how do you do this?

[gemini9525]

what is the digit in the tens place of 7^21?

it was the TMSCA this year.

[peterliang]

7^7 = 823543

3543 x 7^7 = 2917812849

2849 x 7^7 = 2346274007

answer is 0

[blahblah]

I put it into my TI-89, and i got the answer.

[suchenzang]

Or you can use the same concept for finding the units digit but instead taking the last two digits each time. 7^1 = 07, 7 x 7 = 49, 49 x 7 = 343, 43 x 7= 301 The pattern cycles every fourth power, and so 7^21 will have a tens digit of 21mod4 = 1 which is 0. This is sort of like peterliang's solution, except I'm not sure why he's taking the last four digits each time...

[bradp]

[7] [^] [21] [ENTER]

Anyone get the trig identity correct on the test? That thing took a bunch of different steps.

[peterliang]

i dont think that works on a ti 83

[stupidityismygam]

[The3point14]

I have some problems working some of the questions on this test, and rather than make a new thread, I will just post them here. Any help will be greatly apprecaited:

#16 : If the matrix equation -3 [x y] = [ x 4 ] -[ 8 -2y], then x +y equals?
* no idea what this is, full explaination needed :-/

#19 : Rancher Roy wants to build a rectangular shaped pig pen with three parallel fences creating four sections as shown in the drawing. If he has 500 feet of fencing, what is the maximum area of his pen?

#33: let C be the centroid and line(XM) be a median of triangle (XYZ). If XC =9 and CM = n +3, then n = ?
* if anyone has a webpage, or a list of formulas using the special triangle points ( circumcenter, incenter, etc.) I could use it.

Thanks,
Joshua

[stupidityismygam]

[fht]

Brad, I did the trig identity one. It was annoying, it's the only question on the MA test that I seriously double checked. I'm usually not good with those problems, there would always be messups, as I did have one on that problem, but all came out fine and I'm purty happy about it.

The3.14, on the matrix-looking one, it's two simple equations beguiled--
-3x=x-8 & -3y=4-(-2y) x=2, y=-.8, x+y=1.2

#33, the centroid divide the median into segments whose length are 2:1. Hope that helps.
Best Wishes

[AuSmith]

[Monster Physics]

Yes, but you're lucky it worked out that time, if the problem were any different, you would have to do the whole optimization derv = 0 thing. I think thats funny that it turned out you could allot half and half.

[bradp]

His solution always works.. a square has the maximum area of all rectangles with a constant perimeter

A related problem was on the B test...
a rectangular solid with a square base is made out of 36 sq ft of metal and has an open top. What is the height of the solid with maximized volume

hs^2 = maximum
s = 2h (proving this is left as a task... an opportunity for the reader)
SA = 4sh + s^2 = 8h^2 + 4h^2 = 12h^2 = 36[/i]

[peterliang]

sry but this is probably a stupid question but, if you want to make it a square, how come you dont set L = (500-5L) / 2

[AuSmith]

[peterliang]

mhm i see

[The3point14]

Thank you guys for the many solutions to my questions. I can't tell you how much I need help, and you guys never let me down! Hopefully, I will be able to make it out of regionals this year.