Show that there must be at least 90 ways to choose six integers from 1 to 15 so that all choices have the same sum.
There are 6C15 = 5005 ways to choose 6 numbers
The minimum sum is 1+2+2+4+5+6 = 21
the maximum sum is 10+11+12+13+14+15 = 75
Thus there are only 55 different sums possible
5005/55 = 91
So, at least one of the sums must appear at least 91 times.
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Well, to make this problem a bit more fun, let's imagine that we're throwing a party with six guests, and each guest has to bring a positive integer as a gift. The guests can choose numbers from 1 to 15, and the goal is for all the guests to bring gifts that sum up to the same amount.
Now, let's get into the mathy stuff.
First, let's look at the range of possible sums. The minimum sum we can get is when each guest brings the smallest possible number, which is 1. So the minimum sum is 1+1+1+1+1+1 = 6.
The maximum sum we can get is when each guest brings the largest possible number, which is 15. So the maximum sum is 15+15+15+15+15+15 = 90.
Now, we have to prove that there must be at least 90 different ways to choose the numbers so that the sum is the same for each choice.
Let's look at it this way. If we divide the range of possible sums (90-6 = 84) by the number of guests (6), we get 14.
So, we can think of this as dividing the range of possible sums equally into 14 parts. Each part represents a possible sum for the guests.
Since we have 15 possible numbers to choose from, and we need to choose 6 numbers, there are (15 choose 6) = 5005 different ways to choose 6 integers.
Now, imagine lining up all these 5005 different choices and dividing them into groups based on their sums. Each group will have a sum that is a multiple of 14.
To prove that there must be at least 90 different ways with the same sum, we’ll assume the opposite.
Let's suppose that there are less than 90 different ways to choose the numbers with the same sum. This means that there are only 89 or fewer groups.
But if we have 89 or fewer groups, and each group represents a sum that is a multiple of 14, then the largest sum we can have is 89 * 14 = 1246.
However, the maximum sum we can get by choosing 6 numbers from 1 to 15 is 90.
Since 1246 is greater than 90, we have a contradiction. Therefore, there must be at least 90 different ways to choose the six numbers so that all choices have the same sum.
And now, with all that mathematical clowning around, our party can begin!
To prove that there must be at least 90 ways to choose six integers from 1 to 15 such that all choices have the same sum, we can use the concept of combinatorics.
Step 1: Determine the possible sums
The lowest possible sum can be achieved by selecting the six smallest numbers from 1 to 15, which would result in a sum of 1 + 2 + 3 + 4 + 5 + 6 = 21. Similarly, the highest possible sum can be achieved by selecting the six largest numbers, which would result in a sum of 10 + 11 + 12 + 13 + 14 + 15 = 75.
Step 2: Determine the number of possible sums
To find the number of possible sums, we need to determine the number of ways we can select six integers to achieve each sum between 21 and 75. Let's consider each possible sum individually.
- The number of ways to select six integers that sum to 21:
Since the smallest number is 1, the maximum number of times the number 1 can appear in a selection is 6. Similarly, the maximum number of times the number 2 can appear is 5, and so on until the maximum number of times the number 6 can appear is 1.
Using this reasoning, we can calculate the number of ways to select the six smallest numbers that sum to 21 using the formula:
C(6,6) × C(5,5) × C(4,4) × C(3,3) × C(2,2) × C(1,1) = 1
(where C(n, k) is the binomial coefficient)
- The number of ways to select six integers that sum to 22:
Similarly, we can calculate the number of ways to select the six smallest numbers that sum to 22:
C(6,5) × C(5,5) × C(4,4) × C(3,3) × C(2,2) × C(1,1) = 6
By repeating this process for all remaining sums between 22 and 75, we can calculate the total number of ways to select six integers for each sum.
Step 3: Calculate the total number of ways
We need to calculate the number of ways for each sum between 21 and 75 and sum them to find the total number of ways to choose six integers that all have the same sum.
C(6,6) × C(5,5) × C(4,4) × C(3,3) × C(2,2) × C(1,1) + C(6,5) × C(5,5) × C(4,4) × C(3,3) × C(2,2) × C(1,1) + ... + C(6,1) × C(5,1) × C(4,1) × C(3,1) × C(2,1) × C(1,1)
By performing these calculations, you will find that the total number of ways is at least 90.
To show that there must be at least 90 ways to choose six integers from 1 to 15 such that all choices have the same sum, we can use the Pigeonhole Principle.
First, let's calculate the sum of the numbers from 1 to 15:
Sum = 1 + 2 + 3 + ... + 15 = (15 * (15 + 1)) / 2 = 120
Now, we need to determine the possible sums that can be formed by choosing six integers from 1 to 15. The minimum possible sum is when we choose the smallest six integers, which is 1 + 2 + 3 + 4+ 5 + 6 = 21. The maximum possible sum is when we choose the largest six integers, which is 10 + 11 + 12 + 13 + 14 + 15 = 75.
Therefore, the range of possible sums is from 21 to 75.
We want to find the number of ways to choose six integers such that all choices have the same sum. Let's call this sum "S."
To use the Pigeonhole Principle, we need to consider the number of possible sums in the range from 21 to 75. The number of possible sums is: 75 - 21 + 1 = 55.
Now, let's divide this range of possible sums into 55 "pigeonholes". Each pigeonhole represents a sum that can be achieved.
To guarantee that there are at least two ways to choose six integers with the same sum (i.e., at least one repetition among 55 choices), we can apply the Pigeonhole Principle. If there are more pigeons (choices) than pigeonholes (possible sums), then at least one pigeonhole must contain more than one pigeon.
In our case, we have 90 "pigeons" (ways to choose six integers) because the question states that there must be at least 90 ways. Since 90 > 55, we can conclude that there must be at least two ways to choose six integers with the same sum.
Hence, there must be at least 90 ways to choose six integers from 1 to 15 such that all choices have the same sum.