WHAT COOL MATH! JULY CURRICULUM INSPIRATIONS: Math for America DC:

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1 CURRICULUM INSPIRATIONS: Math for America DC: Iovative Olie Courses: Tato Tidbits: WHAT COOL MATH! CURIOUS MATHEMATICS FOR FUN AND JOY PROMOTIONAL CORNER: Have you a evet, a workshop, a website, some materials you would like to share with the world? Let me kow! If the work is about deep ad joyous ad real mathematical doig I would be delighted to metio it here. *** For some reaso i my high-school teachig career I was assiged to teach multiple sectios of geometry each ad every year. (Geometry. Geometry. Always geometry.) Fortuately I was give the freedom to teach the subject i my ow Tato-way. Ad ow that way is a course with THE GREAT COURSES compay! Check out es/course_detail.aspx?cid0 JULY 0 PUZZLER: a) Give a example of oe hudred cosecutive itegers with each a composite umber. b) Show that there exist oe-hudred cosecutive itegers each divisible by at least oe-hudred distict primes. WARNING: Some parts of this essay assume familiarity with ifiite sums ad results from calculus.

2 James Tato 0 AN UNUSUAL QUESTION Here are the prime factorizatios of the first te positive itegers: L L L L L L L L L L The average expoet of the prime is: The average expoet of the prime is 0., of 5 its 0., of 7its 0., ad its zero thereafter. So the average factorizatio for these te itegers is: L COUNTS OF SQUARE FACTORS Cosider the set of square umbers: S,,9,,5,.... { } Side Commet: I like the square umbers because the ifiite sum of their reciprocals, L, equals 9. I fid this coectio betwee squares ad circles delightfully quirky. (See for a proof of this astoudig result.) Let Sq( ) deote the umber of square factors possesses. For example, Sq 8 because 8 has just the squares ad 9 as factors, Sq ( 7), ad Sq ( 00). The followig table shows the square umber factors of the first 0 itegers. I it we have that Sq( ) is the umber of Xs i the th colum. (The 8th colum Sq 8.) cotais two X s ad (This equals about.59 if that meas aythig.) Exercise: What is the average prime factorizatio of the first twety positive itegers? The first oe hudred? Here s our questio: What is the average prime factorizatio of all the positive itegers? Summig over all twety colums ad dividig by 0 gives the average cout of square factors each of the umbers through 0 possesses. This is the total umber of X s i the table divided by 0: Ave Square ( 0) ( ) + + L+ Sq( 0) Sq Sq ad 0

3 James Tato 0 We could cout the total umber of X s i the table summig by rows of X s istead. This would give us a secod way to compute the average cout of square factors. How may X s are there i a particular row of the table? There are two multiples of ie 9 ad 9 each less tha or equal to 0 ad so there are two X s o the row for 9. I geeral, for a table with colums, the umber of X s i the row for the umber a is the largest k so that ka is less tha or equal to : ka. That is, k is the largest iteger less tha or equal to. This is a, the quatity rouded dow to the a earest iteger. ( 0 0,, ad 0 5, for example.) The total umber of X s i the table of twety colums for the first 0itegers is: L L L 8 just as we have see before. But this secod approach gives us a geeral formula for the average cout of square factors of the first itegers: AveSquare L (Notice that oly the first few terms i this sum are o-zero: 0 as soo as a>.) Questio: differs from oly by a a fractio smaller tha. Is it reasoable to write AveSquare L L L 9 5? Could it be that, o average, a umber possesses square factors? Exercise: Did you compute the average cout of square factors of the first 00 itegers? Is that aswer close to.5? ESTIMATING THE ERROR: Write a for the fractioal part of a. 0 0 (For example, 0..., ad 0.) We have: 5 a + a. ad

4 James Tato 0 Let k be the umber of o-zero terms i the sum: AveSquare L This k is give by the largest iteger satisfyig k. That is, k. So we ca write: AveSquare L k L+ k k L+ 9 k k fractios betwee 0 ad L+ Error 9 k kis the largest iteger just uder, or equal to (ad so k grows as is chose to be larger ad larger) ad Error the k fractios betwee 0 ad is the error term quatifyig how AveSquare differs from the sum of reciprocals L +. 9 k Sice each fractio is betwee 0 ad we have: k 0 Error. k Also, which goes to zero as grows. This meas that the error term teds to zero. We have: lim AveSquare lim L+ Error 9 k L We ca iterpret this as ideed sayig: O average, over all itegers, each coutig umber possesses square factors. IN GENERAL Oe ca repeat the previous argumet for ay set of coutig umbers: We ca say: {,,,...} S a a a. Let S( ) be the umber of etries of S S that are. If 0 as the, o average, each iteger possesses + + +L a a a factors from the set S. ad

5 James Tato 0 Example: If S is the set of square umbers, the S, ad S 0 as grows. Example: If S is the set of fourth powers, S the S ad 0 as. O average, each iteger has L fourthpower factors. Example: O average each iteger has L cube factors No oe o this plaet curretly kows the value of this sum of cube reciprocals. Research Corer : For world fame, fid some other meas to compute the average cout of cube factors of the itegers ad hece evaluate the sum: ζ ( ) L Example: The set S could be a fiite. For S, the for we istace, if { } S have 0 as grows. We coclude that, o average, a iteger possesses as a factor of the time. Questio: How do you iterpret our result S,? for the set { } Exercise: Show that, o average, a iteger possesses two triagular umbers as factors. (The triagular umbers are the umbers,,,0,5,, 8,...) THE MOST AVERAGE PRIME FACTORIZATION: Cosider the set: S,,8,,,, L, { } the o-trivial powers of two. For a iteger, let s fid S( ) for this set. This is the umber of powers of two that are less tha or equal to. Fidig this cout is equivalet to fidig the largest iteger ksatisfyig: k. That is, we seek the largest k less tha or equal to log. Thus S log. log S log Now 0 as grows. (Use L Hopital s rule from calculus.) Thus by our result we ca say that, o average, each iteger possesses: L powers of two as factors. The geometric series formula + x + x + x + L x shows that this sum is: L Thus, o average, each iteger has oe power of two as a factor. ad

6 James Tato 0 But let s iterpret matters this way: If a umber has prime factorizatio: a b c 5 7 d L the it has a powers of two as factors. The average value of a must be. I the same way, lookig at the o-trivial S,9, 7,8,... shows powers of three { } that the average value of the expoet b must be: L The average value expoet c is: L Ad i geeral, the average expoet of a prime p is: L p p p p p p p We have: The most average prime factorizatio is: L Exercise: For each iteger cosider the largest value k so that k! divides. Show that, o average, this value equals: e RESEARCH: I ve bee playig with this little result for several years ow. I would be very iterested i ay hearig ay thoughts you might have o this topic. Research Corer : Cosider the expoet of the largest power of te that divides each iteger. Show that, o average, this expoet is / 9. (So the average power of 9 te factor of a iteger is 0? How does this statemet gibe with the most average prime factorizatio?) Research Corer : Show that, o average, a umber has l more odd factors tha eve factors. Use + + L l. (Sice we are ow allowig egative factors, oe eeds to refie delicately the aalysis of the error term.) Research Corer : We showed that if S 0 as grows, the the average cout of factors from the set S is the sum of the reciprocals of the itegers i S. Is this the oly possible coditio that guaratees this result? Is there a problem with our result if the sum of reciprocals happes to be ifiite? ad

7 James Tato 0 Research Corer : For each iteger, look at a b c d e f its prime factorizatio 5 7 L ad work out the product a b c L, but oly iclude the o-zero terms i this product. Is the average value of this product? a b c The product L? THE OPENING PUZZLER: Cosider the oe-hudred cosecutive umbers 0! + 0! + M 0! + 0 The first umber is divisible by, the secod by, the third by, all the way up to the 00 th divisible by 0. We have a list of oe-hudred cosecutive composite itegers. Ufortuately the oly way I kow to aswer the secod puzzler is to make use of a famous result from umber theory. Let m be the product of the first oehudred primes, m the product of the ext oe-hudred primes, m the product of the hudred after that, ad so o. By the Chiese Remaider Theorem there is sure to be a iteger N that is simultaeously less tha a multiple of m less tha a multiple of m less tha a multiple of m M 00 less tha a multiple of m 00. This meas that each of the itegers N +, N +, N +, K, N + 00 is divisible by at least oe-hudred distict primes. 0 James Tato tato.math@gmail.com ad