Can anyone help me answer these math questions? I am having a lot of trouble with it?(It is very hard)?

Question

1. Each of the students in a class writes a different 2-digit number on the whiteboard. The teacher claims that no matter what the students write, there will be at least three numbers of the whiteboard whose digits have the same sum. What is the smallest number of students in the class for the teacher to be correct and why?

2. What is the sum of the digits of all 2-digits numbers from 10 to 99 and why? The answer can't be anything above 1000.

3. We will call a number N green if 6 x N contains none of the digits 0,1,2,3 and 4. There are two digits such that every green contains at least one of them. One of these two digit is 1. What is the other and why?

Answer 1

#1: The sum of the digits of a two-digit number must be at least 1 (since the first digit must be nonzero, or else we don't have a two-digit number), and cannot be more than 18 (given by 99). There are 18 integers in the range [1, 18] and thus 18 possible sums. Therefore, there must be at least 37 students, because you can't assign 37 numbers to 18 sums without assigning at least 3 numbers to one sum.

Edited to add: actually, that answer might be too high - does your teacher actually mean to say that every student _must_ write a different number? If that is so, then the minimum is only 35, since the assumption that you can assign 36 students to 18 sums without having at least 3 numbers with the same sum depends on being able to assign 2 numbers to every sum. For most of the sums this is true, however there is only one two-digit number with a sum of 1, and only one with a sum of 18 (specifically, 10 and 99, respectively). If the students are not allowed to write the same number twice, then only one number can be assigned each of those sums, and so you have to add a third number to another sum two students earlier than you would otherwise.

#2: Each of the first digits 1-9 gets used exactly 10 times, and thus the first digits contribute (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) * 10 or 450 to the sum. Each of the last digits 1-9 get used exactly 9 times (plus zero getting used nine times, but that doesn't add anything), contributing 45*9 or 405 to the sum. Thus the final answer is 450+405, or 855.

#3: There is no such digit. Consider that 113*6=678 and 128*6=768, so both 113 and 128 are green, but have no digits in common besides 1. 1 isn't a required digit either, because 96*6=576, so 96 is green, but does not contain 1.

Answer 2

Hii..!! :D perhaps u ought to take a destroy or dont study that plenty i propose dont rigidity urself to do some thing, u ought to sleep and relax reason ur techniques needs somes situations to sit down down back and ask for help and constantly if u dont understand some thing dont overlook to ask ur instructors or family contributors ... good success : )

Answer 3

2. 855
10, 11, 12...
20, 21, 22...
30, 31, 32...
and so on
ignore the 0's, there are 19 sets of 1,2,3,4...9.
1+2+3+4+5+6+7+8+9=45
45*19=855

i dunno the other two, maybe later.

Answer 4

What did you do to upset your teacher so much? I am not ashamed to say I really don't understand number 3 AT ALL!
I'll wait and see what others make of it.......