Beautiful Liars

Firstable, find the ways to select who told the truth:

$\text{Alan,Blas,David}$

$\text{Alan,Blas,Ronnie}$

$\text{Alan, Blas,Jane}$

$\text{Alan,David,Jane}$

$\text{Alan,David,Ronnie}$

$\text{Alan Jane,Ronnie}$

$\text{Blas,David,Jane}$

$\text{Blas,David,Ronnie}$

$\text{Blas, Jane, Ronnie}$

$\text{David,Jane, Ronnie}$

Then analyse these cases:

- $\text{Alan}$ and $\text{David}$ could not tell the truth at the same time because if $27$ is divides $abc$ $\Rightarrow$ $9$ divides $abc$ so $a+b+c$ must be divisible by $9$ and $15$ is not divisble by $9$

- $\text{Jane}$ and $\text{Blas}$ , both could not tell the truth because $11$ does not divide $648,000$ .

-If $\text{Ronnie}$ told the truth $\text{David}$ could not tell the truth because $abc$ only would be divisible by $3$ not by $9$ .

Discarding cases only remain two: $\text{Alan,Blas,Ronnie}$ and $\text{Alan Jane,Ronnie}$ .

If $\text{Alan,Blas}$ and $\text{Ronnie}$ told the truth $\Rightarrow$ the least number that satisfy this conditions is $27*11*11*3=9801$ but it is not a number of three digits.

If $\text{Alan Jane}$ and $\text{Ronnie}$ told the truth $\Rightarrow abc$ i a divisor of $648,000=2^6\times 3^4\times5^3$ so $abc$ would be divisible by $3^4$ and combining is the only number $3^4 \times 2^2=324$ which satisfies all the conditions.

$\boxed{n=324}$

Observe that Blas and Jane cannot both tell the truth, since $11 \not| \, 648000$ .

Observe that David implies that $\overline{abc} \equiv 6 \pmod 9$ . But this contradicts both Alan and Ronnie (Alan implies that $\overline{abc} \equiv 0 \pmod 9$ , while Ronnie says $\overline{abc} \equiv 0,1,4,7 \pmod 9$ ). Since we have accounted for one liar among Blas and Jane, it cannot be the case that both Alan and Ronnie lie, so David lies. Thus the number is both divisible by $27$ and is a perfect square, which means it's divisible by $81$ .

There aren't many such numbers: $81, 324, 729$ , and in fact $81$ is impossible as it's not three digits. So the number is either $324$ or $729$ . None of them is divisible by $11$ , so Blas must necessarily lie, so Jane tells the truth. But $729 \not| \, 648000$ , so the number is $\boxed{324}$ ; Alan, Jane, and Ronnie told the truth.

1st Lemma - A and B cannot be both false.

(PROOF: Suppose A and B both false. Then D J and R are necessarily all true. But all the perfect squares less or equal than 999 dividing 648,000 ( $=2^6 5^3 3^4$ and not divided by 27 are: 1,4,16,64,25,100,400,9,36,144,576,225,900. The 3-digit possibilities are 100,400,144,576,225 and 900. None of them satisfy D that is then false - absurd!)

2nd Lemma - A and B cannot be both true

(PROOF: If A and B are both true, then we have D, J and R are all false, when at least one of them must be true. In fact : A true implies that 9 divides a+b+c that cannot be 15 (D false); B true implies that 11 divides 648,00 that is clearly impossible (J false) A, B and R true implies that 81*121 (i.e. 9801) would divide abc (a 3 digit number) that is impossible, so A and B true means R is false.)

3rd Lemma - B is false

(PROOF: If B is true, J is clearly false and for 2nd Lemma A is false. So R is true, but the only 3 digit perfect squares divided by 11 are 121 and 484, none of them satisfy D that must be true - absurd!)

4th Lemma - A is true

(PROOF: By 1st and 3rd Lemma.)

5th Corollary - D is false.

6th Corollary - A, R and J are true.

FIND abc.

648,000 = $2^6 5^3 3^4$

We can search for abg among the divisors of 648,000. The perfect squares divided by 27 are necessarily divided by 81, and the ones that are less or equal than 999 are 81 and 324. The only 3 digit number among them is 324.

I programmed it the hit and trial method but it was faster........in java......

    for(int i=100;i<=999;i++)
    {
        if(  648000%i==0 && isSquare(i) && i%27==0)
        {
            System.out.println("number="+i);
        }
    }