Question 1172321: The integers 1, 2, 3, ...., 40 are written on white board. An operation is then repeated 39 times. In each repetition any two numbers say a and b currently on white board are erased and new number a + b -1 is written. What will be the number left on white board at the end?
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! The integers 1, 2, 3, ...., 40 are written on white board. An operation is then
repeated 39 times. In each repetition any two numbers say a and b currently on
white board are erased and new number a + b -1 is written. What will be the
number left on white board at the end?
The sum of the numbers on the board is 1+2+3+4+...+40 = 820
When I erase a and b, the sum of the 38 numbers on the board is 820-a-b.
When I replace them with a+b-1, the sum of the 39 numbers on the board is
820-a-b+a+b-1 or 819.
So the sum goes down 1 each time and so does the number of numbers on the board
so when the operation is repeated the 40th time the sum of the numbers on the
board will have gone down by 40 and the sum will be 820-40 or 780.
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Ikleyn is right. I got careless once I saw how to do it. It should have been
820-39 = 781. You can't do a 40th operation. lol. Happy Holidays!
Edwin
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
Edwin provided a brilliant starting idea in his post.
I came to complete/(to fix) his solution accurately.
If the operation was repeated 39 times, then the SINGLE number which left on the board, is 820 - 39.
After the 39-th step/operation, there is NOTHING to do; the algorithm does not work further,
and the 40-th step can not be done (and is not assumed to be done).
So, the final number is 820-39 = 781. ANSWER
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