## Sunday, September 14, 2014

### More Solutions to Twitter CCSS/SAT Questions

The following is part of what everyone on my Twitter Problems mailing list has been receiving every day or two for the past 3 weeks. For free... You have 2 weeks left to sign up. Free...

As you can see I go beyond the answers. Way way beyond...
Just as the Math Practices of CCSS suggest we do...

1. I walked my daily path 25% slower than usual and took 5 min longer. How  many min does it usually take?

Solution:
Let R=usual rate (mi/min); T=usual time (min)
One can infer that distances are equal from the phrase "daily path". Using D=RT and equating:
((3/4)R)•(T+5)=R•T
R's " cancel" leaving
15/4=1/4•T or T=15.

Generalization:
If slower rate is k•R (0<k<1) and extra time is m min then
kR(T+m)=RT --->
km=T(1-k) --->
T=m(k/(1-k))
Test it: k=3/4,m=5 ---> T=5((3/4)/(1/4))=5•3=15
Special case: If k=1/2 then T=m or if one walks half as fast trip will take double the usual time!

2. Test this "rule":
3 more than the square of an odd integer is a mult of 4.
Now prove it!
Devise a rule if "more" is repl'd by "less!"

Solution:
An odd integer can be expressed as 1 more than even or 2n+1.
"3 more than the square of an odd integer" translates to
3+(2n+1)^2 = 3+4n^2+4n+1=4(n^2+n+1), a mult of 4.

Three less than the square of an odd becomes
(4n^2+4n+1)-3 = 4n^2+4n-2 which represents 2 less than a multiple of 4.
Thus "three less than the square of an odd" cannot be a multiple of 4 and in fact will always leave a remainder of 2. Why?