Download Problem of the Month

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?