Solution:
Let us compute 2010∗10 using the given properties.
Let f(y)=2010∗y.
We are given:
f(0)=0
f(y+1)=f(y)+(2010−y)
Let us compute step by step:
f(0)=0
f(1)=f(0)+(2010−0)=0+2010=2010
f(2)=f(1)+(2010−1)=2010+2009=4019
f(3)=f(2)+(2010−2)=4019+2008=6027
f(4)=f(3)+(2010−3)=6027+2007=8034
f(5)=f(4)+(2010−4)=8034+2006=10040
f(6)=f(5)+(2010−5)=10040+2005=12045
f(7)=f(6)+(2010−6)=12045+2004=14049
f(8)=f(7)+(2010−7)=14049+2003=16052
f(9)=f(8)+(2010−8)=16052+2002=18054
f(10)=f(9)+(2010−9)=18054+2001=20055
Therefore, 2010∗10=20055.
Alternatively, we can find a closed formula.
Let us try to find a general formula for x∗y.
Let f(y)=x∗y.
We have:
f(0)=0
f(y+1)=f(y)+(x−y)
So,
f(1)=f(0)+(x−0)=x
f(2)=f(1)+(x−1)=x+(x−1)=2x−1
f(3)=f(2)+(x−2)=(2x−1)+(x−2)=3x−3
f(4)=f(3)+(x−3)=(3x−3)+(x−3)=4x−6
So, the pattern is:
f(y)=yx−2y(y−1)
Let us check for y=4:
f(4)=4x−24×3=4x−6
Correct.
Therefore,
2010∗10=10×2010−210×9=20100−45=20055
Final answer:
20055