Diophantine equations with three variables

Since $x,y,z$ are all positive integers we have that $1 \le x \le 12, 1 \le y \le 12$ and $1 \le z \le 11$ in order for the LHS of the given equation to not exceed $366$ . Now the given equation can be rewritten as

$30(x + y + z) + (z - x) = 366$ .

Given the above ranges for $x$ and $z$ we know that $-11 \le (z - x) \le 10$ . This in turn implies that

$356 \le 30(x + y + z) \le 377$ .

The only multiple of $30$ that satisfies this inequality is $30*12 = 360$ , and thus $x + y + z = 12$ and $z - x = 6$ . This last condition is satisfied for $(x,z) = (1,7), (2,8), (3,9), (4,10)$ and $(5,11)$ , but since $x + z$ cannot exceed $11$ (in order for $y$ to remain positive) the only possibilities for $(x,y,z)$ are $(1,7,4)$ and $(2,8,2)$ . As both of these solution triples satisfy the original equation we can conclude that $n = 2$ . Also, the only one of these two solutions for which $x,y,z$ are distinct is $(1,4,7)$ , and thus $X = 1, Y = 4$ and $Z = 7$ .

Thus the desired answer is $n + X + Y + Z = 2 + 1 + 4 + 7 = \boxed{14}$ .

Note that since both solution triples are such that their components add to $12$ we didn't really have to worry about determining $X,Y,Z$ specifically, but this one solution is interesting because it represents the days in the months of a leap year, in that there is $1$ $29$ -day month, $4$ $30$ -day months and $7$ $31$ -day months.

Uh oh... lets see...

(Warning : very long due to combination generalization, proofing and the discusser genuine idiocy for scraching head, having no idea what the question or even the title is about (like... I found 3 legit solutions, seriously.. huehuehue??))

29x + 30y + 31z = 366

29 * (x+y+z) + (y +2z) = 366

29 * (x+y+z) = 366 - (y +2z)

Since 29 is a big number, we can assume that (x+y+z) is not big

In terms of narrowing the range, we can use 29 times 10, 15, 20 as a standard. Regardless, the multiplication of 29 is less than 400 (rounded up)

29 * 10 = 290 ; <366

29 * 15 = 435 ; >366

29 * 20 = 580 ; >366

it seems like (x+y+z) is below 15, but upper 10 since (29*10) still leaves alot of room (366-290 = 76)

so : 29 * (a) ;366 - (29 * a) = y + 2z

29 * 11= 319 ;366 - 319 = 47

29 * 12= 348 ;366 - 348 = 18

29 * 13= 377 ;366 - 377 = -11

To this point, (x+y+z) is either 11 or 12, notice that even if i were to use (0) in the sum of 11, it wont be enough to cover 47( 11 = 0+1+10 ; y +2z = 47; y= 1 or 0; z=10 ; 2*10 + (1 or 0) = 21 or 20 )

(x+y+z) = 12 (g)

y + 2z = 18 (j)

(j) - (g) = z - x = 6

z = 6 + x

In terms of sum combination, because z is bigger than half (12/2 = 6 ; z>6), then : z > 6 ; x< 6 , y < 6

Then to generate "manually" (sad life...)

12 = r = z + x + y ; z>6

12 = r(0) = z(0) + x(0) + y(0)
( lowest “acceptable” z combo )

12 = r(0) = r(f)

12 = r(f) = z(0) + x(0) + y(0) + 2f - 2f

12 = r(f) = (x(0)+f) + (x(0)+f) + (y(0) - 2f)

Since z "must be" bigger than 6, therefore :

(6 + 1) ≤ z ≤ 12 ; (7) ≤ z ≤ 12

Lowest z (z(0)) is 7, therefore:

z - 6 = x ; 7 - 6 = 1 ;( x(0))

12 - ( x+z ) = y

12 - ( 7+1 ) = 12 - 8 = 4 ;( y(0))

12 = z + x + y ; 12 = r ; ( x=z-6 )

12 = r(0) = z(0) + x(0) + y(0)

12 = r(0) = 7 + 1 + 4 (valid-ness confirm)

12 = r(1) = 8 + 2 + 2 (8-2 =6, valid) ; twin

12 = r(2) = 9 + 3 + 0 (9-3 = 6, valid) ; em "0" ?

z > 9 is invalid ....

There are 2 ways to determine the "valid-ness"

Way 1 = all numbers must be (+) Following the pattern of r(f), r(3) will give us a negative number.

To be exact, following the pattern

y(1) - y(0) = 2 - 4 = -2

y(2) - y(1) = 0 - 2 = -2

y(3) - y(2) = y(3) - 0 = -2 ; y(3) = -2

With y(3) being -2 and the constant change of (-2), every y(f) in which f > 2 is going to be (-), thus is not counted in (n(+)). "HOWEVER" that doesn’t mean r(f) (f > 2) are not solutions

Ehm, if the condition of the solution accepts "negative numbers" like r(3) of (10, 4, -2), then the "solution (n)" could be "infinitely many"...

This is caused by the formula is able to meet all condition except being all (+), such as :

× > 6 ( as the value keeps going up (x+(f))

2x + y = 18 ( notice that even if (y) is (-), as (x) keep increasing, the value of any 2x(f) + y (f) will always be 18)

Lets try a random input into the formula (actual math included.. huehuehue .-. .... ) :

r(f) = 12 = z(f) + x(f) + y(f) ; ( f) > 2

366 = (12 *29) + (2 *z(f)) + y(f)

r(40) = 12 = z(40) + x(40) + y(40)

r(40) = 12 = 47 + 41 - 76

366 = (r(40) * (29)) + ( 2 * z(40) ) + y (40)

366 = (12 * 29 ) + ( 94 + ( -76) )

366 = 348 + 18

result obtained

Another one :

r(239) = 12 = z(239) + x(239) + y(239)

r(239) = 12 = 246 + 240 - 474

366 = (r(239) * (29)) + ( 2 * z(239)) + y(239)

366 = (12 * 29 ) + ( 492+ ( - 474) )

366 = 348 + 18

result obtained

Okay.... so every r(f) from f=3 (x>9) to possibly f = infinite(+) all have the result of 366.. .... what about r(-f) ? ( f< 0 ; x < 7)?

hmm.. z will be below 7... lets try:

12 = r(f) = (z(0)+f) + (x(0)+f) + (y(0) - 2f)

12 = r(-9) = ( 7 -9) + (1 -9 ) + ( 4 - (-18))

12 = r(-9) = -2 - 8 + 22

366 = (r( -1) * (29)) + ( 2 * z( -1)) + y ( -1)

366 = (12 * 29 ) + ( -4 + 22 )

366 = 348 + 18

..... huh... result obtained.... freaky....

Hmm so.., this mean that the equation in question can be answered even when it broke 3 rules :

1. all (+)

2. z > 9

3. z < 7

and that (n) it is infinitely many if included ((-)n) sets

Second way : z-x = 6... yeah..re-use, re-hash...

A.K.A. DA EZZZ WAIII!!!

So, in this case we are just going to use simple combinations of 12 without anything fancy.

rules :

12 = x + y + z

(+) = x, z, y

z-x = 6

also, z ranges :

( 9 < z) ; (9+1) ≤ x ≤ 12 ;10 ≤ x ≤ 12

z = 10, 11 , 12

(to optimize the deduction, the second highest number will be assume as "x")

12 = 10 + 2 +0 ; z-x =6 ;10-2 ≠ 6

12 = 11 + 1 +0 ; z-x =6 ;11-1 ≠ 6

12 = 12 + 0 +0 ; z-x =6 ;12-0 ≠ 6

(Is 0 even a number ? Ask God)

It broke the rule, and if inputted will gain less than 366 (proofing skipped), so x > 9 is invalid

(Forgive my idiocy but I’m not sure what OP means by integers or even if 0 is valid with given condition or even WTF does the question title means (TT--TT) ,I don’t learn math in English (hiks.. ;;_;;)

after all the hassle, it’s imputing time

29x + 30y + 31z = 366 ; or

29 * (x+y+z) + (2z+y) = 366

( r(0) ; y=4 ; z=7; x=1)

29 * (12)+(4 +14) = 366 (valid)

( r(1) ; y=2; z=8; x=2)

29* (12)+(2+16)=366 (invalid/twin numbers)

(r(2) ; y=0; z=9; x=3)

29*(12) + (0+18) = 366 ( valid (?))

So.....

As you can see....

In total there are 3 (all +) solutions, however we will be excluding 1 of them because of the twin numbers ....

n (all +) = 3 - 1 = 2

n (all +) = ( x, y, z )

n (+) = ( 3, 0, 9) and (1, 4, 7)

Excluded = (2,2,8)

(x+y+z) + n (+) = 12 + 2 =14

the answer is 14 for (+) numbers solutions(or 15 with a twin (or more depending on the condition ( n (-) ) ))

P.S. OP is Original Poster

(P.S.)^2 Yes I'm a 9gagger, by using banana for scale.... I'm a disgrace for being asian and not a doctor... huehuehue...

(P.S.)^3 I found 3 variables by adding 5 new variables.... logic got reversed!¿