Not too reduced

We have, $17n + 12 = xz$ and $n-25=yz$ . For simplification, we make the coefficient of $n$ same in both the equations. We get,

$17n+12=xz$

$17n-425=17yz$

Subtracting, we get

$437=z(x-17y)$

Factors of $437$ are $1, 437; 19$ and $23$ . We cannot take $z=1$ because z is greater than1 and also we cannot take $z = 437$ as it will make xz and yz large. So we take $z=19$ and $z=23$ . Finding minimum value of $n$ by taking least possible value of a and b in both the cases, we get $n= 71$ at $z=23$ and $n=63$ at $z=19$ . Therefore minimum value of n is $\boxed{63}$

Given,

$17n+12=x \times z...........(i)$

$n-25=y \times z................(ii)$

putting $n=25+y \times z$ in (i) we get

$x \times z=17(y \times z)+437\Rightarrow x=17y+\frac {437} {z} (z>1)$

since, $x \in Z$

hence, $z|437\Rightarrow z=19$ or $23$ or $437$

then, $x=17y+1.......(iii)$ ,when, $z=437$

or, $x=17y+19.....(iv)$ when, $z=23$

or, $x=17y+23......(v)$ when, $z=19$

for n to be minimum $x \times z$ must be minimum

from equation (iii),(iv),(v) we see that $x \times z$ is minimum for $z=19$

hence,z=19

now, $y$ is integer and $y>1 \Rightarrow y \geq 2$

hence, $y \times z \geq 38 \Rightarrow n-25 \geq 38 \Rightarrow n \geq 63$

putting, $n=63$ in the equation (i) we get

$17\times 63 +12=x \times 19 \Rightarrow x=57$ ,which is integer.

now,we can check further that (x,y)=1,

putting $x=57$ in (v) we get $y=2$

so,(x,y)=(57,2)=1

as, $n=63$ meets all the requirement so smallest value of $n=63$ (ans)

Because $gcd(x,y) = 1$ , let's use the Euclidean Algorithm to find the highest possible $gcd(17n + 12, n - 25)$ : $17n + 12 = 17(n - 25) + 437$ , so 437 is their greatest common divisor. This means that $z | 437 = 19 \times 23$ . As we want the smallest n., let's chose $z = 19$ . Thus we get: $17n + 12 = 19x$ and $n - 25 = 19y$ Considering $(mod 6)$ on the second equation we get: $n = 6 (mod 19)$ . Substitute $n = 19a + 6$ Thus we get: $19a + 6 - 25 = 19y$ ---> $a = y + 1$ and $17(19a + 6) + 12 = 19x$ ---> $17a + 6 = x$ ---> $17y + 23 = x$ y is strictly larger than 1, so taking $y = 2$ gives $x = 57$ , which are both co-prime. It also gives $a = 3$ and therefore $n = \boxed {63}$

Eliminemos a variável "n" da seguinte forma:

$\\ \begin{cases} 17n + 12 = xz \\ n - 25 = xy \; \times (- 17 \end{cases} \\\\ \begin{cases} 17n + 12 = xz \\ - 17n + 425 = - 17xy \end{cases} \\ ---------- \\ xz - 17xy = 437 \\ x(z - 17y) = 19 \cdot 23$

Para a igualdade acima, temos duas possibilidades, isto é, ou $\boxed{x = 19}$ , ou, $\boxed{x = 23}$ . Obviamente, o menor valor que se pode obter para "n" é fazendo x = 19. Segue,

$(z - 17y) = 23 \\ z = 17y + 23$

De acordo com o enunciado, "y" e "z" são maiores que 1, portanto, o menor valor para "y" é 2. Então, $\boxed{y = 2}$

Por fim,

$n - 25 = xy \\ n = 25 + 19 \cdot 2 \\ n = 25 + 38 \\ \boxed{\boxed{n = 63}}$

SINCE Z DIVIDES BOTH , 17N+12 AND N-25 , Z MUST DIVIDE 17N+12-17(N-25)=437=19X23.SO , TO REDUCE N WE TAKE Z=19 . ALSO , N-25=YZ . SO WE GET N=19Y+25 . SUBSTITUTING THIS VALUE OF N IN EQUATION 1 WE GET 17(19Y+25)=19X . SIMPLIFYING THE EQUATION WE GET 17Y+23=X . WE HAVE BEEN GIVEN THAT Y>1 .SO LET Y=2 . SO FOR Y=2 , X=57 WHICH IS A POSITIVE INTEGER COPRIME TO 2 . WE KNOW THAT N=YZ+25 , SO N=2*19+25=63

From the problem, we have $\gcd(17n+12,n-25) = z$ . From Euler's theorem, we can write this as $\gcd(17n+12-17(n-25)=437,n-25)$ which can be written as $\gcd(19*23,n-25$ . So $z$ can be either $1,19,23, 437$ or some multiples of these. But from the problem, $z$ cannot be $1$ . So that leaves $19,23,437$ or some multiples of these. The smallest valid $z$ is 19. If we write $n-25=19y$ . Since $y \ne 1$ , the smallest $y$ is 2. So we have $n=25+19*2=63$ .

17n+12/n-25 let n=m+25,hence it'll become 17m+437/m 437=19*23 numbers to be chosen for the above criteria of the form 19k or 23k,k>1 so the smallest m=38 and hence n=m+25=63

mdc(17n+12;n-25)=z -> z l 17n+12 - 17(n-25) ->z l 437 -> z= 1 or 19 or 23 or 437 -> z>1 -> z = 19 or 23 or 437

y>1 -> y>=2

If z=19 and y=2 ->n-25=38 -> n=63 (17x63+12)/19=57=x mdc(2;57)=1 ok!

x=57 ; y=2 ; z=19 ; n=63

gcd (17n+12, n-25) = gcd (17(n-25)+425 +12, n-25) = gcd (437, n-25) = z Since z >1 and 437 = 19 x 23, z= 19 or 23 or 437. For n to be as small as possible, let z=19. But n-25 cannot be equal to z as y would be 1 which violates the given condition. So let y=2 such that n-25 = 38. Thus n=63.

So we know z divides 17n+12 and n-25; hence n=25 mod z and plugging this in for n in the first expression gives 437=25 mod z. Hence z|19 23; since z!=1, z=19, 23, or 19 23. But clearly 19 will give the smallest value for n; minimizing n with y=2, we find n=63. Checking, 17n+12 gives x=57, and this is indeed coprime with 2. Done: n=63.