Just another cute problem:
Suppose \(xyz\) is a three digit number such that \(xzy + yxz+yzx+zxy+zyx = 3024\), then can you find \(x
\times y \times z\)?

- anonymous

- chestercat

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- ParthKohli

XYZ = 100x + 10y + z
Is it something related to this?

- anonymous

Could be, my approach is somewhat different.

- ParthKohli

100x + 10z + y + 100y + 10x + z ...... = 3024

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## More answers

- ParthKohli

let me think

- ParthKohli

122x + 212y + 221x = 3024
I can't go any further :?

- cwtan

need Permutation?

- ParthKohli

Nope

- anonymous

@cwtan: I used permutation.

- karatechopper

what is permutation

- ParthKohli

Hmm...I'm not sure which permutation to use here.

- anonymous

Number of arrangement.

- ParthKohli

Number of arrangements if order matters.

- ParthKohli

As far as I know, you used permutations only to arrange xyz in different ways

- karatechopper

i think im gettting to it!!

- karatechopper

wait..whats the permutation formula?

- karatechopper

ffm...http://www.mathwords.com/p/permutation_formula.htm
in this liinik what do the dots mean in the formula

- ParthKohli

Permutations formula:
\(\Large \color{Black}{\Rightarrow _nP_r = {n! \over (n - r)!} }\)

- anonymous

... this is not a straight forward permutation problem.

- karatechopper

can u give me a hint..

- cwtan

I like this question when i am unable to solve it......

- anonymous

is the answer:\[x\times y\times z = 126\]?

- anonymous

Yes, approach? :)

- anonymous

very weird, im pretty sure its nowhere near the best way to look at this.

- anonymous

222 * 18 - 3024 = 972 --- Whose sum of digits = 18 too.

- anonymous

Note that 3024 is divisible by 9. Also note that if the three digit number xyz leaves a remainder of r when divided by 9, that any permutation of the digits must leave the same remainder. So adding up those 5 permutations will leave 5r as a remainder. Hence we have this equation:\[5r \equiv 0 \mod 9\]the only solution to this is r = 0 , so the number xyz is divisible by 9, which means the sum of its digits is divisible by 9.
Now use what siddhantsharan posted above, using the fact that x+y+z can only be 0, 9, 18, or 27. Its guess and check from there.

- anonymous

Actually the above post by joemath shortens it down to only 18.
As x + y + z must be 15 at least For it to be > 3024.
And 222*27 - 3024 will obviously not leave a 3 digit no. Hence 18 has to be correct. Without any guess and check.

- anonymous

Interesting approach can we generalize it?

- anonymous

Why do you choose divisibility by 9 in the first place?

- anonymous

because of how divisibility by 9 is sorta tied to the digits of the number. You can find a numbers remainder when you divide by 9 by adding the digits together. it seemed like a lucky break that 3024 was divisible by 9.

- anonymous

hmm...actually i take that back, the problem will still be doable even if that sum wasnt divisible by 9. Since we are adding 5 numbers, we will always be able to solve the equation:\[5r\equiv k \mod 9\]Since 5 is invertible (mod 9).

- anonymous

where k is the remainder of the sum after division by 9.

- cwtan

So the answer is ?
126?

- karatechopper

i googled and got 126 but at first when i was solving problem i thought i had to find out what x y and z were..

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