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Puzzle 6-08

Puzzle winer

Chanthu Pillai
Crystal Seidel
Jennifer Thompson

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?

Solution
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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.

 

Solution
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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

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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

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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

 

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

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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

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