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FoolForMath
Just another cute problem: Suppose \(xyz\) is a three digit number such that \(xzy + yxz+yzx+zxy+zyx = 3024\), then can you find \(x \times y \times z\)?
XYZ = 100x + 10y + z Is it something related to this?
Could be, my approach is somewhat different.
100x + 10z + y + 100y + 10x + z ...... = 3024
122x + 212y + 221x = 3024 I can't go any further :?
@cwtan: I used permutation.
what is permutation
Hmm...I'm not sure which permutation to use here.
Number of arrangement.
Number of arrangements if order matters.
As far as I know, you used permutations only to arrange xyz in different ways
i think im gettting to it!!
wait..whats the permutation formula?
ffm... http://www.mathwords.com/p/permutation_formula.htm in this liinik what do the dots mean in the formula
Permutations formula: \(\Large \color{Black}{\Rightarrow _nP_r = {n! \over (n - r)!} }\)
... this is not a straight forward permutation problem.
can u give me a hint..
I like this question when i am unable to solve it......
is the answer:\[x\times y\times z = 126\]?
very weird, im pretty sure its nowhere near the best way to look at this.
222 * 18 - 3024 = 972 --- Whose sum of digits = 18 too.
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.
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.
Interesting approach can we generalize it?
Why do you choose divisibility by 9 in the first place?
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.
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).
where k is the remainder of the sum after division by 9.
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..