00:01
So we have a chest that can contain up to a thousand coins.
00:04
We have 11 people.
00:05
Let's call the number of coins in this chest n.
00:08
We only know that n is greater than zero and less than 1 ,000.
00:13
We can evenly distribute the coins in the following way.
00:17
If we give them to all 11 people, there's one coin left over.
00:22
If we give all of the coins to only 10 people, there's one coin left over.
00:26
And we give the coins to only eight people, to only nine people.
00:31
There are eight coins left over.
00:33
So what this means is that let's say that we can rewrite n as 11a plus 1.
00:45
This is because a is the maximum number of times that 11 will go into n.
00:53
And we have a remainder of 1.
00:56
That's exactly what this information is telling us.
00:59
Similarly, n is equal to 10b plus 1 for some integer b.
01:05
And then finally, n is also equal to 9c plus 8.
01:12
And with this, we can set up a system of equations, which i have written here, and solve for n.
01:20
We're going to find a general solution and then figure out what the value of n is from there.
01:26
So let's use the first two equations.
01:28
So we get 11a plus 1 is equal to 10b plus 1.
01:33
So we get minus 1.
01:36
We subtract both sides by 1.
01:38
So we get 11a is equal to 10b.
01:41
Dividing by both sides by b, we get b is equal to 11a over 10.
01:47
So that means 10 divides a.
01:50
In other words, a is divisible by 10.
01:54
We'll come back to this and will actually be pretty important.
01:59
Secondly, we can solve for, we can set the first equation equal to the third equation.
02:06
So we get 11a plus 1 is equal to 9c plus 8.
02:14
And we can subtract both sides by 8.
02:19
So we get 11a.
02:21
And let me rewrite that...