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

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ParthKohliBest ResponseYou've already chosen the best response.0
XYZ = 100x + 10y + z Is it something related to this?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Could be, my approach is somewhat different.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
100x + 10z + y + 100y + 10x + z ...... = 3024
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
122x + 212y + 221x = 3024 I can't go any further :?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
@cwtan: I used permutation.
 one year ago

karatechopperBest ResponseYou've already chosen the best response.0
what is permutation
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hmm...I'm not sure which permutation to use here.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Number of arrangement.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Number of arrangements if order matters.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
As far as I know, you used permutations only to arrange xyz in different ways
 one year ago

karatechopperBest ResponseYou've already chosen the best response.0
i think im gettting to it!!
 one year ago

karatechopperBest ResponseYou've already chosen the best response.0
wait..whats the permutation formula?
 one year ago

karatechopperBest ResponseYou've already chosen the best response.0
ffm...http://www.mathwords.com/p/permutation_formula.htm in this liinik what do the dots mean in the formula
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Permutations formula: \(\Large \color{Black}{\Rightarrow _nP_r = {n! \over (n  r)!} }\)
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
... this is not a straight forward permutation problem.
 one year ago

karatechopperBest ResponseYou've already chosen the best response.0
can u give me a hint..
 one year ago

cwtanBest ResponseYou've already chosen the best response.0
I like this question when i am unable to solve it......
 one year ago

joemath314159Best ResponseYou've already chosen the best response.3
is the answer:\[x\times y\times z = 126\]?
 one year ago

joemath314159Best ResponseYou've already chosen the best response.3
very weird, im pretty sure its nowhere near the best way to look at this.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.1
222 * 18  3024 = 972  Whose sum of digits = 18 too.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.3
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.
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.1
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.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Interesting approach can we generalize it?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Why do you choose divisibility by 9 in the first place?
 one year ago

joemath314159Best ResponseYou've already chosen the best response.3
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.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.3
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).
 one year ago

joemath314159Best ResponseYou've already chosen the best response.3
where k is the remainder of the sum after division by 9.
 one year ago

karatechopperBest ResponseYou've already chosen the best response.0
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..
 one year ago
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