FIFTEENTH ANNUAL NORTH CENTRAL SECTION MAA TEAM CONTEST

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1 FIFTEENTH NNUL NORTH ENTRL SETION M TEM ONTEST November 2, 20, 9:00 am to 2:00 noon NO OOKS, NOTES, LULTORS, OMPUTERS OR NON-TEM- MEMERS may be consulted No cell phones or other communication devices PLESE EGIN EH PROLEM ON NEW SHEET OF PPER Team identification and problem number should be given at the top of each sheet of paper submitted Each problem counts 0 points Partial credit for significant but incomplete work For full credit, answers must be fully justified ut in some cases this may simply mean showing all work and reasoning Have fun! Units digit of S 20 * * * * * Let S n be the sum of the squares of the first n positive odd integers Find, with proof, the units digit (in base 0) of S 20 2 rea of a square t the left are two copies of the same isosceles right triangle, each with an inscribed square If the first (leftmost) inscribed square has area S, what is the area of the other inscribed square? Justify your answer 3 Speshul integers positive integer k is called speshul if there exist positive integers m and n such that Find all speshul positive integers 4 Divisor and remainder mn + m + n = k Determine all pairs (d, r) of positive integers with the property that when each of the numbers 904, 259 and 2040 is divided by d, the same remainder r (with 0 <r<d) occurs page of 2

2 NS/M TEM ONTEST 20, page 2 of 2 5 igger than 20 2? Determine whether there exist non-negative numbers x and y such that x + y = 20 and x 2 + y> From to 20 The positive integer n can be replaced by ab, if a and b are positive integers with a + b = n Is it possible starting from to obtain 20 in a finite sequence of such replacements? 7 Not a perfect square For each positive integer n, let a n = (n + 9) 2 +2n + 99 (s usual, x denotes the greatest integer less than or equal to x) Show that a n is never the square of an integer 8 P () = 20 The coefficients a j in the polynomial P (x) =a 0 + a x + + a n x n are integers satisfying 0 a j 0 and a n > 0 If P () = 2 and P () = 20, determine, with proof, the value of P (6) 9 Fractional part of n Let n be a positive integer which is not a perfect square Let x n be the fractional part of n; ie, xn = n n, where u denotes as usual the greatest integer less than or equal to u Prove that x n + 2n < 0 Numerator divisible by 20 Let p and q be positive integers such that = p q Given that 20 is prime, show that p is divisible by 20

3 Solutions, 20 NS/M TEM OMPETITION, page of 4 Each problem number is followed by an -tuple (a 0,a 9,a 8,a 7,a 6,a 5,a 4,a 3,a 2,a,a 0 ), where a k is the number of teams that scored k points on the problem Units digit of S 20 (28,,,0,,0,9,2,5,5,24) The units digit of S 20 is We show this by observing that S = has units digit, and that S n+0 always has the same units digit as does S n Modulo 0, the sequence of squares of positive odd integers begins (,9,5,9,), after which this cycle repeats itself, because (n +0) 2 = n 2 +20n +00 n 2 (mod 0) Then modulo 0, S n+0 S n will always be the sum of 0 consecutive terms in this sequence, which will be , wherever in the sequence one begins Thus S n+0 S n (mod 0) 2 rea of a square (60,3,0,2,0,2,,0,,0,7) 8 The area is S The legs of the right triangle have length 9 2 S,sothehypotenuseis2 2S Thenthesecondinscribed square has side 2 2S, because the smaller triangles at and 3 are also isosceles right triangles Thus the area of the second inscribed square is 2 2 2S 3 = 8S 9 3 Speshul integers (3,0,0,2,0,0,4,7,2,0,30) Well, of course, every positive integer is speshul Here are two different ways to get the m and n we need () To make mn +=km + kn, startbytakingm = k + Thenwewantkn + n += km+ kn; ie,n + = km, son = km doesit Thus: m = k +, n = km =k 2 + k (2) Take m =2k, n =2k + Then mn + m + n = 4k2 4k = k 4 Divisor and remainder (20,0,2,,,,2,0,0,9,40) The unique such pair is (d, r) =(7, 52) From904=ad+r, 259=bd+r and 2040 = cd+r we obtain on subtraction, 355 = (b a)d and 78 = (c b)d Now, 78 = (2)(355) + 7, and both 78 and 355 are multiples of d, so7isamultipleof d ut7isprime,sod can only be 7 (the condition 0 <r<drequires d>), and then 904 = (2)(7) + 52 shows that r must be 52 One checks that 259 = (7)(7) + 52 and 2040 = (28)(7) + 52, so (d, r) =(7, 52) is the unique solution

4 Solutions, 20 NS Team ompetition, page 2 of 4 5 igger than 20 2? (8,0,,0,2,0,2,4,5,6,38) Such numbers do exist Let f(x) =x x for 0 x 20 It suffices to show that there exists x [0, 20] with f(x) > 20 2 We note that f(20) = 20 2,so it suffices to show that f (x) is negative throughout an interval (20, 20) for some > 0 Well, f (x) = 2x /(2 20 x), and for 0 x 20, 2x < 4022 ut /(2 20 x) > 4022 when / 20 x>8044; ie, when 20 x</ Thus, when <x<20, f (x) < 0, so f(x) decreases in this interval to the value 20 2 at x =20,andhence f(x) > 20 2 throughout this interval 6 From to 20 (38,,,,0,0,,2,0,0,32) Yes, it can be done in many ways (infinitely many, in fact) key observation is that with any n>, n =(n ) + can be replaced by (n ) =n, so if we obtain any integer larger than 20, we can step down from there to 20 in steps of Thus, one solution is = = = = , etc, down to 20 7 Not a perfect square (,0,0,0,0,,0,2,2,,59) We show that (n +5) 2 <a n < (n +6) 2 for all n, and the claim follows Using the facts that 4 < 9 < 5and9< 99 < 0 we have a n = n n +9+2n + 99 n 2 +2n +29 = (n +6) 2 7 < (n +6) 2, and a n n 2 +0n +28 = (n +5) 2 +3 > (n +5) 2

5 8 P () = 20 (35,,6,0,,0,0,0,0,0,33) Solutions, 20 NS Team ompetition, page 3 of 4 We show that P (6) = 44 From the fact that 20 = P () a n n n we know that n 3because 3 =33< 20 < 4 lso, 20 < 2 3,soa 3, and thus if n =3 then a 3 = Weshowthatn =3 Ifn 2, then P (x) =a 0 + a x + a 2 x x +0x 2, and 20 = P () < 20, a contradiction Thus n =3,a 3 =,and P (x) =a 0 + a x + a 2 x 2 + x 3 Then 20 = P () = a 0 + a + a ,whence a 0 + a + a 2 2 =20 33 = 680 ecause 0 a 0 + a = 20, we have 680 a = 560, and therefore 6 > 680/2 a 2 560/2 > 4 Hence a 2 =5,and 20 = a 0 + a = a 0 +a +936, so a 0 +a =75 Theconstraintsona 0 and a then determine that a =6anda 0 =9 Thus P (x) =9+6x +5x 2 + x 3 Note that this also satisfies P () = 2 (which was given but not used for the solution) Finally, then, P (6) = = 44 9 Fractional part of n (8,0,2,0,0,0,0,0,0,0,66) Let r be the integer such that r < n<rifr =2,thenn =2or3andonecaneasily verify directly that the required inequality holds So we assume that r 3, in which case (r ) 2 r, andhencen>rletk = r 2 n Thenk>0sok, and we have x n = n n = n (r ), and then qed x n + 2n = n r ++ 2n = r 2 k r ++ 2n r 2 r ++ 2n r 2 r ++ 2r = r r 2 2r = r + r 2 <, 2r

6 0 Numerator divisible by 20 (3,,0,0,0,0,0,0,0,0,72) We rewrite p q = = Solutions, 20 NS Team ompetition, page 4 of = , Now group these terms in pairs as follows: p q = = 67 =20 =20 r s, for some integers r and s, whereallprimefactorsofs are smaller than 20 We now have 20rq = ps, 20isprime,and20doesnotdivides Therefore20 p, qed

PUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime.

PUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime. PUTNAM TRAINING NUMBER THEORY (Last updated: December 11, 2017) Remark. This is a list of exercises on Number Theory. Miguel A. Lerma Exercises 1. Show that the sum of two consecutive primes is never twice