This can be solved by finding the number of permutations of the eight performers who will all perform before the one who insists on being the last.
3. i=129; j=112
4.k=90;l=116;m=64;n=86
5.o=13;p=78;q=91
6.a=120;b=60;c=140
Can't see #7
You did 8 I believe
9.v=39;w=59;x=121
10.y=59;z=93
11.a=42;b=48;c=132
12.a=60;b=50;c=80;d=100
It will be a negative.
for an example: -20 for x would mqke 2x + 1 into -39 and 8x + 5 into -155. that's 116 apart, so maybe -2(8) + 5 = -11
-2(2) + 1 = -3
but that's only 8 apart, so it'd be higher.
-4(2) + 1 = -7
-4(8) + 5 = -27
BOOM
Tadaaaa,
You probably thought positive, but sometimes it's negative, you know what I mean?
The standard algorithm is the way or steps to find an answer.
