Date: Thu, 28 Aug 1997 21:23:04 +0800
Subject: 2 mathematical problems
Dear Math Central,
I am a trainee teacher. I would appreciate it very much if you could help me solve the following two problems which I think are meant for students at middle grade.
 If a hen and a half lays an egg and a half in a day and a half, how many hens are needed to lay a dozen eggs in one day?
 There are fewer than 200 passengers on a train. If they get off in pairs, one passenger will be without a partner. If they get off in groups of 3 or 4, there will still be one passenger left by himself. However, if they get off in groups of 5, no one will be left by themselves. How many passengers are there on the train?
Thank you very much. I look forward to hearing from you.
Yours sincerely,
Faye
Hi Faye
For your first problem it is helpful to make a chart for hens(H), eggs(E) and days(D) as follows:
H  E  D 
1 1/2  1 1/2  1 1/2 
.  .  . 
.  .  . 
.  .  . 
?  12  1 
To make the units easier first try to eliminate some fractions.
H  E  D 
1 1/2  1 1/2  1 1/2 
3  3  1 1/2 
12  12  1 1/2 
.  .  . 
18  12  1 



< Twice the hens gives twice the eggs in the same time.

< Likewise 

< 1 1/2 times as many hens needed to speed up the process. 

There is a problem here with this solution as a valid row in the table is
and there is a legitimate argument to say that no number of hens can lay a dozen eggs is one day since it takes each hen a day and a half to lay an egg.
For your second problem there is a general method known as the Chinese Remainder Theorem but for a single problem you can just look at the arithmetic progressions involved.
Let the number be n then we know n<200 and n is a multiple of 2 plus 1, a multiple of 3 plus 1, a multiple of 4 plus one and a multiple of 5. Thus
n is in  {1, 3, 5, 7, ... 199} 
n is in  {1, 4, 7, 10, ... 199} 
n is in  {1, 5, 9, 13, ... 197} 
and  
n is in  {5, 10, 15, 20, ... 195} 
You want to find the numbers that are in all 4 sets. Let's use the 4th set to
shorten the others as a starter. We see that 5 is in the 1st and 4th set, and one set
goes up by twos and the other by fives, thus 10 being the least common multiple
(lcm) of 2 and 5 we will find 15 in both sets, 25 in both sets, 35 in both sets
etc. The same argument can be used with the 4th set and each of the remaining
two sets to reduce the problem to finding n where
n is in  {5, 15, 25, 35, ... 195} 
n is in  {10, 25, 40, 55, ... 190} 
n is in  {5, 25, 45, 65, ... 185} 
and  
n is in  {5, 10, 15, 20, 25, ... 195} 
It is now easy to see that one answer is 25. To find others add the lcm of {2,
3, 4, 5}, which is 60, to get 85 and 145 as the other possibilities less than 200.
Cheers,
Penny
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