Board problem for solving.

Question

Numbers from 1 to 20 are written on a board. In every move, You erase any 3 numbers present on the board (say a b and c') .You immediately write 2 new numbers (a+b+c)/3 and 3abc/(a+b+c).Answer the 2 sub-questions based on this information given above.

1.After how many steps are you forced to stop.

2. When only 2 numbers are left in this process, it was observed that one number was 19! (19 factorial). Then what must be the other number?

Answer 1

1.

With every move you remove one number.

Therefore it takes 18 moves to end up with only 2 numbers.

2.

Observe that the product of all the numbers on the bord remains the same.

In one move you remove the product term abc.

You add the product term (a+b+c)/3 × 3abc/(a+b+c) = abc

At the beginning the product of all the numbers is 20!.

And that must be the product of the two remaining numbers at the end.

Therefore the other remaining number is 20.

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Consider an example.

Suppose that you remove the numbers 1, 4, 10

You will add the number (1+4+10)/3 = 5, and

the number 3×1×4×10/(1+4+10) = 8.

Notice that the product of the original three numbers 1×4×10 = 40,

and the product of their two replacements 5×8 = 40.

It is not a coincidence that the two product are the same.

The problem is deliberately set up so that the product of the three removed

numbers is always the same as the product of the two added numbers.

Now suppose that at some point in the game, P is the product of all the numbers on the board.

After the removal of a, b, c the product of all the numbers on the board becomes

P/(abc)

After the addition of the two new numbers, the product of all the numbers becomes

P/(abc) × (a+b+c)/3 × 3abc/(a+b+c) = P

So the product of all the numbers on the board does not change by the replacement.

Therefore the product at the beginning, which is 20!, must also be the product

of the last two remaining numbers.