POW 3 MATH 2
11/15/09
Class G
Lauren McCarthy
Pow 3: Eight Bags of Gold
Problem Statement
A king divides his gold among 8 trusted people. One of the trusted people is selling his gold. The king wants to find the thief but only has a pan balance.
Being conservative, he wants to use the pan balance as few times as possible. What is the least number of trials he will have to do in order to guarantee that he has found the lightest bag?
Process
To solve the problem, I accumulated the least number of trials into a table to look for a pattern. of Bags Least of Gold of Trials
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
1
1
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
20
21
22
23
24
25
26
27
28
3
3
3
3
3
3
3
3
4
29
30
4
4
Tallies
1 vs. 1
1 vs. 1 r .1
11 vs. 11 1 vs. 1
11 vs. 11 r .1 1 vs. 1
111 vs. 111
1 vs. 1 r .1
111 vs. 111 r. 1 1 vs. 1 r .1
111 vs. 111 either 11 vs. 11
1 vs. 1 OR 111 vs. 111
1 vs. 1 r .1
A similar pattern continues throughout the table
Their must always be an equal amount of bags on each side, such as :
5 vs. 5
6 vs. 6
6 vs. 6
As the number of bags increased, I started to discover a patter and realized that the "transition" numbers as I call it, such as 10 which requires 3 trials rather than 2, were key in finding a pattern.
The change from 3 to 4 trials occured because the tallies became :
11 111 1111 vs. 11 111 1111 with 10 bags left over, because
10 bags a transition number requires 3 trials, not 2, it became 1 9 vs. 9
3 5 vs. 5 trials 4 trials total
*each vs. counts as one trial on the pan balance. r. = remainder
I had to assume that the pan-balance could hold more than one bag of gold on each side and that their was only one thief.
Using my white-board in my room, I looked for the least number of trials using tally marks, such as 4 bags of gold= 11 vs. 11 -> 1 vs. 1 so two trials.
Printed by Mathematica for Students
Page 1 of 2.
11/15/09
Class G
Lauren McCarthy
Using my white-board in my room, I looked for the least number of trials using tally marks,
Class G
Lauren McCarthy
Pow 3: Eight Bags of Gold
Problem Statement
A king divides his gold among 8 trusted people. One of the trusted people is selling his gold. The king wants to find the thief but only has a pan balance.
Being conservative, he wants to use the pan balance as few times as possible. What is the least number of trials he will have to do in order to guarantee that he has found the lightest bag?
Process
To solve the problem, I accumulated the least number of trials into a table to look for a pattern. of Bags Least of Gold of Trials
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
1
1
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
20
21
22
23
24
25
26
27
28
3
3
3
3
3
3
3
3
4
29
30
4
4
Tallies
1 vs. 1
1 vs. 1 r .1
11 vs. 11 1 vs. 1
11 vs. 11 r .1 1 vs. 1
111 vs. 111
1 vs. 1 r .1
111 vs. 111 r. 1 1 vs. 1 r .1
111 vs. 111 either 11 vs. 11
1 vs. 1 OR 111 vs. 111
1 vs. 1 r .1
A similar pattern continues throughout the table
Their must always be an equal amount of bags on each side, such as :
5 vs. 5
6 vs. 6
6 vs. 6
As the number of bags increased, I started to discover a patter and realized that the "transition" numbers as I call it, such as 10 which requires 3 trials rather than 2, were key in finding a pattern.
The change from 3 to 4 trials occured because the tallies became :
11 111 1111 vs. 11 111 1111 with 10 bags left over, because
10 bags a transition number requires 3 trials, not 2, it became 1 9 vs. 9
3 5 vs. 5 trials 4 trials total
*each vs. counts as one trial on the pan balance. r. = remainder
I had to assume that the pan-balance could hold more than one bag of gold on each side and that their was only one thief.
Using my white-board in my room, I looked for the least number of trials using tally marks, such as 4 bags of gold= 11 vs. 11 -> 1 vs. 1 so two trials.
Printed by Mathematica for Students
Page 1 of 2.
11/15/09
Class G
Lauren McCarthy
Using my white-board in my room, I looked for the least number of trials using tally marks,