The Sherwood Timber Company

 

By Wanting Zhong

 

Timmy Sherwood is a small, smiling middle-aged man with a big mustache and watery blue eyes. Don't be fooled by his appearance--he has a sharp, bright brain in his tiny bald head.

Curty and Sam are two of his young employees. They have just received their latest assignment--a batch of large, smooth, 4.6-meter long logs. Their boss had told them to cut it into two kinds of smaller logs, 0.7-meters and 0.4-meters. The number of 0.7-meter long logs should be twice that of 0.4-meters.

'I don't want to see any leftovers.' Mr. Sherwood said importantly. 'These precious logs are very hard to find. Just saw them up and pile them neatly into pyramids. The FireBlaze Furniture Co. would come soon. Do the job nicely and efficiently, OK?'

'Yes, sir!' Curty said courteously. 'We'll do what you want us to do.'

'There you are, Curty!' The boss exclaimed. 'I'll trust the job to you two.'

After Mr. Sherwood left, Curty became worried.

'Now, Sam.' He muttered. 'What are we gonna do? Sherwood is always presenting us with tough problems.'

'Don't worry, Curty.' Sam smiled kindly, 'I'll do the brain work. Just another easy little task for my super math brain!'

 

 

  广州市农林下路小学  六年（3）班  钟琬婷

 

:==============

Mr. Trotter’s notes:

soln:

let x = # of long pieces (0.7 m)

let y = # of short pieces (0.4 m)

x = 2y

0.7x + 0.4y = 4.6

7x + 4y = 46

7(2y) + 4y = 46

14y + 4y = 46

18y = 46

2*3*3y = 2*23(9)

y = 23

x = 2*23 = 46  [to be continued later]

:============

 

from Wanting, 6/17/03

Hi, Tio Terry,

Thank you for your reply!

Thanks for that suggestion about mince & hamburger. Yes, there are lots of hamburgers in Guangzhou and the children here love them! Especially McDonalds and KFC, you can spot them all the way across the city. However, fast food seemed to be attractive only to kids. Adults don't seem to enjoy them (at least my parents don't take much liking to them! :-))

As for that logs problem, the main thing is this: Curty and Sam didn't need to use the same plan on all of the logs. Maybe there are several very nice plans (for single logs) and they might combine several plans in order to complete this tough task. The 'x = 2y' formula is adapted for the total number of logs. There are nine logs in all.

All the best,

Nena Wanting

 

:========

from Andrei, 6/24/03

Dear Mr. Trotter,

Thank you for the problem, and excuse me for the delay, but bow I don't look the e-mail often.  I understood from the beginning that it is something not fully specified, and finally I understood that different logs must be cut in different ways.

I solved the problem using algebra: first I noted with k the number of 0.4-meter long logs and with 2k (as derived from the text) the number of 0.7-meter long logs. Now, I have the following equation (I do not write the units, as all of them are meters):

0.4 * k + 0.7 * 2k = 4.6

4k + 14k = 46

18k = 46

9k = 23

As 23 isn't divisible by 9, it is impossible to divide the big log in this manner.

Now, I note with x the number of 0.7-meter long logs from one big log, and with y the number of 0.4-meter long logs. I have:

0.7x + 0.4y = 4.6

7x + 4y = 46

I observe that x must be even. I take each possible value for x and calculate y, keeping in mind that x and y are positive integers:
   x =2   4y = 32 => y = 8                           
     (1st possibility)
   x = 4   4y = 18 => y is not an integer
   x = 6   4y = 4 => y = 1                           
     (2nd possibility)

So, a log could be cut in two different ways into 0.4 and 0.7 m pieces. The idea to finally obtain the required ratio of the number of pieces of each length is to cut a quantity of logs in the first manner (let a be the number of logs so cut) and another quantity in the second (let b be the number f logs). Now, I look at the condition:

2a + 6b = 2 * (8a + b)

a + 3b = 8a + b

2b = 7a

So, for every 2 logs cut with two 0.7-meter long logs and eight 0.4-meter long logs, I have to take 7 logs cut with six 0.7-meter long logs and one 0.4-meter long log. A condition to solve exactly the problem is that the number of big logs to be divisible by 9.

The final answer for the problem is the following:

The quantity of 4.6 m long logs available must be divided into two groups: one group contains 2/9 of the number of logs, and here from each log one makes 6 pieces of 0.7 m long logs and 1 piece of 0.4 m long logs, and the second group contains 7/9 of the number of logs, and here 2 pieces of 0.7 m and 6 pieces of 0.4 m long are realized from each log.

Yours,

Andrei

[N.B.  eight vs 6.  -tt]

 

:============

check of Andrei’s work by Mr T.

1^st try:

6 of long 7 dm     1 of short 4 dm

6 of long 7 dm     1 of short 4 dm

 

2 of long 7 cm       6 of short 4 dm

2 of long 7 cm     6 of short 4 dm

2 of long 7 cm       6 of short 4 dm

2 of long 7 cm       6 of short 4 dm

2 of long 7 cm       6 of short 4 dm

2 of long 7 cm       6 of short 4 dm

2 of long 7 cm       6 of short 4 dm

26 long pieces   vs   44 short pieces  à  incorrect

 

2^nd try:

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

 

2 of long 7 cm       6 of short 4 dm

2 of long 7 cm       6 of short 4 dm

46 long pieces   vs   19 short pieces  à incorrect

 

3^rd try:

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

6 of long 7 dm       1 of short 4 dm

 

2 of long 7 cm       8 of short 4 dm

2 of long 7 cm       8 of short 4 dm

46 long pieces    vs   23 short pieces using 9 big logs à YES!

Key point: if the number of uncut logs in the pile is a multiple of 9, the problem can be solved.  Otherwise no.