USA Mathematical Talent Search. PROBLEMS / SOLUTIONS / COMMENTS Round 3 - Year 12 - Academic Year

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1 USA Mathematial Talent Searh PROBLEMS / SOLUTIONS / COMMENTS Round 3 - Year - Aademi Year Gene A. Berg, Editor /3/. Find the smallest positive integer with the property that it has divisors ending with every deimal digit; i.e., divisors ending in 0,,,..., 9 Solution by Rishi Gupta (8/CA): Let us first look at the prime fators of the answer. There must be a 5, beause any number ending in 5 is divisible by 5. There must also be a, for the same reason. Therefore, so far, we have fators: 5 = 0, and the numbers 0,,, and 5 are overed. We have the numbers 3, 4, 6, 7, 8, and 9 left. If we an find multiples of the three odd numbers (3, 7, and 9), then the even numbers are overed beause of the multiple of ( 3 = 6, 7 = 4, 9 = 8 ). Therefore all we need to worry about is 3, 7, and 9. One solution (whih may not be smallest) would be to use the LCM of 3, 7, and 9, whih is 63. So the answer would have to be less than or equal to Let s see if any of the numbers ending in 7 have fators ending in 3 and 9. Seven and 7 are primes, but 7 has 3 and 9 as fators. So now 7 0 = 70 is a smaller solution. Any other possible solution below 7 would have to use a multiple of 7 or 7. The only andidates are 4 and, neither of whih has a fator of 9 or 9. Verifiation: 70 is divisible by 0,,, 3, 54, 5, 6, 7, 8, and 9. Therefore the solution is 7 0 = 70. Solution by Matthew Pel (/PA): Solution 70. Call the desired integer N. All positive integers have as divisor, so it is immediately dealt with. The only way a divisor of N an end in zero is for it to be a multiple of 0, thus and 5 must be prime fators of N, and N is a multiple of 0. Divisors ending in 5 would have to be multiples of 5, but we already have taken are of 5. So 5 overs 0,,, and 5. Any multipliation with 5 just produes another number ending in 0 or 5, so 5 is not really a onsideration in determining the remaining needed fators. Any divisor ending with 7 automatially gives us another ending with 4 (beause is already a fator of N), and similarly any ending with 9 gives us another fator ending with 8. The smallest and most eonomial solution for 7 and 9

2 would be 7 = 3 3 = 3 9. The 4 (7 = 54 ) and 8 (9 = 8 ) are taken are of simultaneously. Notie 3 and 6 are overed sine 3 is a fator and 6 = 3. So 5 7 = 70 is the smallest positive integer with divisors ending in eah deimal digit. If we are not sure, we an hek eah multiple of 0 up to 70; but notie only 70, 40, 70 and 0 have divisors ending with 7. Of these only 0 has a divisor ending in 3, but 0 has no divisor ending in 8. 7 is the best beause it ontains 3 and 9, and with it takes are of 3, 4, 6, 7, and 9 all at one. Solution N = 70. Solution 3 by Mike Churh (/PA): The smallest suh integer is 70. Proof: First, sine the desired integer has a divisor ending in 0, it is lear that this integer must be divisible by 0. Hene we an eliminate all integers not divisible by 0. That our integer is divisible by an integer ending in 7 implies it is divisible by an integer among the set {7, 7,...}. Hene, we an remove all other integers from our onsideration. Thus the five smallest andidates for our desired integer are: 70, 40, 70, 0, 70 Next, it is desired that our integer have a fator ending with 3, so it must be divisible by one among 3, 3, 3, and so on. Thus we an eliminate 70, 40, and 70, and hene there remain 0 and 70 The divisors of 0 are,, 3, 5, 6, 7, 0, 4, 5,, 30, 35, 4, 70, 05, and 0. As this set laks elements whose last digits are 8 or 9, 0 an also be disqualified. Hene, all integers less than 70 have been disqualified. Now evaluate the divisors of 70:,, 3, 5, 6, 9, 0, 5, 8, 7, 30, 45, 54, 90, 35, and 70. On inspetion, 70 indeed does have the desired property. Editor s Comment: We are grateful to Professor Brue Reznik of the University of Illinois for this nie problem, whih he first published (via Martin Gardner) in Siene Fition Puzzle Tales in the early 980 s. /3/. Assume that the irreduible frations between 0 and, with denominators at most 99, 7 are listed in asending order. Determine whih two frations are adjaent to in this listing. 76 Editor s omment: For a disussion of ontinued frations and Farey series see the Editor s Comments following the solutions.

3 Solution by Christopher Lyons (/CA): 7 We onsider the ontinued fration expansion of : = = [0; 48],, To find the fration that is diretly below 7 76, we must realize that hanging the 8 in the expansion to a higher number would, in fat, make the overall number smaller. So we must figure out how muh to add to the 8 in order to turn the overall fration into one diretly below it on the list. Let us write the number below 7 76 as having the ontinued fration expansion x = 04 ;,, , a where a is a positive real number. When we ondense x into its ommon frational form, we have 7a + x = a We know the denominator is less than or equal to 99, so a We also know that both 7a and 76 76a must be positive integers. But sine 7 and 76 are oprime, a must be an integer itself. The only positive integer less than is, so a =, and 9 x = To find the fration diretly above 7 76, we must subtrat some amount from the 8. We all this larger fration y, and write its ontinued fration expansion as y = 0 ; 48,, --, b where b is a positive real number. When we ondense y into its traditional rational form, we find 7b y = b 9 08 Again, we know the denominator annot exeed 99, so b One more, b must be an integer 76 due to the lak of ommon fators between 7 and 76. Therefore, b =, and 5 y = So < <

4 Solution by Lisa Leung (0/MD): The irreduible frations between 0 and with denominators at most 99, listed in asending order, desribes a Farey series of order 99. Two basi theorems that desribe harateristis of suessive terms in a Farey series F n of order n are: h h () If -- and ---- are two suessive terms of the Farey series, then kh hk =. k k h h () If -- and ---- are two suessive terms of the Farey series, then k + k > n. k k By using theorem () with h k set to 7 76 where h is 7 and k is 76, 76h 7k =. This is similar to solving 76h ( mod 7). This is solved when h 5 ( mod 7). Substituting 5 into the first equation yields k = 67, and h k = Thus < Next, by using theorem (), but with h k as 7 76, 7k 76h =. This is similar to solving 7k ( mod 76). This is satisfied when k 9 ( mod 76). However, when k = 9, it ontradits theorem (). Thus k 9. When k = 85, it satisfies both theorems. h 9 When substituted into the equation, -- = k Thus, < < Solution 3 by Jason Chiu (/WY): Answer: < < h h Theorem. If -- and ---- are two suessive terms of the Farey series F n, then kh hk =. k k For several proofs of this well-known theorem, see G.H. Hardy and E.M. Wright, An Introdution to the Theory of Numbers, Fifth Edition, Oxford University Press, London (979). a a Let -- and ---- respetively denote the frations left-adjaent and right-adjaent to 7 76 in F 99. b b By the ontrapositive of the Theorem, 7b 76a = and 7b 76a =. To solve the first Diophantine equation, apply Eulid s Algorithm to obtain 76 = ,

5 7 = 8 +, 8 = Baking up through these equations gives = 7 8 = 7 ( ) = , yielding the solution (a, b) = (, 9) to the equation 7b 76a =. An alternate way to obtain 7 this solution is to ompute the onvergents to the ontinued fration expansion of The on 76 vergents in this expansion are P 0 P P P = 0, = = --, = = --, = = Q 0 Q 4 4 Q 9 Q P = -- also gives the solution (a, b) = (, 9) to the equation 7b 76a =. Q 9 Sine gd(7,76) =, the general solution of the equation 7b 76a = is a = + 7t, b = t. For any solution (a,b) to 7b 76a =, 7 a k = = b 76b so that hoosing the largest value of b minimizes k. Hene the ordered pair (a, b) = (9, 85) gives the smallest value of k and 9 85 is left adjaent to Similarly, the general solution to 7b 76a = is a = + 7t, b = t so the ordered pair (a, b ) = (5, 67) gives the smallest value a 7 k = = b 76 76b and 5 67 is right-adjaent to Editor s Comment: A similar problem appeared in the January 999 issue of Mathematial Digest, an exellent South Afrian mathematial journal for high shool students. This problem does allow for omputer solutions; we welome them, but ommend suh solutions only if they are based on a lever algorithm and deal with auray. We thank our problem editor, Dr. George Berzsenyi, for this problem. This interesting problem gives us an opportunity to expand on two interesting mathematial topis, ontinued frations and Farey series. Continued frations were disussed in the Solutions to Round of Year 0, available on the USAMTS web site. h By a Farey series F n of order n, we mean the set of all frations -- with 0 h k, gd(h,k) =, k k n, and arranged in asending order of magnitude. For example, F 5 is

6 , --, --, --, --, --, --, --, --, --, These series have remarkable properties, some of whih are mentioned in the solutions above. I restate two of the theorems here so you might easily onfirm them for this example: h h () If -- and ---- are two suessive terms of the Farey series, then kh hk =. k k h h () If -- and ---- are two suessive terms of the Farey series, then k + k > n. k k Proofs of these and other properties are given in G.H. Hardy and E.M. Wright, An Introdution to the Theory of Numbers, Fifth Edition, Oxford University Press, London (979). 3/3/. Let px ( ) = x 5 + x + have roots r r r 3 r 4 ( ) = x,,,, r 5. Let qx. Determine the produt q( r )qr ( )qr ( 3 )qr ( 4 )qr ( 5 ). Solution by Sofia Leibman (8/OH): We an write the polynomial x 5 + x + in the form (x r )(x r ) (x r 5 ). The produt q( r )qr ( ) qr ( 5 ) = (r )(r ) (r 5 ). r i = ( r i )( r i ). So the produt an be written as ( r )( r ) ( r 5 )( r )( r ) ( r 5 ) The solution is -3. = p ( ) p ( ) = [( 5 ) + ( 5 ) + ] [( ) + ( ) + ] = 3 Solution by Sarah Emerson (/WA): Any polynomial of degree n with leading oeffiient an be fatored as (x z )(x z ) (x z )(x z ) n n where z, z, z n are the roots of the polynomial, are omplex numbers of the form a+ bi, (a and b are real numbers and either a or b, or both, an be zero). Therefore, px ( ) = x 5 + x + an be expressed as px ( ) = (x r )(x r )(x r 3 )(x r 4 )(x r 5 ). Also, qx ( ) = x has two roots, s and s, so q( x ) = (x s )(x s ). Then qr ( ) = (r s )(r s ) and

7 qr ( )qr ( )qr ( 3)qr ( 4 )qr ( 5) = = (r s )(r s )(r s )(r s )(r 3 s )(r 3 s )(r 4 s )(r 4 s )(r 5 s )(r 5 s ) = (s r )(s r )(s r 3 )(s r 4 )(s r 5 )(s r )(s r )(s r 3 )(s r 4 )(s r 5 ) sine we hanged an even number of signs and reordered terms. This is equivalent to ps ( )ps ( ). Fatoring qx ( ) = x = (x )(x + ) gives the two roots of q(x): s = and s =. Plug these roots of q(x) into the original equation for p(x). qr ( )qr ( )qr ( 3)qr ( 4)qr ( 5) = p( ) p( ) 5 5 = [( ) + ( ) + ] [( ) + ( ) + ] = 3. This method will work with any two polynomials; however, if both polynomials are of orders that are odd numbers, the produt of f (roots of g( x )) will be the negative of g(roots of f( x)) beause the odd number of negative signs will not all anel out. Editor s Comment: We are thankful to Professor Rob Hohberg of the University of Connetiut for this interesting problem. 4/3/. Assume that eah member of the sequene i mine the appropriate sequene of + and signs so that i = is either a + or a sign. Deter- = Also determine what sequene of signs is neessary if the sixes in the nested roots are replaed by sevens. List all integers that work in the plae of sixes and the sequenes of signs that are needed with them. Solution by Ho Seung (Paul) Ryu (9/KS): Firstly, we notie that = 6. So, if we replae the on the right side with the idential expression 6 in infinite number of times, we will have obtained = So the sequene of signs for 6 is simply -,-,-,-,.... For number 7, = 7 3, whih is of little help, but 3 = 7 +, so = So, by replaing on the right side, we get = , just alternating

8 signs -,+,-,+,.... For the number 8, we will prove that 8 annot be used. That is for a simple reason, that = 8 4, but no sequene of radials and 8s an be larger than This quan tity is given by x = = 8 + x. x = 8 + x and thus x x 8 ± 4 ( 8 ) ± 33 x = = This is less than 4, so 8 does not work in the problem. No numbers greater than 8 satisfy the onditions set forth in the previous statement, so now we an investigate numbers less than 6. Now for 5, = 5, = 5 4, 4 = 5 +, = The numbers just never ease to inrease, so we an never have a working sequene of signs. For 4, = 4 0, and we are stuk right away. For 3, = 3 +, = 3, and so = Thus the sequene of signs is +,-,+,-,..... As for, = Lastly, for, = does not work. +, so = + + and we have sequene +,+,+,+, , 3 = + 8, 8 = Thus for the number six, the required sequene of signs is just a string of minus signs -,-,-,-,.... The numbers that work in plae of six, and the sequene of signs needed are: : All plus signs +,+,+,+,.... 3: Alternating plus/minus +,-,+,-,.... 7: Alternating minus/plus -,+,-,+, = 0. 63, and we just keep inreasing. Therefor, Editor s omment: This wonderful problem was proposed by Dr. Rodrigo Gomez of NSA. 5/3/. Three isoseles right triangles are ereted from the larger side of a retangle into the interior of the retangle, as shown on the right, where M is the midpoint of that side. Five irles are insribed tangent to some of the sides and to one another as shown. One of the irles touhes the vertex of the largest triangle. M Find the ratios among the radii of the five irles.

9 Solution by Lisa Leung (0/MD): Sine the triangles are isoseles right triangles that are ereted from the larger side of a retangle, the two irles marked as a are ongruent and the two irles marked as are ongruent. a M a Without loss of generality, let the length of the retangle be unit, and r a, r b, and r be the radius of irle a, b and respetively. b As shown in Figure, r a +r a = -- 4 r = a From the base of figure 3, -- = ( r + r b ) ( r r b ) = r b r + r From the width of the retangle in Figure 3, ( = 4 ( + ) )r = r b () r b = r -- (3) 4 Substitute into (), -- = r --- r - + r 4 () Figure. / r a Figure. / -- r = / r --- r 4 ( r ) = r --- r 4 4r 4r + = (6 + 8 )r 4r r r r = = ( ) 4+ ( ) r b rb Take the positive square root, r = = ( ) = r a (4) Substitute into (3), r b r -r b Figure 3.

10 r b = = = r 4 4 a Thus, the three smallest irles have the same radii. The ratios among the radii of the five irles are: a:a:b:: = ::::. Editor s omments: We are indebted to Professor Hiroshi Okumura of Japan for this wonderful shungaku problem. Professor Okumura is a longtime enthusiasti promoter of the Japanese equivalent of the USAMTS.