Just another cute problem:

Suppose $xyz$ is a three digit number such that $xzy + yxz+yzx+zxy+zyx = 3024$, then can you find $x
\times y \times z$?

Question

Just another cute problem:

Suppose $xyz$ is a three digit number such that $xzy + yxz+yzx+zxy+zyx = 3024$, then can you find $x 
\times y \times z$?

parthkohli · Answer

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

anonymous · Answer

Could be, my approach is somewhat different.

parthkohli · Answer

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

parthkohli · Answer

let me think

parthkohli · Answer

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

cwtan · Answer

need Permutation?

parthkohli · Answer

Nope

anonymous · Answer

@cwtan: I used permutation.

karatechopper · Answer

what is permutation

parthkohli · Answer

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

anonymous · Answer

Number of arrangement.

parthkohli · Answer

Number of arrangements if order matters.

parthkohli · Answer

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

karatechopper · Answer

i think im gettting to it!!

karatechopper · Answer

wait..whats the permutation formula?

karatechopper · Answer

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

parthkohli · Answer

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

anonymous · Answer

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

karatechopper · Answer

can u give me a hint..

cwtan · Answer

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

anonymous · Answer

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

anonymous · Answer

Yes, approach? :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

Interesting approach can we generalize it?

anonymous · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

cwtan · Answer

So the answer is ?
126?

karatechopper · Answer

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..