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March 2014 PATTERNS vs INVARIANTS 142857 142857 2 × 142857 = 285714 142857 2 × 142857 = 285714 3 × 142857 = 428571 142857 2 × 142857 = 285714 3 × 142857 = 428571 4 × 142857 = 571428 142857 2 × 142857 = 285714 3 × 142857 = 428571 4 × 142857 = 571428 5 × 142857 = 714285 142857 2 × 142857 = 285714 3 × 142857 = 428571 4 × 142857 = 571428 5 × 142857 = 714285 6 × 142857 = 857142 142857 2 × 142857 = 285714 3 × 142857 = 428571 4 × 142857 = 571428 5 × 142857 = 714285 6 × 142857 = 857142 7 × 142857 = 999999 1 0.142857142857... 7 Black Holes Give me any whole number (# of even digits)(# of odd digits)(total number of digits) Repeat process For Example: 19500723 19500723 → # of even digits = 3 # of odd digits = 5 Total # of digits = 8 358 → 123 22221111 → 448 → 303 → 123 22222222 → 808 → 303 → 123 WHY ? ooe –––> 123 oeo –––> 123 eoo –––> 123 eee –––> 303 –––> 123 201320142015 → 7512 → 134 → 123 Take any 3-digit number. Rearrange it: Largest – Smallest 674: 764 – 467 = 303 303: 330 – 033 = 297 297: 972 – 279 = 693 693: 963 – 369 = 594 594: 954 – 459 = 495 738 873 – 378 = 495 954 – 459 = 495 215 512 – 125 = 396 963 – 369 = 594 954 – 459 = 495 123 321 – 123 = 198 981 – 189 = 792 972 – 279 = 693 963 – 369 = 594 954 – 459 = 495 1. What is so special about 495? 2. Is there any 3–digit number does not fall into this black hole? 3. Is there an upper limit of number of steps to get to this black hole? 4. Is there any type of numbers require more steps than other types? 5. Is there a black hole for 2–digit numbers? What about 4-digit numbers? What about 5-digit numbers? What about 6-digit numbers? Is there a pattern? Take 2–digit numbers. 87: 87 – 78 = 9 90: 90 – 09 = 81 81 – 18 = 63 63 – 36 = 27 72 – 27 = 45 54 – 45 = 9 80: 80 – 08 = 72 72 – 27 = 45 54 – 45 = 9 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? What about 4–digit numbers? 2341: 4321 – 1234 = 3087 8730 – 0378 = 8352 8532 – 2358 = 6174 7641 – 1467 = 6174 Is 6174 the Black Hole for 4–digit numbers? What about 5–digit numbers? 12345: 54321 – 12345 = 41976 97641 – 14679 = 82962 98622 – 22689 = 75933 97533 – 33579 = 63954 96543 – 34569 = 61974 97641 – 14679 = 82962 Is 82962 the Black Hole for 5–digit numbers? Try 22220. 22220: 22220 – 02222 = 19998 99981 – 18999 = 80982 98820 – 02889 = 95931 99531 – 13599 = 85932 98532 – 23589 = 74943 97443 – 34479 = 62964 96642 – 24669 = 71973 97731 – 13779 = 83952 What happened? No Black Hole for 5–digit numbers? 22220: 22220 – 02222 = 19998 99981 – 18999 = 80982 98820 – 02889 = 95931 99531 – 13599 = 85932 98532 – 23589 = 74943 97443 – 34479 = 62964 96642 – 24669 = 71973 97731 – 13779 = 83952 98532 – 23589 = 74943 However, 74943 is different from 82962. Could it be that 5–digit numbers have more than one Black Holes? What about 6–digit, 7–digit, …, n–digit numbers? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18… 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18… 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 4 9 16 25 36 49 64 81… 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 4 9 16 25 36 49 64 81… 1 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 … 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 … 3 7 12 19 27 37 48 61 75 91… 8 27 64 125 216… 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 1 4 9 16 25 36 49 64 81… 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 … 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 … 1 3 712 19 27 37 48 61 75 91… 1 8 27 64 125 216… 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 … 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 … 1 3 6 11 17 24 33 43 54 67 81… 1 4 15 32 65 108 175 256… 1 16 81 256… Can you generalize this process and extend it? Is it th th same true for 5 power or 6 power or nth power? Give me any number n. Use the following process: n = odd 3 n +1 n = even n /2 Give me any number n. Use the following process: n = odd 3n+1 n = even n/2 n=13…40, 20, 10, 5, 16, 8, 4, 2, 1 n=37…112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13,…,1 n=101…304, 152, 76, 38, 19, 58, 29, 88, 44, 22, 11,…,1 1. Does it always go to 1? 2. In how many steps? 3. What kind of numbers require more step? 4. Is there a maximum number of steps? 5. Is there a pattern to the number of steps required? 1=0 2=1 3=7 4=2 5=5 6=8 7 = 16 8=3 9 = 19 10 = 6 1=0 2=1 3=7 4=2 5=5 6=8 7 = 16 8=3 9 = 19 10 = 6 11 = 14 12 = 9 13 = 9 14 = 17 15 = ? 16 = 4 17 = 12 18 = 20 19 = ? 20 = 7 21 = 6 22 = 15 23 = ? 24 = 10 25 = ? 26 = 10 27 = ? 28 = 18 29 = 17 30 = ? 199 is a prime Rearrange 199 into 919 (a prime). Rearrange 199 into 991 (a prime). 199 is called a “Permutable” or “Absolute” prime How many permutable primes? 1-digit: 2-digit: 3-digit: 1-digit: 4 (2, 3, 5, 7) 2-digit: 9 (11, 13, 17, 31, 37, 71, 73, 79, 97) 5 (11, 13, 17, 37, 79) 3-digit: 8 (113, 131, 199, 311, 337, 733, 919, 991) 3 (113, 199, 337) 19-digit: 1 (R19 = 11111…111) 23-digit: 1 (R23) 317-digit: 1 (R317) 1031-digit: 1 (R1031) 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 digits3 < 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) 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. “Unprimable” Numbers = A composite number that changes any digit and still composite. What is the smallest “unprimable” number? Example: 200 For 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. 1. There are infinitely many prime numbers (Proved by Euclid about 2000 years ago). 2. There are infinitely many twin prime numbers (still not yet proved). 3. There 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.) There are some numbers that are comparable to prime numbers. The great late Indian mathematician Ramanujan had considered some numbers called “highly composite” numbers. Prime numbers have the minimum number of factors…only two…1 and itself. Highly composite numbers have the maximum number of factors. 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. 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. 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 What about 12? 12 has 6 distinct factors …. Yes What is the next highly composite number? 24 has 8 distinct factors …. Yes What is the next highly composite number? 36 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. For instance: 4 = 22 1 1 6=2 x3 2 1 12 = 2 x 3 24 = 23 x 31 2 2 36 = 2 x 3 4 = 22 6 = 21 x 31 12 = 22 x 31 24 = 23 x 31 36 = 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. For example: 332,640 = 25 x 33 x 51 x 71 x 111 43,243,200 = 26 x 33 x 52 x 71 x 111 x 131 2,248,776,129,600 = 26 x 33 x 52 x 72 x 111 x 131 x 171 x 191 x 231 He wrote his thesis on the proof of this to get his college degree from Cambridge. Pythagoras had considered a category of numbers with very interesting properties that are either very difficult to prove or not yet proven….“friendly numbers”. 220 and 284 Factors 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. 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. The second smallest pair of friendly numbers was not discovered until 1866 by a 16 years old Italian schoolboy. 1184 and 1210. Now, 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. More Patterns. What is so important about n2+n+17 ? n2+n+17 n=1 n=2 n=3 n=4 12+1+17 = 19 22+2+17 = 23 32+3+17 = 29 42+4+17 = 37 a prime a prime a prime a prime n=5 n=6 n=7 n=8 n=9 n=10 n=11 n=12 52+5+17 = 47 62+6+17 = 59 72+7+17 = 73 82+8+17 = 89 92+9+17 = 107 2 10 +10+17 = 127 2 11 +11+17 = 149 2 12 +12+17 = 173 a prime a prime a prime a prime a prime a prime a prime a prime You may be tempted to think that you have found a formula that produces only prime numbers. n=13 n=14 n=15 n=16 n=17 132+13+17 = 199 142+14+17 = 227 152+15+17 = 257 162+16+17 = 289 172+17+17 = 323 a prime a prime a prime a prime??? a prime??? Of 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) Euler came up with another 2 formula n +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. 28 ÷ 7 = ? 28 ÷ 7 = 13 13 7 | 28 7 21 21 Check: 13 x 7 21 7 28 Another check by adding: 13 13 13 13 13 13 + 13 Peasant’s Multiplication 37 × 28 = 1036 37 18 9 4 2 1 28 56 112 224 448 896 37 × 28 = 28 + 112 + 896 = 1036 Why does it work? 28 = 111002 = 1×24 + 1×23 + 1×22 + 0×21 + 0×20 37 = 1001012 = 1×25 + 0×24 + 0×23 + 1×22 + 0×21 + 1×20 = 32 + 4 + 1 37 18 9 4 2 1 28 56 112 224 448 896 20 × 28 22 × 28 25 × 28 37 × 28 = (32 + 4 + 1) × 28 = (25 + 22 + 20) × 28 = 896 + 112 + 28) = 1036 What is the sum of all three angles of any triangle? What is the sum of all three angles of any triangle? 180° What is the sum of all four angles of any quadralateral? What is the sum of all four angles of any quadralateral? 360° What is the sum of all five angles of any pentagon? What is the sum of all five angles of any pentagon? 540° In general, what is the sum of all n angles of any n-polygon? In general, what is the sum of all n angles of any n-polygon? In general, what is the sum of all n angles of any n-polygon? 180(n–2)° How do you prove that? 180(n–2)° Interior angle vs Exterior angle What is the sum of all three exteror angles of any triangle? What is the sum of all n exterior angles of any n-polygon? New Project: Best way to build a high speed rail train to go from San Francisco to New York. It takes about 42 minutes to go from anywhere to anywhere. 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. What if there are 2,000,000 points? What if there are 2,000,000 points in 3–space? Can we find a plane that separates them into 2 sets of 1,000,000 each? Monte Hall’s “Let’s Make a Deal” Problem Behind three doors. One of them has a car and each of the other two has a goat. You chose Door #1. If the host asks you if you want to change your mind and switch your choice to another door, would you? If 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? If 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? Not Switch #1 #2 #3 Outcome Car Goat Goat Win Goat Car Goat Lose Goat Goat Car Lose Probability of winning = 1/3 Switch #1 #2 #3 Outcome Car Goat Goat Lose Goat Car Goat Win Goat Goat Car Win Probability of winning = 2/3 Another 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). 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. However, 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. So 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. 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? Obviously a party of 3 is not enough. Obviously a party of 3 is not enough. A–B C A party of 4 is also not enough. A party of 4 is also not enough. A–B–C–D A party of 5 is not enough. A party of 5 is not enough. A–B–D \ / C–E A 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. / 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. 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. Another 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. 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. One 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. π π ≈ 3.1415926535897… π π ≈ 3.1415926535897… Pi Day: March 14, 2015 9:26:53 morning. Facts About π 1. π was calculated to over 10 trillion digits (1013). 2. Record for most number of digits memorized is over 67,000. 3. π is irrational (1761). 4. π is transcendental (1882). 5. 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). 6. A sequence of six consecutive 9’s appeared begins at the 762nd decimal place (Feynman Point). Facts About π Six 9’s starts at 762. Next six 9’s starts at 193,034. Six 8’s starts at 222,299. Six 0’s starts at 1,699,927. Six 6’s starts at 45,681,781. One 9 – 5 Two 9’s – 44 Three 9’s – 762 Four 9’s – 762 Five 9’s – 762 Six 9’s – 762 Seven 9’s – 1,722,776 Eight 9’s – 36,356,642 Nine 9’s – 564,665,206 Joe’s Accomplishments Joe lived over 1550 years ago. Joe calculated 3.1415926 < π < 3.1415927 Joe used two fractions to approximate π 22/7 = 3.142857143… as coarse ratio 355/113 = 3.14159292035… as refined ratio 22/7 was used 2000 years ago by Greek’s Archimedes Archimedes also calculated 223/72 < π < 22/7 3.140845… < π < 3.142857… By slight increase to the denominators from 7 and 72 to Joe’s 113, π’s accuracy increases from 2 to 6 decimals What So Good About 355/113? 1. |31415926/10000000 – π| < 0.00000005 gets 7 decimal accuracy. 31415926/10000000 = 15707963/5000000 gets only 1 more decimal accuracy than 355/113, but 5,000,000 is so much larger than 113. 2. 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 What So Good About 355/113? 4. Even denominator less than 10000 cannot get better estimate than 113 5. In fact, any denominator less than 16604 cannot be better. What so good about 355/113? Why 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 < 0.00000026677 –0.00000026677 < π – q/p < 0.00000026677 Add –0.00000026677 < 355/113 – q/p < 2×0.00000026677 |355/113 – q/p| = |(355p–113q)/113p| < 2×0.00000026677 Since q/p ≠ 355/113, so |355p – 113q| > 0. But p & q are integers, so |355p–113q| > 1. So, 1/113p ≤ |(355p–113q)/113p| < 2×0.00000026677 p > 1/(113×2×0.00000026677) > 16586. In fact, 52163/16604 = 3.141592387… 355/113 = 3.1415922035… π = 3.1415926535897… Using 16604 gives us 6 decimal accuracy just like 113 but 16604 is more than 100 times larger than 113 How 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. π = 3 + 0.141592653… = 3 + a1 = 3 + 1/(1/ a1) 1/a1 = 1/0.141592653 = 7.062513305… = 7 + a2 So, π = 3 + a1 = 3 + 1/(1/ a1) = 3 + 1/(7+a2) ≈ 3 + 1/7 = 22/7 1/a2 = 1/0.062513305 = 15.99659454…= 15 + a3 So, π = 3 + 1/(7+(1/(15+a3))) ≈ 3 + 1/(7+1/15) = 333/106 1/a3 = 1/0.99659454 = 1.003417097…= 1 + a4 So, π = 3 + 1/(7+(1/(15+1/(1+a4)))) ≈ 3 + 1/(7+1/(15+1)) = 355/113 One More Step: 1/a4 = 1/0.003417097 = 292.6460677…= 292 + a5 So, π = 3 + 1/(7+(1/(15+1/(1+(292+a5))))) ≈ 3 + 1/(7+1/(15+1/(1+1/292))) = 103993/33102 However, 103993/33102 = 3.141592653011900… and π = 3.1415926535897… which is 9 decimals accuracy but 33102 is almost 300 times larger than 113. Also, 355/113 is easy to remember: 113|355 Note that 2 = 1.414213562… 1 = 1 2 1 2 1 1 2 2 And the Golden Ratio is: ᵠ= = 1 5 2 1.6180339887… 1 1 = 1 1 1 1 1 1 1 What would be good fractions to approximate 2 and ᵠ Note that 2 = 1.414213562… = 1 1 2 1 2 1 1 2 2 1, 4/3, 10/7, 24/17, 58/41, 140/99, 338/239, 816/577, … 1+1 = 2 1 4 1 1.33333... 1 2 3 1 10 1 1.42857142857... 4 7 1 3 1 24 1 1.4117647... 10 17 1 7 1 58 1 1.4146341463..... 24 41 1 17 ᵠ = 1.6180339887… = 1 1 1 1 1 1 1 1 1 1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, … 1+1 = 2 1 3 1 1.5 2 2 1 5 1 1.66666... 3 3 2 1 8 1 1.6 5 5 3 1 13 1 1.625 8 8 1729 Hardy–Ramanujan Number 3 1 3 12 1729 = + 3 3 = 9 + 10 (1) 1729 is the smallest natural number that has this property. (2) 1729 is a “near–miss” for n n n z =x +y . 3 3 3 1729 = 12 +1 = 9 + 10 Is there a number that can be written as the sum of 4 cubes in at least 10 different ways? It is hard to find such number but it is relatively easy to show that this number exists. Note: 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? Order 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.