# Thread: algebra problem world problem

1. ## algebra problem world problem

The Emperor of Qin secretly counts his soldiers in an enigmatic manner. When he counts his soldiers in groups of three, two soldiers are left. When the emperor counts in groups of five, three soldiers are left, and when the soldiers are counted in groups of seven, two soldiers are left.

How many soldiers are there?

What is the smallest number of soldiers possible?

What are some of the other possible numbers of soldiers that could be there? Try to generalize.

i'm going crazy with this problem..please SHOW ALL WORK, and i relaly need this in a hurry!!!!!!! if you don't want to solve the whole problem then please just give me a little way to head here..and explain clearly ALSO can you please tell me if you'r a 100% sure or not THANK YOU i really appreciate it     2. well, I guess this is a start but not the correct method.
I think you may have to use algebra in this but
through the process of trail and error

I found the number of soldeirs to be:

23

since you now know the answer, maybe someone from there could help fit it into an equation?

Sorry, I can't help much but at least I found a part of the answer

3. Originally Posted by ncbabe The Emperor of Qin secretly counts his soldiers in an enigmatic manner. When he counts his soldiers in groups of three, two soldiers are left. When the emperor counts in groups of five, three soldiers are left, and when the soldiers are counted in groups of seven, two soldiers are left.

How many soldiers are there?

What is the smallest number of soldiers possible?

What are some of the other possible numbers of soldiers that could be there? Try to generalize.
Call N the number of soldiers. Then we know that
N = 3x + 2
N = 5y + 3
N = 7z + 2
where x, y, and z are (positive) integers.

Let's reduce the number of variables a little.
3x + 2 = 7z + 2 ==> z = (3/7)x

From the first condition x must be divisible by 7, else z is not an integer.

So let's try x = 7:
N = 3*7 + 2 = 23

Does this work? This implies y = 4 and z = 3. Since these are integers, this works. Thus we have a solution, N = 23.

Are there any other solutions?
Let's try x = 14:
N = 3*14 + 2 = 44 ==> y = 41/5, so this doesn't work.

x = 21 ==> y = 62/5
x = 28 ==> y = 83/5
x = 35 ==> y = 104/5
x = 42 ==> N = 128, y = 25, z = 18 <-- Another solution!

So there is more than one solution. In fact:
N = 23, x = 7, y = 4, z = 3
N = 128, x = 42, y = 25, z = 18
N = 233, x = 77, y = 46, z = 33
N = 338, x = 112, y = 67, z = 48
etc.

Can you see a pattern here? (Make sure you prove it!)

-Dan

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