Difference between revisions of "2006 AMC 12A Problems/Problem 14"
(→Solution 3) |
Coltsfan10 (talk | contribs) m (→Solution 2) |
||
| Line 15: | Line 15: | ||
===Solution 2=== | ===Solution 2=== | ||
Alternatively, note that <math>300p + 210g = 30(10p + 7g)</math> is divisible by 30 no matter what <math>p</math> and <math>g</math> are, so our answer must be divisible by 30. In addition, three goats minus two pigs gives us <math>630 - 600 = 30</math> exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is <math>\mathrm{(C) \ }</math>. | Alternatively, note that <math>300p + 210g = 30(10p + 7g)</math> is divisible by 30 no matter what <math>p</math> and <math>g</math> are, so our answer must be divisible by 30. In addition, three goats minus two pigs gives us <math>630 - 600 = 30</math> exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is <math>\mathrm{(C) \ }</math>. | ||
| − | |||
===Solution 3=== | ===Solution 3=== | ||
Revision as of 15:32, 17 September 2020
- The following problem is from both the 2006 AMC 12A #14 and 2006 AMC 10A #22, so both problems redirect to this page.
Problem
Two farmers agree that pigs are worth 
dollars and that goats are worth 
dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a 
dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?
Solutions
Solution 1
The problem can be restated as an equation of the form 
, where 
is the number of pigs, 
is the number of goats, and 
is the positive debt. The problem asks us to find the lowest x possible. and must be integers, which makes the equation a Diophantine equation. Bezout’s Identity tells us that the smallest 
for the Diophantine equation 
to have solutions is when is the greatest common divisor of 
and 
. Therefore, the answer is 
, which is 
, 
Solution 2
Alternatively, note that 
is divisible by 30 no matter what and are, so our answer must be divisible by 30. In addition, three goats minus two pigs gives us 
exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is .
Solution 3
Let us simplify this problem. Dividing by , we get a pig to be: 
, and a goat to be 
.
It becomes evident that if you exchange 
pigs for 
goats, we get the smallest positive difference - 
, since we can't made a non-integer with a linear combination of integers.
Since we originally divided by , we need to multiply again, thus getting the answer: 
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
