A math question, need solution?

Question

The teacher asks each one of the student in a class to write a 2-digit number on the board. The teacher claims that at least three numbers whose digits have the same sum. How many students has there to be in the class in order for the teacher to be correct?

it's westpac question, and the answer is 35. but i need solution

Answer 1

The answer can't be less than thirty. All these numbers hve the sum of three. 12, 21, 30. To get the sum of 1 and 2 there would have to be more than thirty students. So the correct answer is thirty!!!!

Answer 2

The sum of the digits of a 2 digit number can be any number in the range 1 - 18. So the probability of writing a 2 digit number of the same sum is 1/18. If you want to have at least 2 numbers of the same sum, there must be 19 students (18+1). And so on, if you want to have at least 3 numbers of the same sum, the student number should be 18*2+1=37
(not 35 as you mentioned)

Answer 3

The answer to the question as posed is 37:

There are 18 different sums of two digit numbers, ranging from 1 (10) to 18 (99). With 36 students, you could have each of these twice, hence the 37th student would have to pick a number with a sum already showing twice.

If you want the answer to be 35, there should be an extra condition: Each child has to pick a *different* two digit number:

As the sums 1 and 18 only correspond to exactly one 2-digit number (10 and 99), it is only possible to write down 34 different numbers before having 3 with the same sum...

Answer 4

This questions assumes that each student must pick a different number.

The possible two digit numbers range from 10 tp 99. If you look at the possible sums of the digits of those numbers they range from 1 to 18.

10 - sum is 1
99 - sum is 18

It is easy to come up with sums for every number 1 thru 18.

Let the first 18 students pick numbers that each have a different sum.

Now let the second group of students pick numbers that haven't been picked before. There are only 16 different sums possible since there is only one number that sums to 1 and one number that sums to 18.

Now we have 18 + 16 = 34 students, but no three numbers have digits with the same sum.

Now no matter what number the 35th student picks, the sum of its digits will match one of the one of the sums that already has two numbers.

So it requires 35 students to be certain that at least three numbers will have digits that total to the same sum.

Answer 5

the sum of the digits of any 2-digit number is in one of the integers from 1 to 18.

the sum number1 number2
1 10 only one
2 11 20 only two
16 79 97 only two
17 89 98 only two
18 99 only one

so there can be 34 different numbers which have no three numbers whose digits have the same sum , the reason is in the other sums(i.e. from 3 to 15) we can just take two of numbers
for instace, there are three or more numbers(such as12,21,and 30 )whose digits have the same sum ,then we can just take two of them.So
(18-5)*2+8=34
if there is only 34 students, the teacher can't absolutely be correct. If the teacher is always correct, there must be 35 students in the class.
I can not express the process in detail, and i hope you think on your own.

Answer 6

i think the answer is 37 because the lowest possible number is 10 (1+0 = 1) and the highest possible number is 99 (9 + 9 = 18) so the sum of any two digit number must be from 1 to 18 and she said that three must be the same so ther must be two of every possible sum and one more to make sure that one sum is repeated three times
18 x 2 + 1 = 37

Answer 7

start with the tens

12 21
13 31
14 41
15 51
16 61
17 71
18 81
19 91

this is so far 8

now do the 20s

we have already had 21 so we must start from 23

23
24
25
26
27
28
29

this is seven

now do the 30s

31 32 and 33 we cannot have

so

34
35
36
37
38
39

this is 6 and so on


so 8 + 7 + 6 + 5 + 4 + 3 + 2 +1 = 36

Answer 8

I think this is what you are looking for:

00
01
02
↓
98
99

With a two digit number you have 100 possibilities.
Picking any one number, you will have a 100th of a chance of selection that number.
For that number to be picked three times :

1/100 * 1/100 * 1/100 = 1/1,000,000

That is, a one million-ant of a chance of the three number being the same.
This implies, 1,000,000 students are needed to pick three numbers that are the same.

Answer: 1,000,000.

Answer 9

If the students all have to write a different number, then the numbers that total 3, like Sabrana has it......

Answer 10

a answer might mean the two lines intersect. simply by fact that, they're boths set equivalent to x, you may set the two equations equivalent and resolve for y 2y+6=3y+10, so 6=y+10, or y=-4. Plug this into the 1st equation to get: 2(-4)+6=-2. hence, they intersect at (-2,-4)