Gauging Interest - Mock MATHCOUNTS Nationals

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BOGTRO

5818 posts

BOGTRO

#1 h Mar 21, 2011, 7:03 PM • 2 Y

Y by Adventure10, Mango247

I'm trying to gauge interest for a mock MATHCOUNTS National Competition. I'm starting to write one, but I'm not going to finish it unless there is some amount of interest. If you would be interested in taking this, post here or PM me please.

EDIT: And I'll probably need a proof-reader, if anyone is interested in that.

Z Y

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AwesomeToad

4535 posts

AwesomeToad

#2 h Mar 21, 2011, 8:06 PM • 2 Y

Y by puppybugs, Adventure10

I might be able to help proofread if you would like.

I'm not in middle school though, so I won't take the test.

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ProblemSolver747

190 posts

ProblemSolver747

#3 h Mar 22, 2011, 1:56 AM • 2 Y

Y by Adventure10, Mango247

I'll take it. Thanks!

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applepi2000

2226 posts

applepi2000

#4 h Mar 22, 2011, 11:13 PM • 2 Y

Y by Adventure10, Mango247

I would also be interested

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anonymous0

455 posts

anonymous0

#5 h Mar 22, 2011, 11:37 PM • 2 Y

Y by Adventure10, Mango247

I'll join.

Z Y

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BOGTRO

5818 posts

BOGTRO

#6 h Mar 22, 2011, 11:48 PM • 2 Y

Y by Adventure10, Mango247

All right, I'll finish working on this. Don't expect it for a while though, I only have about 10 (decent) problems written and I need about 70.

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fractals

3028 posts

fractals

#7 h Mar 23, 2011, 12:29 AM • 2 Y

Y by Adventure10, Mango247

I will join this tournament.

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alkahfi

187 posts

alkahfi

#8 h Mar 23, 2011, 8:48 PM • 2 Y

Y by Adventure10, Mango247

i will join

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btilm305

439 posts

btilm305

#9 h Mar 24, 2011, 3:54 AM • 1 Y

Y by Adventure10

I'll help with the problems. Assuming you will say hi to me at nationals

I'm representing Louisiana

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62861

3564 posts

62861

#10 h Mar 25, 2011, 2:21 PM • 3 Y

Y by centslordm, Adventure10, Mango247

i will join. i am representing michigan

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Nitzuga

600 posts

Nitzuga

#11 h Mar 25, 2011, 8:40 PM • 2 Y

Y by Adventure10, Mango247

I'll join.

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KingSmasher3

1399 posts

KingSmasher3

#12 h Mar 26, 2011, 2:25 PM • 2 Y

Y by Adventure10, Mango247

I will join.

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y3k-9

39 posts

y3k-9

#13 h Mar 27, 2011, 5:31 AM • 2 Y

Y by Adventure10, Mango247

I can join or proofread... both will be fun for me.

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gapoc459

432 posts

gapoc459

#14 h Mar 27, 2011, 2:31 PM • 2 Y

Y by Adventure10, Mango247

I'll proofread!

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ha123

178 posts

ha123

#15 h Mar 27, 2011, 2:39 PM • 2 Y

Y by Adventure10, Mango247

I'll join!!!

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BOGTRO

5818 posts

BOGTRO

#156 h Apr 28, 2011, 4:53 PM • 2 Y

Y by Adventure10, Mango247

Results:

applepi2000 defeats ksun48 4-3
mathocean advances by default. (Against luppleAoPS)
mathocean defeats dragon96 4-3
applepi2000 advances by default. (Against Iggy)
applepi2000 defeats mathocean97 to become the BOGTRO MOCK TEST CHAMPION.

E-mail me at bogtro123@gmail.com for prizes if applicable.

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AwesomeToad

4535 posts

AwesomeToad

#157 h Apr 28, 2011, 4:57 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Darn ksun48

Darn.

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Iggy Iguana

246 posts

Iggy Iguana

#158 h Apr 28, 2011, 9:26 PM • 1 Y

Y by Adventure10

Sorry for not making it.

Took a nap at 5:30 after JMO.

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luppleAOPS

70 posts

luppleAOPS

#159 h Apr 29, 2011, 1:15 AM • 2 Y

Y by Adventure10, Mango247

did you say you were coming out with another test bogtro?

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BOGTRO

5818 posts

BOGTRO

#160 h Apr 29, 2011, 1:33 AM • 2 Y

Y by Adventure10, Mango247

Doubtfully. I expected to end this much earlier than I did.

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g.c.boxd

181 posts

g.c.boxd

#161 h May 4, 2011, 1:51 AM • 2 Y

Y by Adventure10, Mango247

BOGTRO wrote:

17. Note that $2^{n+100}=2^n \pmod {1000}$

Umm why is this true? I know it is (using wolfram)... but why? Im probably missing something really stupid.....

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BOGTRO

5818 posts

BOGTRO

#162 h May 4, 2011, 2:08 AM • 2 Y

Y by Adventure10, Mango247

Since $2^{100} \equiv 0 \pmod 8$ , it suffices to show that $2^{100} \equiv 1 \pmod {125}$ , which is true, although not simple to prove (best method is to go to wolframalpha

).

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applepi2000

2226 posts

applepi2000

#163 h May 4, 2011, 2:21 AM • 2 Y

Y by Adventure10, Mango247

BOGTRO wrote:

Since $2^{100} \equiv 0 \pmod 8$ , it suffices to show that $2^{100} \equiv 1 \pmod {125}$ , which is true, although not simple to prove (best method is to go to wolframalpha

).

Or, use the totient theorem, since $\Phi(125)=100$ gives $2^{100}\equiv 1\mod 125$ .

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dragon96

3212 posts

dragon96

#164 h May 4, 2011, 2:35 AM • 2 Y

Y by Adventure10, Mango247

If you're doing MathCounts in the future, hopefully they won't add any questions about Euler's Theorem. However, you'll probably need it if you're planning on doing well on Olympiads in the future. Or BOGTRO's tests.

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BOGTRO

5818 posts

BOGTRO

#165 h May 4, 2011, 4:34 PM • 2 Y

Y by Adventure10, Mango247

Ha.

It's kind of one of those things you should know, that last two digits repeat every 20 and last 3 repeat every 100 (it's not that much more complicated than knowing that last digits repeat every 4).

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g.c.boxd

181 posts

g.c.boxd

#166 h May 4, 2011, 9:33 PM • 1 Y

Y by Adventure10

BOGTRO wrote:

Since $2^{100} \equiv 0 \pmod 8$ , it suffices to show that $2^{100} \equiv 1 \pmod {125}$ , which is true, although not simple to prove (best method is to go to wolframalpha

).

Ok yeah I get that part... but doesnt $2^{n+100}=2^n \pmod {1000}$ mean that $2^{100} \equiv 1 \pmod {1000}$ ? Which obviously isnt true... what am I mistaken in thinking? Its probably really stupid...

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AIME15

7892 posts

AIME15

#167 h May 4, 2011, 9:52 PM • 2 Y

Y by Adventure10, Mango247

$ad \equiv bd \pmod{m}$ does not imply that $a \equiv b \pmod{m}$ . This only holds if $(d,m) = 1$ .

More generally, $ad \equiv bd \pmod{m} \implies a \equiv b \pmod{(d,m)}$ .