What is the smallest positive integer in the set { $24x+60y+2000z : x,y,z \in \Bbb Z$ }

Question

Problem : What is the smallest positive integer in the set { $24x+60y+2000z : x,y,z \in \Bbb Z$ }

Solution:By hit and trial method answer is $4$

Is there any method to find it ?

Answer 1

1. Since the greatest common divisor (GCD) of 24, 60, 2000 is 4, $24x + 60y + 2000z = 4(6x + 15y + 500z)$ is always multiple of 4. Therefore, the smallest positive integer should be equal to or greater than 4.
2. We can find $x, y, z$ satisfying $6x + 15y + 500z = 1$ like this:
   1. $15 - 6 \times 2 = 3$ shows that $6\times(-2) + 15\times(1) = 3$.
   2. $500 \times 2 - 3 \times 333 = 1$ shows that $6 \times -2 \times -333 + 15 \times 1 \times -333 + 500 \times 2 = 1$.
   3. Therefore, $6 \times 666 + 15 \times -333 + 500 \times 2 = 1$.
3. Therefore, when $x,y,z = 666,-333,2$, $24x + 60y + 2000z = 4$. Therefore, the smallest positive integer $24x + 60y + 2000z$ where $x, y, z \in \mathbb{Z}$ is 4.

Step 2 can be done by calculating GCD by using Euclidean algorithm.

Answer 2

Note that the set $S$ of numbers $s=ax+by+cz$ for fixed integers $a,b,c$ and variable integers $x,y,z$ is an additive set in the sense that if $s_1,s_2\in S$ then $$s_1\pm s_2=a(x_1\pm x_2)+b(y_1\pm y_2)+c(z_1\pm z_2)$$ This implies that $S$ is the set of multiples of the minimum positive value $d$ it contains - keep adding/subtracting $d$ until you get a non-negative integer less than $d$, which must be zero.

Now note that $GCD(a,b,c)$ divides every element of $S$, and hence $d$. Also $a,b,c$ are represented by $x,y,z =(1,0,0), (0,1,0), (0,0,1)$ respectively, so $d|GCD(a,b,c)$.