| * Operations and Matrices |
"*" operates as follows:
a*b = a + (b2 - ab)/(a-b)
"#" operates as follows:
n# = n*([n-1]*([n-2]*...*(3*(2*1))))
Given that
what are the value(s) of x and y? |
Problem Moderated by: Sasha |
| Problem Solution |
(b2 - ab)/(a-b) = -b, so a*b = a-b. This is good to know.
A pattern quickly emerges when you apply the star operation from the inside out.
2 - 1 = 1
3 - 1 = 2
4 - 2 = 2
5 - 2 = 3
6 - 3 = 3
...
So,
6# = 3
7# = 4
13# = 7
and -15# = -8
That is,
Therefore, x = 1 and y = 3/4 |
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