Ms. Washington
Ms. Washington is a math teacher at the Bonsworth High. She loves her job despite its challenges. An interesting aspect of this school is the student diversity. The compul sory math course has students like Delwin who is the star of the school football team and has no interest in math. He spends most of his time in game practice or at the gym. She has to make him feel that he is capable of learning the subject. Rosie gets her kicks by gossiping. Alex thinks that he is the smartest kid ever born. He shows off on every chance he gets. Rosie will tell you that they call him Smart Alec. Sara is a math whiz but minds her own business. She is a top student and loves her boyfriend Johnny.
Ms, Washington challenges them all – Delwin, Alex or Sara. How does she do it ? This is what she did today. She wrote these numbers on the board and said, “Who can add all these numbers ?”
Delwin’s Method
Many students raised their hands but she called on Delwin: why don’t you show everyone to do the summation?
Delwin: Can I use a calculator ?
Ms. Washington thought for a second and then allowed it.
| ① | ② | ③ | ④ | ⑤ | ⑥ | ⑦ | |
| A | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| B | 2 | 4 | 6 | 8 | 10 | 12 | 14 |
| C | 4 | 8 | 12 | 16 | 20 | 24 | 28 |
| D | 8 | 16 | 24 | 32 | 40 | 48 | 56 |
| E | 16 | 32 | 48 | 64 | 80 | 96 | 112 |
| F | 32 | 64 | 96 | 128 | 160 | 192 | 224 |
| G | 64 | 128 | 192 | 256 | 320 | 384 | 448 |
Sara wrote the answer 3556 in her note book even before Delwin pulled out his calculator. Rosie saw this and wanted to tell everyone.
Delwin entered the numbers one at a time on the calculator while getting angry that he had to do all this work. In 15 minutes, he wrote the answer 3556 on the board.
Delwin: Ms. Washington, is this answer correct ?
Ms. Washington nodded to say yes, and Delwin felt proud that he could do the sum.
Alex’s Method
Two more students raised hands. They were Alex and Rosie.
Ms. Washington: Yes, Alex. What do you want to say ?
Alex: Ms Washington, I can do it much faster.
Ms. Washington: Please, tell.
| ① | ② | ③ | ④ | ⑤ | ⑥ | ⑦ | ||
| A | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 28 |
| B | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 56 |
| C | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 112 |
| D | 8 | 16 | 24 | 32 | 40 | 48 | 56 | 224 |
| E | 16 | 32 | 48 | 64 | 80 | 96 | 112 | 448 |
| F | 32 | 64 | 96 | 128 | 160 | 192 | 224 | 896 |
| G | 64 | 128 | 192 | 256 | 320 | 384 | 448 | 1792 |
| 3556 |
Alex: The numbers in each row conform to an arithmetic series.
In row A, the difference between the numbers in the consecutive columns is 1, in row B it is 2. For arithmetic series,
The sum of a row = ((First term + last term)/2) x the number of terms.
This can also be written as:
The mean number of the row x the number of rows.
The row mean which is the number in the middle column ④. I just added all the terms of row ④ which came out to be 508 and multiplied it by the number of rows which is 7. It came out to be 3556 – the same answer as Delwin’s.
Ms. Washington: Thanks Alex. That was smart thinking. You defined an arithmetic series for us. In this the difference between each consecutive numbers is constant. Rosie – you had also raised your hand.
Rosie: Ms. Washington, I saw Sara write this answer in her notebook as soon as you as wrote the numbers – way before Delwin even got his calculator out. I am curious, how she did it.
Sara goes to the board
Ms. Washington: Sara, what did you have up your sleeve ? Come to the board and show the class.
Sara: Alex pointed out that the numbers in each row formed an arithmetic series.
I saw that the numbers in row B were double those in A. In fact, the numbers in column ① were 1, 2, 4, 8, 16, 32, 64. The ratio of numbers from one row to the next was 2 x. So these numbers formed a geometric series.
For a geometric series:
sum = a (rn-1)/(r-1), where a = the first term, r = the common ratio which here is 2, and n = the number of terms which here is 7.
So the sum of the terms in the first column will be 1 x (r7-1)/(r-1).
The last row starts with r6 =64, thus r7 = 2 x 64 =128 and r7-1= 127. If we use Alex’s formula. instead of column ①, I can use
a = 4 x 7 for the sum of the first row to be 28.
Then the final answer will be 28 x 127 = 3556.
Johnny: Hello everybody, let us say three cheers for Sara for teaching us what a geometric series is.
Ms. Washington: I am glad that everyone got the same answer. Alex and Sara, thanks for teaching the class about the arithmetic and the geometric series.
I want you to go home and talk to your family about the song 12 days of Christmas. Figure out, the total cost to the true love. By the way, this way you can also show off with what Sara taught you. The assignment sheet is on the table. Take one and have fun doing it.
Challenge
Twelve Days of Christmas (Slightly modified, cost of the new items for each day is given)
On the first day of Christmas my true love gave to me
A partridge in a pear tree ($100)
On the second day of Christmas my true love gave to me two turtle doves ($100) and a partridge in a pear tree
On the third day of Christmas my true love gave to me three French hens ($200), two turtle doves and a partridge in a pear tree
On the fourth day of Christmas my true love gave to me four calling birds ($400), three French hens, two turtle doves and a partridge in a pear tree
On the fifth day of Christmas my true love gave to me five silver rings ($800),
four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the sixth day of Christmas my true love gave to me six brown geese ($1600), five silver rings, four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the seventh day of Christmas my true love gave to me seven swans a swimming ($3200), six brown geese, five silver rings, four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the eighth day of Christmas my true love gave to me eight maids a cleaning ($6400), seven swans a swimming, six brown geese, five silver rings, four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the ninth day of Christmas my true love gave to me nine ladies dancing ($12800), eight maids a cleaning, seven swans a swimming, six brown geese, five silver rings, four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the tenth day of Christmas my true love gave to me ten lords a leaping ($25,600), nine ladies dancing, eight maids a cleaning Seven swans a swimming, six brown geese, five silver rings, four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the eleventh day of Christmas my true love gave to me eleven pipers piping ($51,200), ten lords a leaping Nine ladies dancing, eight maids a cleaning, seven swans a swimming, six brown geese, five silver rings Four calling birds, three French hens, two turtle doves and a partridge in a pear tree
On the twelfth day of Christmas my true love gave to me twelve drummers drumming ($102,400), eleven pipers piping,ten lords a leaping Nine ladies dancing, eight maids a cleaning, seven swans a swimming, six brown geese, five silver rings Four calling birds, three French hens, two turtle doves and a partridge in a pear tree
Now, let us calculate what it cost the true love on each day (previous day + new items for the day)
| Day of Christmas | Gift to True Love | Cost of the new items | All gifts for the Day |
| First | Partridge in Pear Tree | $100 | $100 |
| Second | Two Turtle Doves | $100 | $200 |
| Third | Three French Hens | $200 | $400 |
| Fourth | Four Calling Birds | $400 | $800 |
| Fifth | Five Silver Rings | $800 | $1,600 |
| Sixth | Six Brown Geese | $1,600 | $3,200 |
| Seventh | Seven Swans-a-Swimming | $3,200 | $6,400 |
| Eighth | Eight Maids- a cleaning | $6,400 | $12,800 |
| Ninth | Nine Ladies Dancing | $12,800 | $25,600 |
| Tenth | Ten Lords-a-Leaping | $25,600 | $51,200 |
| Eleventh | Eleven Pipers Piping | $51,200 | $102,400 |
| Twelfth | Twelve Drummers Drumming | $102,400 | $204,800 |
Note that the cost for the day doubled each time indicating that this is geometric series with a = cost for the first day = $100, r = common ratio = 2, and n = number of terms =12.
Therefore, the total cost for all the days = a (rn-1)/(r-1) = 100 (212-1)/(2-1) = 100 (212-1)
From the data you can see that 211 = the cost for the twelfth day = $204800.
Therefore 100 (212-1) = 2 x $204800 -100 = $409600 -100 = $ 490500.
Math Stories For You 