has 9 distinct factors …. Yes Ramanujan made a list of all highly composite numbers up to 6,746,328,388,800. He examined the highly composite numbers expressed as product of primes.
56For instance: 4 = 22 6 = 21 x 31 12 = 22 x 31 24 = 23 x 31 36 = 22 x 32
574 = 22 6 = 21 x = 22 x = 23 x = 22 x 32 He discovered that the final exponent of all highly composite numbers, except for 4 and 36, is 1. Also, in the prime factorization of any highly composite number, the first exponent always equals or exceeds the second exponent and the second always equals or exceeds the third, and so on.
58For example:332,640 = 25 x 33 x 51 x 71 x 11143,243,200 = 26 x 33 x 52 x 71 x 111 x 1312,248,776,129,600 = 26 x 33 x 52 x 72 x 111x 131 x 171 x 191 x 231He wrote his thesis on the proof of this to get his college degree from Cambridge.
59Pythagoras had considered a category of numbers with very interesting properties that are either very difficult to prove or not yet proven….“friendly numbers”.
61Factors of 220 = 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110 with a sum of 284. Factors of 284 = 1, 2, 4, 71, 142 with a sum of 220.
62This was discovered about 2000 years ago by Greeks This was discovered about 2000 years ago by Greeks. The next pair did not get discovered until 1636 by Fermat. 17,296 and 18,416.
63The second smallest pair of friendly numbers was not discovered until 1866 by a 16 years old Italian schoolboy and 1210.
64Now, with the help of some powerful computers, hundreds of pair of friendly numbers had been discovered. Question: Are there infinitely many pairs of friendly numbers existed? Not yet proven.
65More Patterns. What is so important about n2+n+17 ?
66n2+n+17 n= = 19 a prime n= = 23 a prime n= = 29 a prime n= = 37 a prime
67n= = 47 a prime n= = 59 a prime n= = 73 a prime n= = 89 a prime n= = 107 a prime n= = 127 a prime n= = 149 a prime n= = 173 a prime
68You may be tempted to think that you have found a formula that produces only prime numbers. n= = 199 a prime n= = 227 a prime n= = 257 a prime n= = 289 a prime??? n= = 323 a prime???
69Of course, if you use logic plus patterns, you will see that n2+n+17 = (17m)2+(17m)+17 = = 17(17m2+m+1) not a prime when n is a multiple of 17 such as 17m. So 323 = (17)(19)
70Euler came up with another formula n2+n+41 Euler came up with another formula n2+n+41. It produces prime numbers from n=1 to 39 but it fails at n = 40 and 41. However, people proved that this formula is actually quite good generating prime numbers under 10 million a respectable 47.5% of the time.
89What is the sum of all three exteror angles of any triangle?
90What is the sum of all n exterior angles of any n-polygon?
91New Project: Best way to build a high speed rail train to go from San Francisco to New York.
92It takes about 42 minutes to go from anywhere to anywhere.
93Suppose there are 2 points on a plane Suppose there are 2 points on a plane. Can you find a line that separates these two points? Suppose there are 3 points on a plane. Can you find a line that has the same number of points on either side? Suppose there are 10 points. Suppose there are 11 points.
97Behind three doors. One of them has a car and each of the other two has a goat. You chose Door #1.
98If the host asks you if you want to change your mind and switch your choice to another door, would you?
99If the host opens the other two doors and shows you that there is a goat behind each of these two doors and asks you if you want to change your mind and switch your choice to another door, would you?
100If the host opens the one of the other two doors and shows you that there is a goat behind that door and asks you if you want to change your mind and switch your choice to another door, would you?
101Not Switch #1 #2 #3 Outcome Car Goat Goat Win Goat Car Goat Lose Goat Goat Car Lose Probability of winning = 1/3
102Switch #1 #2 #3 Outcome Car Goat Goat Lose Goat Car Goat Win Goat Goat Car Win Probability of winning = 2/3
103Another problem. During the old days, when people look up on the sky and see 8 stars lining up on a straight line, they think that special phenomenon must be set up artificially on purpose. The former great astronomer Carl Sagan said that this is really nothing special about it. He pointed out that, actually, if you have looked at larger enough of a sample, that phenomenon of 8 stars lining up happens everywhere. He is actually speaking of a very special topic in mathematics (one which is very popular among math competition problems).
104Example. Pick any two students here. You may find one boy and one girl Example. Pick any two students here. You may find one boy and one girl. As you keep picking out two students many times and suddenly you see both of the students are of the same sex. You are so amazed about this find that you proclaimed that someone must have done this for a purpose.
105However, upon examining this, you find that if you widen your pick … picking 3 students at a time, then the phenomenon of having two students of the same gender is nothing special. In fact, every time you do your pick of 3 students, two students of the same gender are among your pick.
106So picking 3 students at a time is the smallest number of students you can pick to guarantee this phenomenon happened. The first person who formalize this theory and process is Frank Ramsey who died in 1930 at the age of 26. We now called this theory the Ramsey Theory in honor of him. It turns out problems from Ramsey Theory are some of the most difficult problems in mathematics.
107Here is a classical Ramsey problem Here is a classical Ramsey problem. What is the minimum number of guests that need to be invited so that either at least 3 guests will know each other or at least 3 will be mutual strangers?
113A party of 5 is not enough. A – B – D \ / C – E
114A party of 6 is enough. Either A knows at least three of the other five or A does not know at least three of the other five.
115/ B Assume A knows B and C and D. A –– C \ D If B–C or C–D or B–D, then either ABC or ACD or ABD know each other.If not, then B, C, and D are total strangers.
116Assume A does not know B and C and D Assume A does not know B and C and D. If B does not know C or C does not know D or B does not know D, then ABC or ACD or ABD are totally strangers. If B knows C and D and C knows D, BCD know each other. If B knows C and D but C does not know D, then ACD are totally strangers.
117Another way to prove this is to list and examine all the combinations of relationships between those 6 people. It turns out these are a total of 32,768 possibilities. This brute force method does not provide any insight. Now if we increase the number of 3 persons to 4, brute force method is impossible to do because the large number of possibilities.
118What about 6? People think the least number is between 102 and 165. It has been proved that at least 18 people need to be invited to guarantee at least 4 persons all know each other or all strangers.What about 5? Nobody knows. People think the least number is between 43 and 49.What about 6? People think the least number is between 102 and 165.
119One of the largest doing with computers to solve math problems occurred in 1993 involving 110 computers running simultaneously to show that the minimum number of guests to invite to a party to guarantee at least 4 mutual friends or at least 5 mutually strangers is 25.
122Facts About π π was calculated to over 10 trillion digits (1013). Record for most number of digits memorized is over 67,000.π is irrational (1761).π is transcendental (1882).π is NOT normal – all possible sequence of digits appear equally often. That means one can use the digits from π to form random numbers (not proven).A sequence of six consecutive 9’s appeared begins at the 762nd decimal place (Feynman Point).
123Facts About π Six 9’s starts at 762. Next six 9’s starts at 193,034. One 9 – 5Two 9’s – 44Three 9’s – 762Four 9’s – 762Five 9’s – 762Six 9’s – 762Seven 9’s – 1,722,776Eight 9’s – 36,356,642Nine 9’s – 564,665,206
124Joe’s Accomplishments Joe lived over 1550 years ago. Joe calculated < π < Joe used two fractions to approximate π 22/7 = … as coarse ratio 355/113 = … as refined ratio 22/7 was used 2000 years ago by Greek’s Archimedes Archimedes also calculated 223/72 < π < 22/ … < π < … By slight increase to the denominators from 7 and 72 to Joe’s 113, π’s accuracy increases from 2 to 6 decimals
125What So Good About 355/113?1. | / – π| < gets 7 decimal accuracy / = / gets only 1 more decimal accuracy than 355/113, but 5,000,000 is so much larger than Any fraction with denominator less than 113 cannot get a better approximation of π. 3. Any fractional approximation of π with denominator less than 1000 cannot get better estimate than 113
126What so good about 355/113? What So Good About 355/113? 4. Even denominator less than cannot get better estimate than 1135. In fact, any denominator less than cannot be better.What so good about 355/113?
127Why 113 Is So Good?Let us try to find a fraction q/p ≠ 355/113 but gives about the same or better estimate.0 < 355/113 – π <– < π – q/p <Add– < 355/113 – q/p < 2×|355/113 – q/p| = |(355p–113q)/113p|< 2×
128Since q/p ≠ 355/113, so |355p – 113q| > 0 Since q/p ≠ 355/113, so |355p – 113q| > 0. But p & q are integers, so |355p–113q| > 1. So, 1/113p ≤ |(355p–113q)/113p| < 2× p > 1/(113×2× ) >
129In fact, 52163/16604 = … 355/113 = … π = … Using gives us 6 decimal accuracy just like 113 but is more than 100 times larger than 113
130How Did Joe Get 355/113?It is a mystery since Joe did not write it down or publish it on internet. Let us find one possible way that he could have done it. Remember: We are not trying to get a good estimate for π. We are trying to find a good fraction to estimate π.
1451729 is the smallest natural number that has this property. 1729 is a “near–miss” forzn = xn + yn.1729 = =
146Is there a number that can be written as the sum of 4 cubes in at least 10 different ways?
147It is hard to find such number but it is relatively easy to show that this number exists.
148Note: There are 1000 cubes that are less than or equal to 1,000,000,000. Is there a number that can be written as the sum of 4 cubes in at least 10 different ways?
149Order the first thousand cubes as a1, a2, a3, a4, …, a1000. There are 1000C4 = 1000×999×998×997/24 > 40×1,000,000,000 ways to pick out any 4 cubes from this sequence.Any sum of 4 cubes from this sequence cannot exceed 4×1,000,000,000.So there are at the most 4,000,000,000 values to choose from by these more than 40,000,000,000 sets of sum of 4 cubes. Therefore, there is at least one value (among these 4,000,000,000) that is equal to 10 or more sets of sum of four cubes.