How many coins are there? (Please verify/confirm my work)

[Math100]

Homework Statement

Solve the puzzle-problem below:
Yen Kung, 1372. We have an unknown number of coins. If you make 77 strings of them, you are 50 coins short; but if you make 78 strings, it is exact. How many coins are there?
[Hint: If N is the number of coins, then N=77x+27=78y for integers x and y.]

Relevant Equations

None.

Proof: Let N be the number of coins.
From the hint, we get the following Diophantine equation:
N=77x+27=78y for integers x and y.
Applying the Euclidean Algorithm produces:
78=1(77)+1
77=77(1)+0.
Now we have gcd(77, 78)=1.
Note that 1$\mid$27.
Since 1$\mid$27, it follows that the Diophantine equation
77x+27=78y can be solved.
Then we have 1=78-1(77).
This means 27=27[78-1(77)]
=27(78)-27(77)
=27(78)+27(-77).
Thus, xo=27 and yo=27.
Substituting xo=27 and yo=27 into the Diophantine equation
77x+27=78y, we get 77(27)+27=78(27)=2106.
Therefore, there are 2106 coins.

Is this the right/correct answer? I couldn't find the answer from my textbook because it didn't provide any. Can someone please review/confirm my work for this problem and see if this is the correct answer? Thank you.

 

[FactChecker]

Why don't you test your answer in the original problem statement?

 

[Math100]

Why don't you test your answer in the original problem statement?

How should I do that?

 

[FactChecker]

Plug your value into the equations and see if it works. I did it in about 3 minutes.
Checking your answers is a step that you should always do until you have completely mastered the problems.

 

[Math100]

I got 77(27)+27=78(27)=2106.

 

[Frabjous]

I think the hint is wrong. Where did the “50“ go?

 

[FactChecker]

It is best to go to the very original problem statement. That way you know that there is no mistake.

 

[Math100]

I think the hint is wrong. Where did the “50“ go?

Really? Let me think. I thought that the hint is correct, so I falsely followed it.

 

[Math100]

But if that's the case, then we should have 77x+50=78y since the original problem mentioned that "you are 50 coins short". And I substituted the value of xo=27, yo=27 into 77(27)+50=78(27) but the equation doesn't make sense at all, because the left side of the equation doesn't match/equal to the right side of the equation.

 

[Frabjous]

1) I think you have misinterpreted 50 coins short. You are 50 coins short of 77 strings.
2) The answer (27,27) you derived with the wrong equation will not solve the correct one.

 

[FactChecker]

The answer you got from the hint gives the correct answer to the original problem statement. So there is some translation from the 50 in the problem to the 27 in the hint. The fact that 50+27=77 may have something to do with it.

 

[FactChecker]

(2106+50)/77=28 and 2106/78=27. So the solution to the hint works in the original problem statement. Now the question is, where did the 50 go to and the 27 come from in the hint?

 

[Math100]

(2106+50)/77=28 and 2106/78=27. So the solution to the hint works in the original problem statement. Now the question is, where did the 50 go to and the 27 come from in the hint?

77-50=27

 

[Frabjous]

(2106+50)/77=28 and 2106/78=27. So the solution to the hint works in the original problem statement. Now the question is, where did the 50 go to and the 27 come from in the hint?

That‘s a cool trick. It took me a while to figure it out.

 

[Orodruin]

It boils down to being more natural to work with 27 than with -50. In 77x+27, x is the number of coins in each non-completed string. If you instead let z be the number of coins in each completed string then x=z-1 and 77x+27 = 77z-50. These are equivalent since x is an integer iff z is an integer.

 

[PeroK]

You can solve it directly as follows. You start laying out the coins in 77 strings (rows), but you are 50 coins short, so the first 27 rows have an extra coin:

27 rows with $k + 1$ coins
50 rows with $k$ coins

Now, take one coin from each of the first 27 rows and make a 78th row to give:

27 rows with $k$ coins
50 rows with $k$ coins
1 row with 27 coins

Therefore: $k = 27$ and $N =78 \times 27$.

 

[PeroK]

How should I do that?

Get hold of 2106 coins. Or, manipulate 2106 boxes on a spreadsheet.

 

[Frabjous]

Why is the ‘27‘ version of the equation preferable to the ’-50‘ version?

If you have an equation of the form,
ax+b=cy {a,b,c>0}{c>a}
then
x=y=b/(c-a) is a solution when b/(c-a) is an integer
This is probably generalizable.

 

[Orodruin]

Why is the ‘27‘ version of the equation preferable to the ’-50‘ version?

If you have an equation of the form,
ax+b=cy {a,b,c>0}{c>a}
then
x=y=b/(c-a) is a solution when b/(c-a) is an integer
This is probably generalizable.

The versions are completely equivalent as I showed in #15. It is only a matter of which people tend to feel more natural to work with.

 

[Frabjous]

The versions are completely equivalent as I showed in #15. It is only a matter of which people tend to feel more natural to work with.

I was aiming for asked and answered. The ‘27’ equation allows you to guess x=y which is simpler to solve.