The ZYX Of Numbers

For the ones digit z + y + z + x + y = x + 2y + 2z = 6 + 10n (eqn 1) where n is a positive integer

For the tens digit y + z + x + z + x + n = 2x + y + 2z + n = 3 + 10m (eqn 2) where m is positive integer

For the hundreds digit x + x + y + y + z + m = 2x + 2y + z + m = 25 (eqn 3)

Now, from (eqn 3) we have m = 25 - 2x - 2y - z

Then substituting m to (eqn 2) 2x + y + 2z + n = 3 + 250 - 20x - 20y - 10z

n = 253 - 22x - 21y - 12z

Finally, substituting n to (eqn 1) x + 2y + 2z = 6 + 2530 - 220x - 210y - 120z

221x + 212y + 122z = 2536

You will arrive at x = 2, y = 7, z = 5

The sum of all 6 numbers is 222(x+y+z)

222*(x+y+z)=2536+zyx

Divide both sides by 222

(x+y+z)=11 94/222+ zyx/222

since x y z are all integer x+y+z must be an integer and the right hand side must also be an integer to satisfy this condition zyx must satisfy (222-94)+222n that is 128+222n where n is any integer greater than or equal to 0

Since zyx must be less than 1000 (since no digit may be greater than 9 for a max of 999, or 987 for unique digits) it is easy to list all the possible candidates

If left hand side = right hand side (LHS=RHS) then the solution is valid, If they do not then the solution is invalid

n=0 zyx=128 LHS=1+2+8=11 RHS=12 11 != 12 so invalid

n=1 zyx=350 LHS=3+5+0=8 RHS=13 8 != 13 so invalid

n=2 zyx=572 LHS=5+7+2=14 RHS=14 14 = 14 thus this is the solution

n=3 zyx=794 LHS=7+9+4=20 RHS=15 20 != 15 so invalid

n=4 zyx=1016 zyx greater than 999 so all n greater than 3 are invalid

100x+10y+z + 100x+10z+y + 100y+10x+z + 100y+10z+x + 100z+10x+y = 2536

221x+212y+122z = 2536

222x+222y+222z = 2536 + zyx

222(x+y+z) = 2536 + zyx

zyx = 222(x+y+z) - 2536

Trying multiplication of 222 which should be greater than 2536

222*14 = 3108

=>

zyx = 222(2+7+5) - 2536

zyx = 572

Let $XYZ$ be a three digit number represented as $100X+10Y+X$

now it can easily be shown that

$XYZ+XZY+YXZ+YZX+ZXY+ZYX=222(X+Y+Z)$

Given the fact

$XYZ+XZY+YXZ+YZX+ZXY = 2536$

$\Rightarrow XYZ+XZY+YXZ+YZX+ZXY+ZYX= 2536+ZYX = 222(X+Y+Z)$ $\Rightarrow ZYX = 222(X+Y+Z) - 2536$

Considering $ZYX > 0$ we have $X+Y+X \ge 12$ And Considering $ZYX$ is a 3 digit number we have $X+Y+X \le 15$

So we need to find a number $222(X+Y+Z) - 2536$ where $12 \le X+Y+X \le 15$

Using enumeration its easy to determine that $ZYX = 572$

Ans:572; By verification method. On adding all MSBs, 2x+2y+z+carry = 25.

222(x+y+z)=2536+zyx (by adding zyx both sides) 2536/222=11.4...so check 222 12-2536=128,1+2+8=11 not 12 222 13-2536=350;3+5+0=8 not 13 222*14-2536=572,5+7+2=14 hence ans is 572;

X+2y +2z=26 (eq.1) 2x + y +2z =21(eq.2) 2x +2y +z =23 (eq.3) Solving for all unknown values leads to z=5, y=7,& x=2 , so the number is 572

you can easily get to the situation where 222(x+y+z)=2536+zyx or, x+y+z=11+(94+zyx)/222 or, so 94+zyx=222n now you can easily get the value of ZYX=572
upvote if you think its the easiest method. thank you

Add all digits we get three equations : X+2Y+2Z>=6--(1),2X+Y+2Z>=3--(2),2X+2Y+Z<=25--(3). Add left part of all eqns we get 5(X+Y+Z) so right hand equality must be a multiple of 5.So the inequalities when converted to equalities we get eqn(1)=26.eqn(2)=21,eqn(3)=23 and solving for X,Y,Z we get ZYX=572

let x+y+z=k; k is a positive integer

3 possibilities arise then

<b>case 1</b> :

                          2k-x=26;

                          2k-y=21;

                            2k-z=23

=>5k=70(adding the 3 equations ) hence x+y+z=14 from case 1 ...now substitute and solve

<b>case 2 </b>:2k-x=16,,,,2k-y=12,,,,2k-z=24 ===>5k=52 ,so INVALID !

case 3:k-x=6,,,,,2k-y=3,,,,,2k-z=25 ===>5k=34, so INVALID !

other 2 cases are INVALID because k is integer:;