each of the students in a class writes a dirrerent 2 digit number on the whiteboard. the teacher claims that no matter what the students write, there will be at least three numbers on the whiteboard whose digits have the same sum. what is the smallest number of students in the class for the teacher to be correct?
I got the answer of 31 students. am i right? if i am not,could someone please show me how to figure this out?
Look at the sums:
ways to get sum of 1...10
ways to get sum of 2....20,11
How can three students get 3> 2,0;1,1, disallow 0,2
How can three students get 4> 4,0;22, 13, 31
get 5? 32, 23, 14, 41, 50
get 6 60, 51,15,24,42, 33
get 7 25, 16, 52, 61, 70, 43,34
get 8 17,71, 80, 62, 26, 35, 53,44
get 9 45, 54, 63,36, 27,72,`18,81,90
get 10 19,91,28,82,37,73,46,64, 55
get 11 29,92,38,83,74,47,56,65
notice the pattern...the ways to get a sum is always Sum until 9, then it is receeding in order.
get 12 ...7 ways (39,93,48,84,57,75, 66
get 13 six ways
get 14 five ways
get 15 four ways
get 16 three ways
get 17 two ways
get 18 one way.
Ok, then to have three in one sum, you will have two single sums, 15 of two sums, and one of three sums.
Number kids: 2 + 15*2 + 1*3
check my thinking.
In 18 ways get the sum from 1 to 18, I think if 3 students choose one any way of these sums, the teacher is correct. So the number of the students is 17x2+1x3=37
Your thinking is correct! In order to ensure that there will be at least three numbers on the whiteboard whose digits have the same sum, you need to have students write different 2-digit numbers.
Let's analyze the possible sums that can be obtained from the digits of these 2-digit numbers.
- For a sum of 1, there is only one possible combination: 0 + 1.
- For a sum of 2, there are two possible combinations: 0 + 2 and 1 + 1.
- For a sum of 3, there are two possible combinations: 0 + 3 and 1 + 2.
To guarantee that there will be at least three numbers with the same sum, we need to focus on the sums that have multiple combinations.
Starting from a sum of 4, there are increasing numbers of possible combinations.
- Sum of 4: 0 + 4, 1 + 3, 2 + 2 (3 combinations)
- Sum of 5: 0 + 5, 1 + 4, 2 + 3 (3 combinations)
- Sum of 6: 0 + 6, 1 + 5, 2 + 4, 3 + 3 (4 combinations)
- Sum of 7: 0 + 7, 1 + 6, 2 + 5, 3 + 4 (4 combinations)
- Sum of 8: 0 + 8, 1 + 7, 2 + 6, 3 + 5, 4 + 4 (5 combinations)
- Sum of 9: 0 + 9, 1 + 8, 2 + 7, 3 + 6, 4 + 5 (5 combinations)
- Sum of 10: 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5 (5 combinations)
- Sum of 11: 2 + 9, 3 + 8, 4 + 7, 5 + 6 (4 combinations)
- Sum of 12: 3 + 9, 4 + 8, 5 + 7, 6 + 6 (4 combinations)
- Sum of 13: 4 + 9, 5 + 8, 6 + 7 (3 combinations)
- Sum of 14: 5 + 9, 6 + 8, 7 + 7 (3 combinations)
- Sum of 15: 6 + 9, 7 + 8 (2 combinations)
- Sum of 16: 7 + 9, 8 + 8 (2 combinations)
- Sum of 17: 8 + 9 (1 combination)
To find the smallest number of students needed, we add up the number of combinations for each sum:
2 + (3 * 2) + (4 * 2) + (5 * 2) + (5 * 2) + (5 * 2) + (4 * 2) + (4 * 2) + (3 * 2) + (3 * 2) + (2 * 2) + (2 * 2) + 1 = 2 + 6 + 8 + 10 + 10 + 8 + 8 + 6 + 6 + 4 + 4 + 1 = 31
Therefore, the smallest number of students needed for the teacher to be correct is 31.
Your thinking is correct! To find the minimum number of students needed for the teacher to be correct, we can analyze the sums of the digits of the numbers written by the students.
Starting with the sum of 1, there is only one possible number (10). For the sum of 2, there are two possible numbers (11, 20). We continue this pattern and find that for each increasing sum, the number of possible numbers grows by 1 until we reach the sum of 9.
Starting from the sum of 9, the number of possible numbers starts decreasing by 1. So for the sum of 10, there is only one possible number, for the sum of 11, there are two possible numbers, and so on.
To have three numbers with the same sum of digits, we need to have at least one sum with three or more possible numbers. From the calculation above, we can see that the first sum that has three possible numbers is 12. After that, the number of possible numbers for each sum decreases by 1.
To find the minimum number of students required, we add up the numbers of possible numbers for each sum.
1 (sum) + 2 (sum) + 3 (sum) + ... + 9 (sum) + 3 (sum) + 2 (sum) + 1 (sum)
This can be simplified to:
1 + 2 + 3 + ... + 9 + 3 + 2 + 1
The sum of the numbers from 1 to 9 is given by the formula: n * (n + 1) / 2, where n is the number of terms. So, in this case, the sum of the numbers from 1 to 9 is 9 * (9 + 1) / 2 = 45.
Therefore, the minimum number of students needed for the teacher to be correct is:
45 + 3 + 2 + 1 = 51
So your initial answer of 31 students is not correct. The correct answer is 51 students.