U S A Mathematical Talent Search. PROBLEMS / SOLUTIONS / COMMENTS Round 4 - Year 11 - Academic Year

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1 U S A Mathematial Talent Searh PROBLEMS / SOLUTIONS / COMMENTS Round 4 - Year 11 - Aademi Year Gene A. Berg, Editor 1/4/11. Determine the unique 9-digit integer M that has the following properties: (1) its digits are all distint and nonzero; and () for every positive integer m =, 3, 4,..., 9, the integer formed y the leftmost m digits of M is divisile y m. Solution 1 y Lisa Leung (9/MD): Let the 9-digit integer M e represented y adefghi where eah letter represents a unique digit from the set of {1,,...9}. Sine ade must e divisile y 5, e an only e 5 or 0. However the digits are non-zero, so e = 5., d, f, and h are even numers sine they are the last digits of numers that are divisile y, 4, 6, and 8. When examining ominations of d and fgh, where d will e divisile y 4 and fgh will e divisile y 8, one an onlude from Tale 1 (at right) that d is or 6, and h is or 6, sine and g annot e even. The possile values for d are 1, 16, 3, 36, 5, 56, 7, 76, 9, and 96. The possile values for fgh are 416, 43, 47, 496, 816, 83, 87, 896. Tale (elow) shows the possile digits of eah position after the initial analysis. a++ must e divisile y 3 and d+e+f must e divisile y 3. The only possiilities of a,,, d, e, and f left are given in Tale 3. a d e f g h i Tale : Among the five possiilities, only is divisile y 7. a d e f ghi Tale 3: d Thus, M = Tale 1: Solution y Zhihao Liu (10/IL): Answer: Let adefghi =M, where letters a through i represent digits 1 through 9, let ade represent a five digit numer, and so on. Sine 5 ade (read as, 5 divides ade ), we onlude e=5. Note that, d, f, and h are all even, thus a,, g, and i are odd. Sine ad and adefgh are divisile y 4 and their ten s digits are odd, we onlude that d and h are

2 and 6 (in some order), and and f are 4 and 8 (in some order). Also, 3 a++, 3 a+++d+e+f, and 3 a+++d+e+f+g+h+i = 45, so 3 d+e+f, and 3 g+h+i. From this we see that def is either 58 or 654. Sine 8 fgh and f is even, we onlude that 8 gh. With these restritions, we eliminated all 9-digit numers exept: , , , , , , and However, only possesses the property that the numer formed y its first seven digits is divisile y 7. Therefore, M = Editor s Comment: We express thanks for this prolem to Sándor Róka, the founding editor of Aaus, Hungary s mathematial journal for students age /4/11. The Fionai numers are defined y F 1 =F = 1 and F n =F n-1 +F n- for n>. It is well-known that the sum of any 10 onseutive Fionai numers is divisile y 11. Determine the smallest integer N so that the sum of any N onseutive Fionai numers is divisile y 1. Solution 1 y Chi Cao Minh (1/TX): Let x and y represent two onseutive Fionai numers. Continuing the sequene: Fionai Numer Sum of Sequene Fionai Numer Sum of Sequene 1 x x 13 89x + 144y 33x +376 y y x + y x + 33y 377x + 609y 3 x + y x + y 15 33x +377 y 610x + 986y 4 x + y 3x + 4y x + 610y 987x y 5 x + 3y 5x + 7y x + 987y 1597x + 583y 6 3x + 5y 8x + 1y x y 584x y 7 5x + 8y 13x + 0y x + 584y 4181x y 8 8x + 13y 1x + 33y 0 584x y 6765x y 9 13x + 1y 34x + 54y x y 10946x y 10 1x + 34y 55x + 88y 6765x y 17711x y 11 34x + 55y 89x + 143y x y 8657x y 1 55x + 89y 144x + 3y x y 46368x y Dividing the Sum of the Sequene y 1 shows that the smallest numer N is 4.

3 Solution y Riky Liu (10/MA): Suppose we have found N suh that F k + F k F k + N 1 is always divisile y 1. Then ( F k F k F k + N ) (F k + F k F k + N 1 ) = F k + N F k must e divisile y 1. In other words, given any two Fionai numers with indies greater than one and separated y N, their differene must e divisile y 1. To simplify things, notie that if F (mod1) and if F 3 (mod1), then F k (mod1) for all k > 1. This F N + F N + 3 F N + k F k + F k F N + k + 1 F N + k + is easily shown y indution: F k + F N k + (mod1). So, we need only find the seond pair of onseutive Fionai numers equivalent to 1 and (mod 1) respetively. These our at F 6 and F 7. Thus, N = 6 - = 4. N = 4. - Editor s Comment: This prolem was reated y the editor. 3/4/11. Determine the value of S = Solution y Emily Kendall (11/IN): 1999 S = a (a + 1) a = a + a 3 + 3a + a + 1 = a (a + 1) a = a + a + 1 = a = 1 a + a = a + a a = = , a telesoping sum. a a + 1 a = 1 1 = = Editor s Comment: This timely prolem was adapted y our Prolem Editor, Dr. George Berzsenyi, from MatLap, Transylvania s Hungarian language mathematis journal for students at the middle and high shool levels.

4 4/4/11. We will say that an otagon is integral if it is equiangular, its verties are lattie points (i.e., points with integer oordinates), and its area is an integer. For example, the figure on the right shows an integral otagon of area 1. Determine, with proof, the smallest positive integer K so that for every positive integer k K, there is an integral otagon of area k. Solution 1 y Sawyer Taony (11/VA): The integer K that I ame up with is K = 13, and I will show you how I ame up with this. First, I will show you that there is an integral otagon of area k for all k greater than or equal 13. Here are the diagrams of integral otagons of areas 13 through 1: y 1 1 x Notie that for the last five (17-1) there is a olumn shaded within eah otagon exatly five units tall and one wide. To find an otagon of area, one would only need to widen the olumn in the otagon of area 17 y one unit. This would reate the extra five square units. Sine we have five onseutive otagons with this feature, it is lear that y expanding the width of these olumns we an eventually reah any area.

5 So now I must show that there does not exist an integral otagon of area 1. The smallest otagon whose height is five is the one of area 13 on the previous page, so learly oth height and width of an otagon of area 1 must e less than 5 (height and width are the dimensions of the smallest retangle with vertial and horizontral sides into whih the integral otagon fits). The largest four y four integral otagon is shown aove and has an area of 14. The only other one is to the right, and it has area Next we move down to four y three otagons. There is only one otagon with dimensions four y three; it is shown at right (with an area of 10). 10 The final otagon is one of height and width three that has area of seven. This is shown at the right. So there is no integral otagon of area 1. This makes K = Comment y Ahijit Mehta (10/OH): The smallest integral otagon, none of whose sides are parallel to the oordinate axes, is shown elow. Its area is 35, so it does not alter the solution. 35 Editor s Comment: Dr. Berzsenyi ased this prolem on a similar, ut more diffiult prolem, proposed y A. C. Heath in the Marh 1976 issue of The Mathematial Gazette. 5/4/11. (Revised ) Let P e a point interior to square ABCD so that PA = a, PB =, PC =, and =a +. Given only the lengths a,, and, and using only a ompass and straightedge, onstrut a square ongruent to square ABCD.

6 Solution 1 y Nyssa Thompson (10/HI): I am assuming asi onstrution knowledge: on struting a right angle and a square with a given side length. Conneting point P to points A, B, and C reates two trian- gles, ΔPAB and ΔPBC, that share the same side PB. Sine ABCD is a square, AB = BC = CD = DA. We an refer to any of the sides as s. Furthermore, we an see that ABP + CBP = 90 eause they sum to ABC whih is a orner of the square ABCD. Also, sine it was given that AP = a, BP =, and CP =, we an refer to these lengths as suh. Thus ΔPAB has sides of lengths a, s, and, and ΔPBC has sides of lengths, s, and. A D a P B C We an rotate these triangles so that instead of sharing side, they share the s side. Then instead of the sides of length s forming a right angle, the sides of length form a right angle (see elow). C A a s B A a P or B or D C C A a s B Thus, we have a polygon with two adjaent sides of length forming a right angle at B, and remaining sides a and. The segment from the vertex at B to the vertex at A and C is of length s. With this knowledge we an onstrut the square ABCD if we are given a,, and using only ompass and straight edge. 1. Construt a right angle.. Set the ompass to length, plae its point on the vertex of the right angle, and mark off length on eah side of the angle. 3. Next set the ompass to length a or, plae the ompass point on the end of one of the lengths

7 (not the vertex), and make a irle with the radius a or. 4. Repeat step 3 with the remaining segment length a or aordingly. Construt the irle on the opposite s endpoint. 5. Now, these two irles should either interset at two points or e tangent. 6. Choose the point whih lies within the right angle and, using the straightedge, onnet it to the vertex of the right angle. This length is s. 7. Construt a square with side length s. It will e ongruent to square ABCD! Solution y Megan Guihard (1/WA): Let s e the length of the sides of ABCD and let ABP = θ. Then, y the law of osines, we have a = s + sos ( θ), s + a + a os ( θ) = , and θ = os 1 s We also know, again y the law of s s osines, that = s + sos ( CBP) = s + sos (90 θ) = s + ssin ( θ). Sine + a we know that θ = os 1 s , s sin ( θ ) = 1 os ( θ) s s + a s a s + a = s a s + s + a s a = s Then, ssin ( θ ) = a s + s + a s a. Sine = s + ssin ( θ), ssin ( θ) = s + = s + a = s a Squaring oth sides of this equation gives a s + s + a s a = s a s + a s + a s 4a s + a 4s + 4 = s (a + )s + (a 4 + ) = 0 4 s (a + )s + (a + ) = a (s (a + )) = a s (a + ) = a s = a + + a

8 s = a + a s = a + a os (135 ). Notie that this is simply the law of osines for a triangle with sides a,, and s, with the angle opposite s measuring 135. Sine sides and their inluded angle determine a triangle, we an onstrut s y onstruting a triangle with two sides a and suh that the angle etween a and is 135. (We an onstrut an angle of 135 y onstruting a 90 angle, iseting it to get a 45 angle, and then adding another 90 angle.) One we have s, we an onstrut a square ongruent to ABCD fairly easily y onstruting 90 angles and new sides of length s until we have a square with four sides of length s. Editor s Comment: Notie that the first solution does not use = a +. Given that the square ABCD exists, does this onstrution work for any point P (interior or exterior) and set of distanes PA=a,PB=, PC =? We are indeted to Professor Gregory Galperin of Eastern Illinois University for suggesting this wonderful prolem.