The Mathematical Circus project was created in 2011 by the LUDUS association with the main goal of promoting the interest and motivation for learning mathematics. The Mathematical Circus team performs mathematical magic shows where complementary skills are merged to produce a high intervention capacity within a wide geographical range in Portugal. Some tricks are performed by a mathematical clown, a unique
character in the circus world, who brings together the usual foolish characteristics of a clown and the rigorous mathematical knowledge. In this paper we will describe and explain some of the tricks of the Mathematical Circus repertoire involving the clown.info:eu-repo/semantics/publishedVersio

Almeida, P.

Hall, A.

Pais, S.

Vettori, P.

English

IC-online

Mathematical Circus Clown TricksA. Hall∗, P. Almeida∗, P. Vettori∗and S. Pais†3 de Março de 2019ResumoThe Mathematical Circus project was created in 2011 by the LUDUSassociation with the main goal of promoting the interest and motiva-tion for learning mathematics. The Mathematical Circus team performsmathematical magic shows where complementary skills are merged to pro-duce a high intervention capacity within a wide geographical range inPortugal. Some tricks are performed by a mathematical clown, a uniquecharacter in the circus world, who brings together the usual foolish cha-racteristics of a clown and the rigorous mathematical knowledge. In thispaper we will describe and explain some of the tricks of the MathematicalCircus repertoire involving the clown.Keywords: mathematical magic, recreational mathematics, mathemati-cal circus.1 Introduction2 Who wants to be a millionaire trickThe mathematical circus’s trick “Who wants to be a millionaire”starts with theproblem of finding “which object is the odd one out”and is connected mathema-tical logic. This kind of problems is one of the important topics in competitiveexams or employment tests in which mathematics logical questions are beingasked and the solution is expected to be unique. Nevertheless, while the cri-teria one uses to get the correct solution is the most obvious one, many timeswe could use different criteria and obtain another answer that is still based onlogical thought.For example, if one asks∗CIDMA – Center for Research and Development in Mathematics and Applications,Department of Mathematics, University of Aveiro, Aveiro, Portugal; andreia.hall@ua.ptThe authors were supported by Fundação para a Ciência e a Tecnologia (FCT), withinproject UID/MAT/04106/2019 (CIDMA).†CITUR - Centre for Tourism Research, Development and Innovation, PolytechnicInstitute of Leiria; sonia.i.pais@ipleiria.pt1Which one of the numbers 28, 49, 63, 23 and 56 is the odd one outThe most obvious answer would be 23 because all the other numbers are divisibleby 7, but any of the following answers would also be logically correct.Only 63 has less units than tens.Only 23 is primeOnly 49 has a O in its spellingOnly 28 has an HOnly 56 has two Fs (or two Is) in its spellingOnly 49 has two Ns in its spellingOnly 49 is a squareOnly 23 has no digit with a closed shape (8, 6, 9, 0, 4) in itOnly 49 has 2 digits with closed shapesOnly 49 has no digit in common with any of the other numbers.In this trick, each volunteer is convinced that his is the correct answer, butafter each of them to explains his reasoning they all (and all the members ofthe audience) become aware of the other possible logical explanations.2.1 Clown’s DivisionsThe second part of the trick (clown’s elementary arithmetic operations) servesto explain why writing an algorithm in the correct form is so important. Thisis one of our most successful tricks and everyone at the end smiles, even thestudents of ages between 7 and 10 who start by shouting “that is wrong!”, “youdon’t know how to do it!”, “No! The correct answer is ...”.When the clown “proves”that 24 divided by four is 15, he uses the followingreasoning“four doesn’t go into two, but four goes into four, once. One timesfour is four, subtracted from four is zero. Next, we pull down thetwo. Now, four doesn?t go into two neither does it go into zero. Butfour goes into twenty, five times. Finally, subtracting twenty fromtwenty leaves us with a zero remainder”So, one question that comes to mind is: For which numbers can we apply asimilar reasoning and obtain a zero remainder? Any such example we will calla clown division.Suppose we want to divide a two-digits number n, i.e. n = n1 × 10 + n0,with n1 > 0 by a one-digit number d < 10. If we follow the clown reasoning, weobtain the following conditions:• d > n1;• d ≤ n0 and n0 = dq1 + r, where q1 and r are positive integers and r < d;• after pulling down n1 one realizes that d can’t go into n1 or into r, but itcan go into n1 × 10 + r, q0 times, that is n1 × 10 + r = dq0.2Therefore, in order to perform the clown’s “division algorithm”, we musthaven1 × 10 + n0 = n1 × 10 + d× q1 + r= d× q0 + d× q1= d× (q1 + q0)with the conditions10 > n0 ≥ d > n1 > 0 and n0 = d× q1 + r where r < d.The only numbers that satisfy all these conditions and so, for which we canperform the clown’s division, are the following:12÷ 2 = 15 16÷ 4 = 1314÷ 2 = 25 24÷ 4 = 1516÷ 2 = 35 28÷ 4 = 2518÷ 2 = 45 36÷ 4 = 1815÷ 3 = 14 15÷ 5 = 1218÷ 3 = 24 25÷ 5 = 1424÷ 3 = 17 35÷ 5 = 1627÷ 3 = 27 45÷ 5 = 18and18÷ 6 = 1236÷ 6 = 1548÷ 6 = 1728÷ 7 = 1349÷ 7 = 1648÷ 8 = 152.2 Number of clown’s division in a basis bSuppose we want to perform this trick using a basis b different from 10, i.e. wewant to “divide”n1 × b+ n0 by d and obtain q1q0 such thatb > n0 ≥ d > n1 > 0, n0 = d× q1 + r where r < dandn1 × b+ n0 = d(q1 + q0).For example we get the clown divisions:Basis 12: 14÷ 4 = 13 and 13÷ 3 = 14Basis 4: 12÷ 2 = 12Basis 5: 13÷ 2 = 13.So, given a basis b, how many clown’s divisions do we have?Let b > d ≥ 1 be integers and denote by Ωb,d the number of clown’s divisionsone can perform in basis b, when the quotient is d and denote by Γb the totalnumber of clown’s divisions that are possible in basis b. From above, we can seethat Ω10,i = 4 for i = 2, 3, 4, 5 and for i = 6. Also Γ10 = 22.If b = 2 there exists no integers that satisfy 2 > n0 > n1 > 0, If b = 3 thenwe would need to have n1 = 1, d = 2 and n0 = 2, but 1 × 3 + 2 = 5 which is3not a multiple of 2. So, it is not possible to make clown divisions in basis twoor three, that is, Γ2 = Gamma3 = 0.Let x be a real number and denote by bxc hhe greatest integer that is smalleror equal to x and by dxe the smallest integer that is greater or equal to x.Now, suppose b ≥ 4 and take d such that 2 ≤ d < b. We know that there areb multiples of d in the sequence 1, 2, 3, . . . bd and there are⌊bd⌋multiples of din the sequence 1, 2, 3, . . . , b. Therefore, we have b−⌊bd⌋multiples of d betweenb+ 1 and db. Below, we will see how many of these multiples, when written inbasis b, have the units-digit greater or equal to d but before that we need torecall the following elementary result about congruences.Proposition 2.1 ([3][Theorem 57). ] Let a,m and n be integers and considerδ = gcd(a, n). Then the congruence ax ≡ m mod n is soluble if and only if δdivides m, in which case it has δ solutions.Let’s proceed in our quest to find how many clown divisions exists in basisb. Take δ = gcd(d, b). In the sequence 0, 1, . . . , d− 1 there are dδ multiples of δ.Let v be a multiple of δ in this sequence, then the congruence dx ≡ v mod bhas δ solutions. Since dδ × δ = d there are d multiples of d between 1 and dbthat have a units-digit (when written in basis b) smaller than d.Now, if d is not a divisor of b then all the multiples of d smaller or equal tob have a units-digit greater or equal to d, so the d multiples of d that have aunits-digit smaller than d are all greater than b. SoTherefore, if b is not a multiple of d we haveb−⌊bd⌋− dclown divisions for the divisor d.If d divides b then we have exactly one multiple of d which has its units-digitsmaller than d, namely 1× b+ 0. So the number of clown divisions in this case,isb− bd− (d− 1).Note that, if b is not a multiple of d thenb−⌈bd⌉− (d− 1) = b−⌊bd⌋− d,and if d divides b thenb−⌈bd⌉− (d− 1) = b− bd− (d− 1).We just proved the following result4Theorem 2.2. Let b > 1 be a positive integer and 1 ≤ d ≤ b. Then there areexactlyΩb,d = b−⌈bd⌉− (d− 1)multiples of d for which it is possible to do a clown division. Or in another words,there are exactly Ωb, d quadruples (n0, n1, q0, q1) such that n1×b+n0 = d(q0+q1),10 > n0 ≥ d > n1 > 0 and n0 = d× q1 + r where r < d.For example, we obtain Ω10,4 = 10−3−3 = 4 or Ωb,b−1 = b−2−(b−2) = 0,for any b > 1.In order to obtain an approximation for the number of clown divisions in abasis b we recall a result connected with the divisors summatory funtion andDirichlet divisor problem.Theorem 2.3 ([4][section 3.2). ] Let x be a real number and γ be the Euler-Mascheroni constant. Let τ(n) be the number of divisors of n. Then∑1≤n≤xτ(n) =∑1≤n≤x⌊xn⌋= x log(x) + (2γ − 1)x+ ∆(x),where ∆(x) = O(x13 ).Theorem 2.4. Let b > 1 be a positive integer. Then there are exactlyΓb =(b− 2)(b+ 1)2−∑2≤d≤b−1⌈bd⌉.Moreover, Γb is approximately(b− 2)(b+ 1)2− b log(b) + (2γ − 1)b+ τ(b),where γ is the Euler-Mascheroni constant.Demonstração. We haveΓb =∑2≤d≤b−1Ωb,d=∑2≤d≤b−1(b−⌈bd⌉− (d− 1))= b(b− 2)− (b− 1)(b− 2)2−∑2≤d≤b−1⌈bd⌉=(b 2)(b+ 1)2+∑2≤d≤b−1⌈bd⌉5The second statement follows immediately from theorem?? and the fact that∑2≤d≤b−1⌈bd⌉= [∑1≤d≤b−1⌊bd⌋− τ(b)Foe example Γ10 = 44 − 22 = 22 and, for hexadecimal numbers, Γ16 =119− 44 = 75. For the first values of b we obtain the following values of Γb:b 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16Γb 0 0 1 2 5 7 12 16 22 28 37 43 54 64 75Note that Γb = a(b − 1), where a(n) is the sequence A161664 in Sloane’s TheOnline Encyclopedia of Integer Sequences (https://oeis.org/A161664).2.3 Clown’s greedy divisionsSuppose that in the “Who wants to be a millionaire”trick, the clown has 27sweets and there are three winners, if the clown uses his “division”algorithm,he would be really greedy and would get all the 27 sweets. If we have a clown’sdivision for which the dividend is equal to the quotient we will call that “di-vision”a clown’s greedy division. We saw above that we can also have clown’sgreedy divisions in other basis, e.g. in basis 4, 12 ÷ 2 = 12 and in basis 5,13 ÷ 3 = 13. So, now we will focus on what are the necessary conditions for abasis b to have a clown’s greedy division.Let b, d, n0, n1 and r be integers such that b > n0 ≥ d > n1 > 0, n0 =d× n1 + r, 0 ≤ r < d and n1 × b+ n0 = d(n1 + n0). From the last condition weobtain(b− d)n1 = (d− 1)n0.Since n0 = d × n1 + r and r ≥ 0, then n0 ≥ dn1. So, (b − d)n1 ≥ d(d − 1)n1.Hence,b ≥ d2.Let v = b − d2. Then n1 × b + n0 = d(n1 + n0) and n0 = d × n1 + r impliesn1(d2 + v) + dn1 + r = dn1 + d(dn1 + r). Therefore,n1v = (d− 1)r.Let δ = gcd(v, d− 1). Then gcd(vδ ,d−1δ)= 1. Now, since d−1δ divides n1vδ andd−1δ is coprime withvδ , thend−1δ must divide n1. Similarly,vδ divides r.Moreover, since d > r, δ ≤ d− 1 and vδ ≤ r, we must havev ≤ (d− 1)2.We can now state and prove a result that identifies all the clown’s greedydivisions for a basis b.6Theorem 2.5. Let b > 1 be an integer. If there is no positive integer d suchthat d2 ≤ b ≤ d2 + (d− 1)2, then there is no clown’s greedy division for the basisb. Otherwise take d such that d2 ≤ b and d2 + (d− 1)2 ≥ b, take v = b− d2 andtake δ = gcd(v, d− 1). If there exists a multiple, of vδ , say r, such thata) 0 ≤ r ≤ d− 1,b) (d−1)rv ≤ d− 1,then, taking n1 =(d−1)rv and n0 = dn1 +r, we obtain the clown’s greedy divisionn1 × b+ n0 ÷ d = n1n0.Demonstração. Clearly, we just need to see if dn0 = n1 × b+ r. Butn1 × b+ r = n1d2 + n1v + r= n1d2 + (d− 1)r + r= dn0.If b = 10 then d must be 3, v = 1 and δ = 1. Therefore r = 2 would implyn1 = 4 > d− 1, so r = 1, n1 = 2 and n0 = 7. Hence we get the clown division2× 10 + 7÷ 3 = 27.Suppose b = 18, for example. Then d = 4, v = 2 and δ = 1. Therefore r = 2,n1 = 3 and n0 = 14. Let’s denote n0 by E. Then 3× b+ E ÷ 4 = 3E.There are many basis b for which we have no clown greedy division. Forexample if b = 14 or b = 15 we have no clown greedy divisions because thereexists no d for which d2 ≤ b ≤ d2 + (d− 1)2. Also, if b = 20, b = 21, b = 23 orb = 24, then d would have to be 4, and v would be 4, 5, 7 or 8 respectively. Butin each case we would get δ = 1 which would imply r ≥ d.One question remains “Is there a value b0, such that if b ≥ b0 there is alwaysone clown’s greedy division for the basis b?”.AcknowledgementsThis work was supported by Fundação para a Ciência e a Tecnologia (FCT),within project UID/MAT/04106/2019 (CIDMA).Referências[1] B. Abbot and L. Costello, In The Navy - 7 into 28 goes 13 times, 1941.https://www.youtube.com/watch?v=MS2aEfbEi7s[2] I. Bastos, Magia Matemática com Números, Msc Dissertation, Universi-dade de Aveiro, 2015.7[3] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers,Oxford, 4th Edition, 1960.[4] G. Tenenbaum, Introduction to analytical and probabilistic number the-ory, AMS, 3rd Edition, 2015.8

Mathematical circus clown tricks

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Mathematical circus clown tricks

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