This is a direct application of the Euclidean Algorithm. $\begin{aligned} \gcd(2n-33,5n+73) &= \gcd(5n+73,2n-33) \\ &= \gcd(n+139,2n-33) \\ &= \gcd(n+139,-n+172) \\ &= \gcd(n+139,311) \end{aligned}$

Since $311$ is prime, have have to have $311|n+139$ for $\gcd(n+139,311)\ne 1$ .The smallest multiple of $311$ that makes $n>0$ is, obviously, $311$ .

Therefore, our answer is $311-139=\boxed{172}$ .

Since $gcd(2n-33, 5n+73)$ $\neq 1$ , they must be both multiples of some integers. Say one of their divisors is $a$ . Since they are both multiples of $a$ , their difference should also be a multiple of $a$ .

We take $5n + 73 - ( 2n - 33 )$ = $3n + 106$ .

Then, we further reduce the result by taking away $2n - 33$ from it. $3n + 106 - (2n - 33)$ = $n + 139$ .

After that, we multiply the result by 2 and take $2n - 33$ from it. $2(n + 139) - (2n - 33) = 311$ .

Note that 311 is still a multiple of a. Since 311 is a prime number, 311 has no other divisors other than 1 and itself. We cannot take 1 as the value of a because it is stated that $gcd(2n-33, 5n+73)$ $\neq 1$ . Since the smallest $n$ is required, the value of $a$ should be as small as possible. In this case, we take $a = 311$ and let $2n - 33 = a$ in order to get the smallest $n$ .

$2n - 33 = 311$

$n = 172$

We substitute $n = 172$ into $5n + 73$ : $5(172) + 73 = 933$ . Since 933 is a multiple of 311, $n = 172$ is a valid answer.

In conclusion, $n = \boxed{172}$ .

For $\gcd(2n-33,5n+73)\neq1, 2n-33$ and $5n+73$ must be multiples of each other. SInce $5n+73>2n-33$ , let $5n+73=k(2n-33)$ where $k$ is a positive integer. Isolating $n$ we have $n=\frac {73+33k}{2k-5}$ . For $n$ to be the smallest positive integer, $2k-5=1$ and $k=3$ . Therefore, $n=172$ .

Let a prime number $p$ be the greatest common divisor of the numbers $2n-33$ and $5n+73$

$p|2n-33$ and $p|5n+73 \Rightarrow p|3n+106 \Rightarrow p|n+139 \Rightarrow p|2n+278$

So $p|2n-33$ $p|2n+278$

This implies that $278+33 =pk, \: k\geq1$ As it turns out, $331$ is prime so our $GCD$ is $331$

Now we have $311| 2n-33$ $311 | 5n+73$ $\Rightarrow 311 | 7n+40$

This implies $7n = 271 \left(mod \: 311\right)$

Therefore we can say $7n = 271 + 311k, \: k \geq 0$

Obviously if we want to minimise the value of $n$ we must also minimise the value of $k$

$271 = 5 \left(mod \: 7\right) \Rightarrow 311k = 2 \left(mod \: 7 \right)$

$311 = 3 \left(mod \: 7 \right)$ so $k$ is of the form $3 +7l, \: l \geq 0$

If we take the minimum value of $k=3$ we get $n=172$

If we plug $n = 172$ back into the original two expressions we get $2n-33 = 311, \: 5n+73 = 933$ both of which are clearly divisible by $311$ so $n = 172$ is the answer.

Since n $\in\,$ $\mathbb{Z^+}$ so $2n-33\,<\,5n+73$ .

So it is wise to say $5n+73\;=k\cdot(2n-33)$ for some k $\in\,$ $\mathbb{R}$ .

$\rightarrow\;k\;=\frac{5n+73}{2n-33}$

$\rightarrow\;k\;=\frac{2\cdot(2n-33)+\frac{1}{2}(\cdot(2n-33)+311)}{2n-33}$

$\rightarrow\;k\;=2+\frac{1}{2}+\frac{311}{2n-33}$

Since we want gcd $(2n-33,5n+73)\neq1$ , so we require $n$ to be least & in turn $k$ as least.

$\rightarrow\;311\;=2n-33$

$\rightarrow\;n\;=\boxed{172}$ .

(5n + 73)/(2n - 33) ≠ 1

so for example we assume that their quotient is equal to 2

(5n + 73) - (2)(2n - 33) = r

n + 139 = r

making r = 0 is not correct since n will be equal to -139

Then we try the quotient 3

(5n + 73) - (3)(2n - 33) = r

-n + 172 = r

Making r = 0

-n + 172 = 0

n = 172

Let the $gcd(2n - 33, 5n + 73$ be k, for a positive integer $k$ with $k$ is not equal to $1$ . Then, we have the following equations: $2n - 33 = kx$ and $5n + 73 = ky$ for a positive integers $x, y$ .

Multiply the first equation with 5 and the second one by 2, subtract the second equation with the first one give us: $k(2y - 5x ) = 311$ Since $311$ is a prime number, it might be suggested that the only positive integer $k$ is $k = 311$ . Noting n as an element of positive integer, the inequality $2n - 33 < 5n + 73$ holds and in order to obtain the smallest $n$ we should have the smallest $2n - 33$ , in another words, it forces $2n - 33 = 311 -> n=172$ .

Hence, the smallest positive integer $n$ is $\boxed{172}$

From Euclid's algorithm we get the maximum gcd is 311, which is prime. Therefore both numbers must be multiples of 311, and the minimum value of n is 172

(#)include <stdio.h> (#)include <string.h>

main() { int a,b,i,j; for(i=0; i<999;i++) { a=2 i-33; b=5 i+73; for(j=2; j<=a; j++) if(a%j==0 && b%j==0) { printf("\n %d %d %d", a,b, i); }

} }

This will give i= 172 i= 483 i= 794

172 being the smallest.

Using Euclidean algorithm we get $\begin{aligned} \text{gcd}(2n - 33,5n + 73) &= \text{gcd}(2n - 33,n + 139) \\ &= \text{gcd}(-311,n + 139) \\ &= \text{gcd}(311,n + 139) \end{aligned}$ Note that $311$ is prime, hence we need $n + 139 = 311$ , i.e. $n = \boxed{172}$

for taking n from 1 to 1000 and checking gcd(m) from 2 to 1000 ... we get n=172 and m=311(minimum value for n) here is its c programme

include<stdio.h>

int main() { int n,m; for(m=2;m<1000;m++) { for(n=1;n<1000;n++) { if((2 n-33)%m==0 && (5 n+73)%m==0){

         printf("%d %d\n", n, m);}
    }
}

}

put n=k+17

gcd(2k+1,5k+158)

let 5k+158/2k+1=p/q

=>k=[311(2p+5q/2p-5q)-321]/20;

for k to be positive integer (2p+5q/2p-5q)=11,31,51,.......

gives smallest value of k=155

and hence n=172

let gcd(2n-33,5n+73) is not equal to 1.let it is "d"...now we got d|2n-33 and d|5n+73...now we can interpret that d|2(2n-33),hence can say d|(5n+73)-2(2n-33).... i.e d|n+139...again we can find that d|n-172...hence we can say d|311..

now as 311 is a prime number so d=311.. now we have to find the least n for which gcd of that 2 no. is 311...hence we got the least n is (172)