
Class game
The natural numbers from 1 to 10 are each written on the blackboard 10 times. The student in the class than play the following game: a student deletes 2 of the numbers and instead of them writes down on the blackboard their sum decreased by 1;after that another student do the same; and so on. The game continues until only one number remains on the blackboard. The remaining number is:
a) Less than 440
b)451
c)460
d)488
e)more than 500

It should not matter what order you add them in so I will go strait down on a row of 10x1
(1
2)=1+1=2
(3
4)=3+3=6
(5
6)=5+5=10
(7
8)=7+7=14
(9
10)=9+9=18
You are really only adding every odd number to itself which goes up by four past 2.
So the sum should equal added twice all the odd numbers from 1 to the number of numbers divided by two.
So the equation for a 10x1 should be:
$\displaystyle \sum_{n=1}^{5}(2n1)+(2n1)=50$
So you should now have to repeat this 10 time right?
SO 10(50)=500

OK OK Integral; why does this one equal 10 ?
1 2 3
1 2 3
1 2 3
The answer to the original puzzle is 451, not 500:
original sum of the 100 numbers = 550
since 1 is deducted 99 times, then 550  99 = 451
Or you can 1st do a row 1 to 10: you'll get 46 as final number; 55  9
Do other rows: end up with 46 ten times; 46*10  9 = 451
Formula (n by n square):
last number = [n^2(n  1) + 2] / 2