Answer:
Player II should remove 14 coins from the heap of size 22.
Step-by-step explanation:
To properly answer this this question, we need to understand the principle and what it is exactly is being asked.
This question revolves round a game of Nim
What is a game of Nim: This is a strategic mathematical game whereby, two opposing sides or opponent take turns taking away objects from a load or piles. On each turn, a player remove at least an object and may remove any number of objects provided they all come from the same heap/pile.
Now, referring back to the question, we should first understand that:
22₂ = 1 0 1 1 0
19₂= 1 0 0 1 1
14₂= 0 1 1 1 0
11₂= 0 1 0 1 1
and also that the “bit sums” are all even, so this is a balanced game.
However, after Player I removes 6 coins from the heap of size 19, Player II should remove 14 coins from the heap of size 22.
So I'll walk through the first 2. Then please try the other ones and let me know if you run into more problems.
5m-6
If m is equal to 7, then replace m with 7.
5(7)-6
5x7=35 so the expression is now 35-6. Solve.
35-6=29
So that was the first one.
4m+t
We just do what we do with the first problem.
4(7)+(2)
4x7=28 so the expression is 28+2.
28+2=28
---
hope it helps
p.s. When there is subtraction, add. When there is division, multiply to cancel that out. When there is multiplication, divide to cancel. etc.
The same as the diagonal of a square.
Hope this helped!
Answer:
Step-by-step explanation:
you can go to a calculator
Do 14 times n which give you 14n. and divide by 3.
Answer: (14n)/3
