Mathcounts marathon!!!!

[wiseidiot]

Solutions to math154's previous probs:

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1) A perfect number is a number whose divisors (other than itself) add to that number.

I personally just memorized the first few: 6,28,496....then they get big

so there are only two perfect numbers less than 100

phew! that was hard!

2) For this problem, I counted the number of ways to get 0,1,2,3,4 heads then subtracted from total possibilities.

Total possibilities:

0 heads: 1 (obviously TTTTTTT....)
1 head: 20 (the H can be anywhere in the 20 flips)
2 heads: 20C2=190
3 heads: 20C3-1140
4 heads: 20C4=4845


is the answer, to lazy to simplify

[wiseidiot]

oops, forgot new probs.

Here they are:

1)A painting was sold at an auction for 442500 dollars. This was 8850% of its estimated value before the auction. How many dollars was the pre-auction value?

2)Let How many natural number factors does have?

srry, if they're too easy

[math154]

1)

Assuming the estimated value is the same as the pre-auction value: solve the equation:

2)

EDIT: I hope I didn't screw up the answers this time.

1)Bob and Joe are playing a game in which they can earn 234032432234524543 or 24758947189023453254347893240 points per turn. What is the greatest score they CANNOT obtain?(You are allowed to not simplify your answer.)

2)Starting from and going to (consecutive powers of 2), what percent of the powers have a 1 as the "first" digit?

[wiseidiot]

check number 1:

Hint:
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8850% is samething as multiplying by 885

For number 2:
Note that I did not write the prime factorization of

[wiseidiot]

As for your new problems:

Solutions:

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1) (First number) x (second number) -(first number) - (second number) (did you have to make the numbers so big?)

It's hard to explain why this works just search "Frobenius number" in wikipedia, it explains it.

2)I assume calculators aren't allowed on this one.

Lets list a few to see if there's pattern,

all have their first digit as 1 (It's useful to have you powers of 2 memorized up to )

Listing only the powers: 0,4,7,10

I see a small pattern, the powers either increase by 3 or 4

Then I calculated that also has first digit as 1

which gives us 0,4,7,10,14

NOW there is definately a pattern (0,4,7,0,4,7....)

Listing the rest gives: 0,4,7,10,14,17,20,24,27,30,34,37,40,44,47,50 which is a total of powers of 2

edit: didnt see that question asked for percents. 16/51 = 31%(nearest whole number)

[math154]

[wiseidiot]

you're answers were all correct

[math154]

Impersonating somebody: It's actually 'your'.

Here's another question.
Replace my question with 3^0...
What is the answer?

[RunpengFAILS]

lol i bet it's around 30% again.

[math154]

And another one: how many squares with all four vertices on the grid points are on a 6x6 grid where there are 6 grid points on each side?

[wiseidiot]

After trying out a bunch of numbers I found that all have their first digit as 1 so from onwards 50% of powers of three have their first digit as 1

[wiseidiot]

solution to math154's problem:

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Sorry there's not an in depth solution but after doing a bunch of casework I got 105 squares as the answer. Is that right? I probably undercounted, AGAIN!!!

[math154]

For the powers of 3, it's actually still around 30%. lol.
You assumed too quickly. It actually sort of cycles, with a period of almost no "1's", then a period of every other one as a "1", then over and over again.

For the square grid one, I would actually reduce the problem to a smaller grid case, and find a function/pattern (like starting with a 2x2, 3x3,...).
Unless you know the nifty formula...
The formula (I think) is . So...you were right !!! Nice job not under-counting.

lol we're the only ones posting here.
and i like ur calculator

[wiseidiot]

2 new problems:

1) I have 162 coins in my collection of nickels, dimes, and quarters which has a total value of 22 dollars. If I have 12 more quarters than nickels, how many dimes do I have?

2) How many natural numbers less than 1000 have exactly three different distinct positive integer factors

edit: That's a cool formula. I wonder why that would work...

[Ihatepie]

1.
quarters are x, nickels are y, dimes are z.
x+y+z=162
x+(x-12)+z=162
2x+z=174
25x+5(x-12) + 10z=2200
30x+10z=2260
3x+z=226
-2x+z=174
x=52
z=70

2.
4,9,25,49,121,169,289,361,529,841,961

so 11. Basically, the squares of prime numbers.

Here is one:

A 3x3x3 cube is painted on all sides and then cut into 27 equal sized smaller cubes. A cube is randomly chosen and then rolled. What is the probability that when the cube stops moving, only one painted side can be seen?

btw, the only reason you are the only guys posting here is all of the posts were
done in one day.

[uldivad9]

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8/27 * 0
12/27 * 1/3 = 4/27
6/27 * 5/6 = 5/27
Sum = 1/3

[math154]

EDIT:uldivad, for the center pieces (6/27), you did the possibility that the cube would have no painted faces shown (1/6). It would be 6/27*5/6 instead.

Unless I screwed up somewhere,
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Center piece selected:

Edge piece selected:

Total probability:

[wiseidiot]

moogra is right, i did only the scalene triangles.

[moogra]

@Wiseidiot: you forgot about 1,1,1 and 1,2,2 etc
ANSWER

There are 4 equilateral triangles
There are 8 isosceles non equilateral triangles
There are 6 scalene triangles

[math154]

No...the problem stated ONE side showing. With the corners, it would be impossible for 2 of the 3 painted sides to be covered.