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Puzzle 6-08
Puzzle winer |
Chanthu Pillai
Crystal Seidel
Jennifer Thompson
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Problem |
topic: Close out sale.
A store has a closeout sale on all remaining 40-watt fluorescent tubes. The first customer purchased half of all of the tubes plus one half a tube more. The second customer purchased half of the remaining tubes plus one half a tube more. The third customer purchased half of the remaining tubes plus one half a tube more, Finally the last customer purchased the 50 remaining tubes. How many tubes were on sale at the store? |
Solution |
F denotes the total number of 40-watt fluorescent tubes in closeout sale.
Customer one purchased
C1 = F/2 + 1/2 tubes.
Number of tubes remaining is
L1 = F - C1
L1 = F/2 - 1/2 .
Customer two purchased
C2 = (L1)/2 + 1/2
C2 = [F/2 - 1/2]/2 + 1/2
C2 = F/4 - 1/4 + 1/2
C3 = F/4 + 1/4 tubes.
Tubes remaining
L2 = L1 - C2
L2 = [F/2 - 1/2] - [F/4 + 1/4]
L2 = F/4 - 1/4 - 1/2
L2 = F/4 - 3/4.
Customer three purchased
C3 = (L2)/2 + 1/2
C3 = [F/4 - 3/4] /2 + 1/2
C3 = F/8 - 3/8 + 1/2
C3 = F/8 + 1/8 tubes.
Tubes remaining
L3 = L2 - C3
L3 = [F/4 - 3/4] - [F/8 + 1/8]
L3 = F/8 - 7/8.
Last customer purchased all remaining 40-watt fluorescent tubes
C4 = L3
C4 = 50 tubes.
Therefore,
F/8 - 7/8 = 50
F – 7 = 400
and finally,
F = 407.
The number of tubes on sale was 407, C1 = 204, C2 = 102, C3 = 51 and C4 = 50. |
Puzzle 5-08
Puzzle winer |
Mr Daniel Duan |
Problem |
topic: Heavy and heavier bricks.
We have ten rectangular bricks of exactly the same dimensions that are labeled one
through ten. Each brick is of uniform density but the mass of the brick labeled 1 is M,
the mass of the brick labeled 2 is 2M, . . . , mass of the brick labeled K is KM and
finally the mass of the brick labeled 10 is 10M. How may these bricks be stacked, only
one upon the other, such that the top brick is at a maximal horizontal distance from the
bottom brick and what is this distance?
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Solution |
Download solution here. |
Puzzle 4-08
Puzzle winer |
There is no winner. |
Problem |
topic: Integers
Show that if N is a positive integer, then one of N - 1, N or N + 1 may be written as a positive integer M + the sum of the digits composing M i.e., if M = m3 m2 m1 = (m3 )(100) + ( m2 )(10) + ( m1 )(1), then N could be written as M + (m3 + m3 + m1). As an example, if N = 34, then N = 26 + (2 + 6) = M + (m2 + m1) where M = 26. Another example, if N = 21, the N = 15 + (1 + 5) = M + (m2 + m1) . where M = 15. Finally, if K = 31, then there is no such integer M.
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Solution |
Download solution here. |
Puzzle 3-08
Puzzle winer |
Mr. Ricky J. Girardi |
Problem |
topic: Positive Integers
Determine, if possible, eleven consecutive integers such that the sum of their squares is 2266. |
Solution |
Suppose n is an integer.
The sum of 11 consecutive integers, starting at n, may be written as
n^2 + (n + 1)^2 + (n + 2)^2 + . . . + (n + 10)^2 .
Which may be written
n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) + (n^2 +6n + 9) + . . . + (n^2 +20n + 100).
11n^2 + 2n + 4n + 6n + . . . + 20n + 1 + 4 + 9 + , , , + 100.
11n^2 + 2n(1 + 2 + 3 + . . . + 10) + (1 + 4 + 9 + , , , + 100).
11 n^2 + 110n + 385.
Can we determine an integer n such that
11 n^2 + 110n + 385 = 2266.
11 n^2 + 110n - 1881 = 0.
n^2 + 10n - 171 = 0.
(n -- 9)(n + 19) = 0.
Thus we have that n = 9 OR n = - 19. |
Puzzle 2-08
Puzzle winer |
There is no winner. |
Problem |
topic: Coffee in the mug
A mug of coffee is at a temperature of 192 degrees F. in a large room at a temperature of 72 degrees F. After exactly five minutes the coffee cools to 162 degrees F. Write the cooling function for this situation and determine the temperature of the coffee at exactly15 minutes. Initially the boiling hot coffee was poured into a boiling hot mug. How long had the coffee in the mug been in the room when the temperature of 192 degrees F was measured? |
Solution |
A mug of coffee is at a temperature of 192 degrees F. in a large room at a temperature of 72 degrees F. After exactly five minutes the coffee cools to 162 degrees F.
(1.) Write the cooling function for this situation.
(2.) Determine the temperature of the coffee at exactly 15 minutes after the temperature
of 162 was recorded.
Initially the boiling hot coffee was poured into a boiling hot mug.
(3.) How long had the coffee in the mug
been in the room when the temperature of 192 degrees F was measured?
(1.) USING NEWTON'S LAW OF COOLING
T(t) = Aexp(-kt) + 72.
Using the fact that initially the temperature was 192 degrees F, we have that T(0) is 192.
192 = Aexp( 0 ) + 72
192 = A + 72
We now have that A is 120.
The equation now stands as T(t) = 120exp(-kt) + 72
with k yet to be determined.
After 5 minutes we have that the temperature is 162 degrees F, I.e., T(5) is 162. T(5) = 162
This results in our having the expression
162 = 120exp(-5k) + 72.
or
90 = 120exp(-5k).
From this we find that k may be determined b (using the natural logarithm ln).
k = (1/5)ln(120/90)
= 0.057536.
Using this value for k we have that
T(t) = 120exp(-0.057536t) + 72 Which answers part 1.
(2.) At t = 15 minutes
T(15) is 120exp(-15*0.057536) + 72. = 122.625
The temperature would be 122.65 degrees F 15 minutes later, which answers part 2.
(3.) Now the last portion, how long was the coffee sitting before temperature
was 192 degrees F ? We shall assume initial temperature was 212 degrees F.
T(t) = Aexp( -kt ) + 72
where A is 120, k is 0.057536 and to is the time such that 212 = 120exp( -0.057536 to) + 72.
From which it follows that
exp(-0.057536*to) = (212 - 72)/120.
and hence that
- 0.057536*to = ln[140/120]
= 0.154151 to = (ln[140/120])/(-0.057536) =0.154151/(-0.057536 ) =
-2.6792
which indicates that the coffee was poured 2.6792 minutes earlier than the time of the
first temperature measurement of 162 degrees F which answers part 3. |
Puzzle 1-08
Puzzle winer |
There is no winner. |
Problem |
topic: Rational and Irrational
First describe every rational number x such that x + 1/x is an integer and then describe every irrational number y such that y + 1/y is an integer |
Solution |
Download solution here.
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Puzzle 4-07
Puzzle winer |
Eugene Chaung, Justin Hansen, Chris Howard |
Problem |
topic: Distance
From each of two doorways, on opposite sides of a room, a person walks without stopping and at a constant speed to the doorway on the opposite side. At more than one third of the way across and 50 feet from starting doorway the slower walker meets the other walker. Upon reaching the opposite doorway, each reverses direction and walks, at the same constant speed as before and without stopping, to the doorway from which they had originally started. They pass one another at a distance of 20 feet from one of the doorways. What is the distance between the doorways? |
Solution |
Download solution here |
Puzzle 3-07
Puzzle winer |
There is no winner. |
Problem |
Topic: Age
Problem: Describe all the circumstances and conditions under which two given people can have ages, in years, whose digits are exactly reversed from each other. An example could be 29 and 92. We assume that the ages are less than 100 and also that 00, 01, 02, 03, … and 09 are the digits for age 0, 1, 2, 3, … and 9 respectively. |
Solution |
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Puzzle 2-07
Puzzle winer |
There is no winner. |
Problem |
Topic: Roots
Problem: Determine all real numbers A such that Ax3 + (2 – A)x2 – x – 1 has three real roots r1, r2 and r3, two of which have the same value, that is, r1= r2, r2 = r3 or r1= r3. Create, for each such A, (A: r1, r2, r3), where A is a real number and r1, r2 and r3 are the associated real roots, two of which have the same value. |
Solution |
Download solution here |
Puzzle 1-07
Puzzle winer |
Ms. Melody Lam |
Problem |
topic: prime numbers
Determine if there is a positive integer n such that both 2n + 1 and 2n – 1 are prime numbers. If there is such a number, then determine all such. Carefully justify your answer. |
Solution |
Download solution here.
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