In a set of 5 numbers you can't find any 3 numbers that doesn't up to a multiple of 3.
Any integer can be divided in 3 groups which are 3n, 3n+1 or 3n+2. N can be any whole number.
Given that we have to choose 5 numbers, we have 2 options of choosing 3 numbers of the numbers from any of the 3 groups.
The 1st option is we have at least 1 number from each group. we can choose 3 numbers of the 5 numbers at least 1 from each different group. Like for example 3 from 3n, 4 from 3n+1 and 5 from 3n+2. We can use the letters n, m and l to show the different numbers. So when we add up the numbers it'll look something like this, 3n+3m+1+3l+2= 3(n+m+l)+3= 3(n+m+l+1). As you can see this number is definitely a multiple of 3 because we can look at the numbers in the bracket as 1 integer number and 3 has to multiply that number. Therefore 3(n+m+l+1) can be divided by 3.
The second option is we have at least 3 numbers from 1 group. we can choose 3 or more numbers of the 5 numbers from 2 of the groups. For example, we choose 2 numbers from 1 group and 3 numbers from another or we can choose 1 number from 1 group and 4 numbers from another group. Either way, if we choose the 3 numbers from 3m+1, it'll look something like this: 3m1+1+3m2+1+3m3+1 (the number next to the letter m shows that the numbers are all different). Now we simplify 3m1+1+3m2+1+3m3+1 to 3(m1+m2+m3+1). We know that 3(n+m+l+1) can be divided by 3 so 3(m1+m2+m3+1) can be divided by 3 as well.
If we choose the 3 numbers from 3l+2, it'll look like this: 3l1+2+3l2+2+3l3+2= 3(l1+l2+l3+2) which is also divisible by 3.
In conclusion, there's always 3 numbers from any 5 numbers that can adds up together to get a multiple of 3.