How number nifty are you?

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thesurge

Golden Member
Dec 11, 2004
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The first few are pretty easy (note that AMC8 means for 8th graders, AMC10 for 10th graders, etc.); however, it gets harder.

forbes- math quiz

also note, that for the real AIME:
1) You aren't allowed to use a calculator
2) It isn't multiple choice.
 
H

HN

Diamond Member
Jan 19, 2001
8,186
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0
Originally posted by: thirdeye
That's way to much work for a Friday afternoon.
Click to expand...
a friday afternoon where 1/4 of the office is PTO, 1/4 is "working from home", 1/4 is leaving at noon, 1/4 leaving ~3pm, and 1/4 are just plain hungover ;)
 
S

ShadowOfMyself

Diamond Member
Jun 22, 2006
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Originally posted by: HN
Originally posted by: thirdeye
That's way to much work for a Friday afternoon.
Click to expand...
a friday afternoon where 1/4 of the office is PTO, 1/4 is "working from home", 1/4 is leaving at noon, 1/4 leaving ~3pm, and 1/4 are just plain hungover ;)
Click to expand...

Hm... 5/4 total? I guess ghosts are walking among us :D

Oh and that gives me headaches, logic problems are much better
 
E

eLiu

Diamond Member
Jun 4, 2001
6,407
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lol, I did these contests in HS. After 3 yrs I'm a lot slower than I used to be, but I can still do it... yay.

They should post like putnam problems; those are fvckloads harder.
 
TuxDave

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
Edit: Ok I know where I screwed up. "My idea is that your three digit number can never contain a zero because you would end up with non-unique digits. "

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

I concluded that you can only end up with these forms.

O = Odd
E = Even

OOE
OEO
EOO
EEO
EOE
OEE

Where the different type is the average of the other two. (eg OOE, the even number is the average of the two odd numbers)

Looking at the combination of OO I get 20. 1,3,5,7,9 for the first number and the second number can only have 4 choices.

Then you multiply 3 since OOE,OEO,EOO have the same number of choices to get 60.

EE (where the first digit isn't in the hundred's place) you get 0,2,4,6,8 so 5x4=20
EE (where the first digit is in the hundred's place) you get 4*4=16
So OEE,EOE,EEO you get 20+16+16=52

52+60=112. Yay.

<-- nerd
 
H

HN

Diamond Member
Jan 19, 2001
8,186
4
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Originally posted by: ShadowOfMyself
Originally posted by: HN
Originally posted by: thirdeye
That's way to much work for a Friday afternoon.
Click to expand...
a friday afternoon where 1/4 of the office is PTO, 1/4 is "working from home", 1/4 is leaving at noon, 1/4 leaving ~3pm, and 1/4 are just plain hungover ;)
Click to expand...

Hm... 5/4 total? I guess ghosts are walking among us :D

Oh and that gives me headaches, logic problems are much better
Click to expand...
2 purposes -- one was to make someone do math, the other was to indicate who among us in the office is hungover :)
 
T

tfinch2

Lifer
Feb 3, 2004
22,114
1
0
Originally posted by: TuxDave
Someone mind working this out for me?

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
a) 96 b) 104 c) 112 d) 120 e) 256

I ended up with 96. My idea is that your three digit number can never contain a zero because you would end up with non-unique digits. I also concluded that you can only end up with these forms.

O = Odd
E = Even

OOE
OEO
EOO
EEO
EOE
OEE

Where the different type is the average of the other two. (eg OOE, the even number is the average of the two odd numbers)

Looking at the combination of OO I get 20. 1,3,5,7,9 for the first number and the second number can only have 4 choices.

Then you multiply 3 since OOE,OEO,EOO have the same number of choices to get 60.

Doing the same for the combinations of EE I get 4*3 = 12 and times 3 for the EEO,EOE,OEE cases I get 36.

36+60=96. But that's wrong. :(
Click to expand...

120 Average of 2 and 0 is 1
 
TuxDave

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
Originally posted by: tfinch2

120 Average of 2 and 0 is 1
Click to expand...

Yah.. i saw that example. So if I included 0s I get 112 now. I like my answer better than the one they gave.
 
T

thesurge

Golden Member
Dec 11, 2004
1,745
0
0
Originally posted by: TuxDave
Originally posted by: tfinch2

120 Average of 2 and 0 is 1
Click to expand...

Yah.. i saw that example. So if I included 0s I get 112 now. I like my answer better than the one they gave.
Click to expand...

How about this solution?

Since the number must be the same mod 2 to have an average, we have:
(2)(5C2) ways to pick even and odd digits.
There are 3! ways to order them.
(2)(5C2)(3)=120
However, we over-counted the numbers that start with zero. There are (2!)(2)=8 of them.

Therefore, 120-8=112.
 
JujuFish

JujuFish

Lifer
Feb 3, 2005
11,448
1,056
136
I got to the end, and Forbes said I got 7 of 8 questions right. There were only seven questions. :confused:
 
Jeff7

Jeff7

Lifer
Jan 4, 2001
41,596
20
81
Originally posted by: thirdeye
That's way to much work for a Friday afternoon.
Click to expand...
Indeed. I just took a quiz that involved finding eigenvalues and eigenfunctions, and one Fourier Series problem. Brain not want work now. Brain want rest.
 
A

andylawcc

Lifer
Mar 9, 2000
18,183
3
81
...... urg. my head really is hurting. no joke

I got the first two right. the third one messed up a bit but got the concept right.
the fourth one I tried but failed (at least I understood the question and the means to get it right)
all the other ones just eluded me.
 
DrPizza

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
167
111
www.slatebrookfarm.com
Originally posted by: TuxDave
Edit: Ok I know where I screwed up. "My idea is that your three digit number can never contain a zero because you would end up with non-unique digits. "

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

I concluded that you can only end up with these forms.

O = Odd
E = Even

OOE
OEO
EOO
EEO
EOE
OEE

Where the different type is the average of the other two. (eg OOE, the even number is the average of the two odd numbers)

Looking at the combination of OO I get 20. 1,3,5,7,9 for the first number and the second number can only have 4 choices.

Then you multiply 3 since OOE,OEO,EOO have the same number of choices to get 60.

EE (where the first digit isn't in the hundred's place) you get 0,2,4,6,8 so 5x4=20
EE (where the first digit is in the hundred's place) you get 4*4=16
So OEE,EOE,EEO you get 20+16+16=52

52+60=112. Yay.

<-- nerd
Click to expand...

I think you looked at it the hard way...
I started with 0, and didn't look for what you concluded, although it sort of appears.
0&1? nope
0&2? Yes, avg = 1
0 & 3? Nope.

At this point, I started counting: distinct two digit pairs such that the 3rd digit will be their average.
02, 04, 06, 08, 24, 26, 28, 46, 48, 68
That takes care of the even ones;
And, for the odds,
13, 15, 17, 19, 35, 37, 39, 57, 59, 79

(note: I quickly realized there would be 4+3+2+1 combinations of evens and the same number of combinations of two odds, hence I was at 20 without actually thinking out all of the pairs, or writing them all down)

With the third digit being the average of the two existing digits, it created 20 different three digit sets. There are 6 different orders for almost each of these (in a moment)
i.e. for 39, the third digit is 6. You can have 369, 396, 639, 693, 936, 963

So, 20*6 = 120.
But, then, 024 really isn't a three digit number. 24 is a two digit number. For each of the pairs that contain a 0, there are two ways to arrange them so that the 0 is first. That's 4*2=8. So, my final total is 112 also.

btw. Since I agree with Tuxdave on the answer of 112, the test is incorrect. :p :)
 
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