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494 Solution by the Mayhem Sta. Consider a two-digit temperature C = 10a + b in degrees Celsius, where a and b are integers with 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. The equivalent temperature in degrees Fahrenheit is F = 9 9 9 C + 32 = (10a + b) + 32 = 18a + b + 32 . 5 5 5 We want the rounded version of this real number to equal 10b+a. Therefore, 1 10b + a − 2 100b + 10a − 5 −325 315 9 ≤ 18a + b + 32 5 ≤ 180a + 18b + 320 ≤ 170a − 82b < 82b − 170a < < < ≤ 10b + a + ; 2 100b + 10a + 5 ; −315 ; 325 . 1 Sine b ≤ 9, then 82b ≤ 738. Sine 82b − 170a > 315, then 170a < 423 83 = 2 . Sine a is an 82b − 315 < 738 − 315 = 423, whene a < 170 170 integer, then a ≤ 2. Therefore, we only need to try a = 1 and a = 2. If a = 1, the inequalities beome 315 + 170(1) < 82b ≤ 325 + 170(1) 3 or 485 < 82b ≤ 495 or 5 75 < b ≤ 6 ; sine b is an integer, then b = 6. 82 82 Similarly, if a = 2, then b = 8. Hene, the two possibilities are C = 16 (giving F = 60.8, whih rounds to F ≈ 61) and C = 28 (giving F = 82.4, whih rounds to F ≈ 82). Also solved by MATTHEW BABBITT, student, Albany Area Math Cirle, Fort Edward, NY, USA; JACLYN CHANG, student, Western Canada High Shool, Calgary, AB; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Ranho Palos Verdes, CA, USA; RICARD PEIRO, IES \Abastos", Valenia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; There was one inomplete solution submitted. Most submitted solutions involved an expliit or impliit omplete enumeration of ases from C = 10 to C = 39 after some examination of bounds. Problem of the Month Ian VanderBurgh What's in a denition? Mathematis is littered with them. Often, we pay attention to them; sometimes we treat them a bit avalierly. Here are two problems involving geometri sequenes. In the seond of these problems, the preision of our denition turns out to aet the answer. Problem 1 (2009 Amerian Invitational Mathematis Examination) Call a 3-digit number geometri if it has 3 distint digits whih, when read from left to right, form a geometri sequene. Find the dierene between the largest and smallest geometri numbers. 495 Problem 2 (2009 Eulid Contest) If log2 x, (1 + log4 x), and log8 4x are onseutive terms of a geometri sequene, determine the possible values of x. So what's a geometri sequene? Many of you will know this already, but by way of reminder, here is our rst attempt at a denition: Denition 1: A geometri sequene is a sequene of numbers in whih eah term after the rst is obtained from the previous term by multiplying by a onstant. Often, we would all the rst term in the sequene a and the multiplying fator r, whih gives the sequene a, ar, ar2 , ar3 , . . . . (You may notie that I've deliberately avoided the phrase \ommon ratio" { stay tuned!) Let's use this version of the denition to solve the rst problem. Solution to Problem 1 The smallest 3-digit integers have hundreds digit 1. Let's see if any of these integers is geometri. Call the tens digit of our andidate number r (note that r is an integer). Sine the hundreds digit is 1, the tens digit is r, and the digits form a geometri sequene, then the units digit is r2 . The andidate 3-digit integer is as small as possible when r is as small as possible. Sine the digits are distint, then r 6= 1 (otherwise r = 1 would give 111) and r 6= 0 (otherwise r = 0 would give 100). So the smallest andidate ours when r = 2, whih yields the integer 124, whih must be the smallest 3-digit integer that is geometri. The largest 3-digit integers have hundreds digit 9. Let's see if any of these integers are geometri. Consider a andidate integer and suppose that the multiplying fator between onseutive digits is R. Then the tens digit is 9R and the units digit is 9R2 . Sine we want this integer to be as large as possible, we try the dierent possibilities for 9R. If 9R = 9, then R = 1, whih would give the integer 999, whih violates the ondition of distint digits. If 9R = 8, then R = 89 ; in this ase, 9R2 = 64 , whih is not an 9 7 2 , whih is not integer. If 9R = 7, then R = 9 ; in this ase, 9R = 49 9 2 an integer. If 9R = 6, then R = 3 , whene 9R2 = (9R)R = 6R = 4, whih yields the integer 964, whih is thus the largest 3-digit integer that is geometri. Thus, the dierene between the largest and smallest 3-digit integers that are geometri is 964 − 124 = 840. At this point, you're probably wondering about the preamble { the definition doesn't seem to be aeting anything so far. Here's another rak at the denition of a geometri sequene: Denition 2: A geometri sequene is a sequene of numbers with the property that if a, b, c are onseutive terms, then b2 = ac. 496 And another one: Denition 3: A geometri sequene is a sequene of numbers with the property that if a, b, c are onseutive terms, then ab = bc . Again, you may wonder what the big deal is all about. So I have a question for you: is 1, 0, 0 a geometri sequene? What do the dierent versions of the denition tell you? Solution to Problem 2 First, we express the logarithms in the three terms using a ommon base, namely the base 2. We obtain: 1 + log4 x = log8 4x = log2 x 1 log2 x ; log2 4 2 log2 4x log2 4 + log2 x 2 1 = = + log2 x . log2 8 3 3 3 1+ = 1+ Next, we make the substitution u = log2 x to make the next alulations less umbersome. In terms of u our sequene is thus u, 1 + 12 u, 23 + 13 u. Sine this sequene is geometri, then 2 1 1+ u = u 3(2 + u)2 12 + 12u + 3u2 0 0 = = = = 4u(2 + u) (multiplying 2 4u + 8u ; u2 − 4u − 12 ; (u − 6)(u + 2) ; 2 2 1 + u 3 3 ; by 12) ; and so u = log2 x = 6 or u = log2 x = −2, hene x = 64 or x = 41 . So what's the big deal? Let's look at what the sequenes are for the two possible values of x. If x = 64 (or u = 6), the sequene is 6, 4, 83 , whih is geometri and seems pretty innouous. If x = 14 (or u = −2), the sequene is −2, 0, 0. Oh dear! Why is this a problem? Whih denition are you using? This sequene is geometri by Denition 1 and Denition 2, but aording to Denition 3 it is NOT geometri. So the hoie of denition (that is, one's partiular onvention) hanges the answer to Problem 2. Using Denition 1 or Denition 2, the answer is x = 64 or x = 41 ; but using Denition 3, the answer is x = 64 only. There is a happy ending to this saga, though. Lukily, as the 2009 Eulid Contest was being pre-marked, the markers were alerted to this dilemma of dierenes of denitions and both versions, with proper justiation, were aepted as orret. So pay attention to denitions { are they ompletely preise? And think ritially about seemingly equivalent denitions { are they really equivalent?