Numbers are drawn from the 27 integers 1 to 27. At least how many number(s) is / are drawn at random to ensure that there are two numbers whose difference is divisible by 4?

Please help me to solve this oneπŸ˜©πŸ™πŸ™

I experimented with a list of integers 1-27. For example, if we choose 1,2,3,4......then we'd have to jump to 9,10,11,12 to ensure no differences of 4.....then to 17,18,19,20.....and lastly to 25,26,27. That is 15 numbers altogether. Any other number drawn after that would produce a difference of 4 with one of these, so we would have to draw at least 16 numbers.

This was the max number I could find after trying other combinations as well.

Because two numbers whose difference is divisible by 4, we can sort the numbers from 1 - 27 like this

1 5 9 13 ... 25
2 6 10 14 ... 26
3 7 11 15 ... 27
4 8 12 16 ...
Now if we draw 2 random numbers from a random line, the difference of the 2 numbers we have drawn will be divisible by 4.
Because of that, we can draw 4 random numbers from 4 different lines
The 5th will be the last number to ensure there are two numbers whose difference is divisible by 4.
So the final answer is 5 numbers

Well, well, well! Looks like we have a mathematical question to tickle my funny bone. Alright, let's give this a whirl!

To ensure that we have at least two numbers whose difference is divisible by 4, let's consider the worst-case scenario.

We could take the numbers: 1, 5, 9, 13, 17, 21, and 25. In this case, no two numbers have a difference divisible by 4.

But as soon as we add one more number, let's say 29 (just to show off), we can form two pairs: (1, 5) and (9, 13), where both pairs have a difference divisible by 4.

So, in this case, we needed to draw at least 8 numbers to make sure we have two numbers whose difference is divisible by 4.

I hope that puts a smile on your face and clarifies things for you!

To determine the minimum number of randomly drawn integers needed to ensure that there are two numbers whose difference is divisible by 4, we need to consider the worst-case scenario.

Since we want to ensure that there are two numbers with a difference divisible by 4, we can look at the remainders when the numbers are divided by 4.

In the range of 1 to 27, we have the following possible remainders:
- Remainder 0: 4, 8, 12, 16, 20, 24
- Remainder 1: 1, 5, 9, 13, 17, 21, 25
- Remainder 2: 2, 6, 10, 14, 18, 22, 26
- Remainder 3: 3, 7, 11, 15, 19, 23, 27

The worst-case scenario would be drawing numbers one by one without any repetition until we have drawn one number from each of the four possible remainders.

In this scenario, we would need to draw a minimum of:
- 1 number with remainder 0
- 1 number with remainder 1
- 1 number with remainder 2
- 1 number with remainder 3

Thus, we need a minimum of 4 numbers to ensure that there are two numbers whose difference is divisible by 4.

To determine the minimum number of integers that need to be drawn to ensure there are two numbers whose difference is divisible by 4, we can analyze the worst-case scenario.

First, let's consider the remainders of the numbers between 1 and 27 when divided by 4:

Numbers: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Remainders: 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3

As you can see, there are four different remainder values: 0, 1, 2, and 3. If we draw five integers, based on the pigeonhole principle, at least two of them must have the same remainder when divided by 4.

For example, if we pick the numbers 5, 9, 14, 18, and 23:

Numbers: 5 9 14 18 23
Remainders: 1 1 2 2 3

Among those numbers, 9 and 23 have the same remainder (1) when divided by 4. Thus, their difference (23 - 9 = 14) is divisible by 4.

Therefore, at least five numbers need to be drawn to ensure that there are two numbers whose difference is divisible by 4.