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Question

go to http://mathgod.com/mathcomps/interschools/fall2007/fall2007interschool-test.pdf and see if you can figure these out?!!!
I need to do them for Mu Alpha Theta, and its not cheating cuz it says I can use any resource (see for yourself) that includes internet and other ppl!. I'm a freshman and let me tell ya, if I try these I will be going in way over my head. Computer programs help, especially for the last questions that aren't really questions, don't ask me which ones! I have no idea...I got a few answers though.

wow....smart answer lol

actually I thought 99/19 was the answer. Its reciprocal is 19/99 which as a repeating decimal would be 0.191919... right? idk

oh thats true... number 14 looks IMPOSSIBLE to figure out. The only one I'm completely sure about is number 24, but that isn'e even a math problem. lol

I got---or rather someone else-- got number 7.
a. =293 (where q is 293 and p is 20)
b. =10885 (where q is 10885 and p is 743)

with that you could figure out number sixteen... if you have some program that could that in like seconds. Otherwise prepare to give up, I don't even understand the problem that well.

Answer 1

Great set of questions.
#29. Dice 1 is labelled 1,3,5,7,9,11
Dice 2 is labelled 0,0,0,1,1,1
Maybe I interpretted this question differently from manjyomen.

#7. See the previous question I worked out.
Note that I corrected an error in the comments.
q(1) = 293
q(2007) = 13937

#16.
Using hte same program, the first value of p for which multiple q's come up is p = 163, for which q = 2388, 2389

The range of possible q values is 2007p/137 to 2008p/137 not inclusive. So q spans p/137. This explains why the first value of p is greater than 137. That's the only way the range of q can encompass 2 integers.

Similarly, the values of p which possess 2007 q's must be greater than 137*2006.
Now we go to the program, and I'm getting
p = 274871 has 2007 values of q ranging from 4026760 to 4028766.

#13
It's 1/2.
We are given that the first 20 draws were all red. Implicit is the fact that we are also given that there are at least 20 red balls in the original set. So there are 81 possible no. of red balls (20, 21, .., 79, 80) each with probability 1/81.

Suppose there are n red balls.
After 20 draws, there are now 80 balls, n-20 red.
P(next one is red) = (n-20)/80
We now need to sum these probs from n = 20 to 100.
We end up with an arithmetic series whose sum is 1/2.

*SOME MORE*

#26
The factors of 10^n are 2^n and 5^n. 2 and 5 are the only prime factors, and any combination of the 2 and 5 will result in a factor which ends in zero. E.g. 10^2 = 2^2 * 5^2 = 20*5 = 2*50 etc.

n = 8 is the first value where a zero comes up (in the 5^8 term).

The two digit values are 18 and 33, such that x + y = 51

10^18 = 2^18 * 5^18
= 262144 * 3814697265625

10^33 = 2^33 * 5^33
= 8589934592 * 116415321826934814453125

I was able to do this in maple. As far as I see, this is nothing but trial and error until you get it. I can’t think of any mathematical technique that can help you figure out which factors have a zero somewhere in its digits. That’s insane. It would take way too long to go through all the 3 digit n’s.



#21
There are constraints which we must observe:
1)Every child gets exactly 12 candies
2)Every child has more chocolates than caramels and more caramels than cookies. No ties.
3)Only one child (Bob) has 4 caramels.
4)No combination is repeated.
5)Scott (who is different from Bob) has more than 1 cookie.
6)All chocolates add up to 26.

If there are n children, then there are 12n candies total. The total no. of candies must be less than 26*3 so 36 <= candies <= 72.
Ie. 3 <= n <= 6.

Here are the possible combinations of candies:
a)11,1,0
b)10,2,0
c)9,3,0
d)9,2,1
e)8,4,0
f)8,3,1
g)7,5,0
h)7,4,1
i)7,3,2
j)6,5,1
k)6,4,2
l)5,4,3

Combination a,i,e works. There will be 3 children, such that Bob gets 8,4,0; Scot gets 7,3,2; and Josh gets 11,1,0.

Now although not spelled out clearly, my guess is that no child got zero cookies, because it said “they just had the right number”, whatever that means.
There is one combination of 4 children that works – f,i,j,l.
Bob gets 5,4,3; Scott gets 7,3,2; Josh gets 8,3,1; and Emanon gets 6,5,1.


#20
I believe this integral diverges. The answer is ∞.


#26
Solving in maple, there are 2 complex roots which evaluate to
0.6951 +/- i*0.7815

The exact expression is shown in the link below.
http://i16.tinypic.com/6nubmg8.jpg

Answer 2

Well, the answer to #15 is 1111111111111111111 (nineteen 1s). The prime has to be a factor of (10^19 - 1) = 9999999999999999999 (nineteen 9s). Factoring out the 9 gives this number, which is a "repunit prime".

http://primes.utm.edu/glossary/page.php?sort=Repunit

I'm interpreting "period 19" to mean the cycle repeats every 19 digits. For example 1/7 = .142857142857... repeats
every 6 digits. 19/99 = .191919.... repeats every two digits. Anyway, 99/19 is not a prime number.

* * * *

#36 is a straightforward cryptogram. However, when you decipher it, the message asks you to write the number of digits in the factorial of 2007. Hmm... that's going to be a lot of digits. I think the answer is 5,759, but I could be off by 1 or 2.

* * * *

OK, looking at these questions a day later, I think I can do #32. Writing this as (2000 + 7)^k and visualizing the expansion of the polynomial, it is clear that only the 7^k term contributes to the last three digits. Therefore, the last three digits of 2007^2007^2007^2007 are the same as the last three digits of 7^7^7^7.

Now, the powers of 7 are nice in that the last two digits cycle after only four iterations: 01, 07, 49, 43, and then back to 01. In other words, 7^(4k+a) has the same last two digits as 7^a, where k is any positive integer. The last three digits cycle after twenty iterations, i.e. 7^20 ends in 001. We can use this to reduce 7^7^7^7 further:

Since 7^7 ends in 43, the mod base 20 of 7^7 is 3. Therefore 7^7^7^7 must end in the same three digits as 7^7^3. Similarly, 7^3 ends in 43, so the mod base 20 of 7^3 is also 3. That means 7^7^3 must end in the same three digits as 7^3, and the last three digits of 7^3 (actually the only three digits) are 343. So, the answer is 343.

Interestingly, from the way this power tower reduced, any number of exponents could have been used, i.e. 2007^2007^... (repeat as many times as you want) still has 343 as the last three digits.

* * * *

Ah, one more! #22 is √63

Answer 3

q35
alice's speed = v, A to B, 30 mins, and
bob's speed = 50, B to A, 40 mins.
both start at noon.

let them meet h hour after noon.

distance between A to B
= v(h + 1/2)
= 50(h + 2/3)
= 50*2/3 + v*1/2

v/2 = 50h, v = 100h
vh = 50*2/3 = 100/3 = 100h^2
h^2 = 1/3, h = √3/3
v = 100√3/3

distance AB
= vh + v/2
= 100/3 + 50√3/3
= 50(2+√3)/3

and both will pass each other in every 2h time.

let n be days that passed before they pass each other the 2007th time.
time after noon on 2007th passing
= (2*2007-1)*h - 24n
= 4013(√3/3) - 72n/3
= (6950.71989 - 72n)/3

for an answer of 0<=time<24, n should be 96.
answer
= (6950.71989 - 72*96)/3
= 38.71989/3
= 12.90663
= 12 hours 54 minutes and nearly 24 seconds.
(or simply nearly 1 hour after midnight)

q29
for conventional dices
2 and 12 = 1/36
3 and 11 = 2/36
4 and 10 = 3/36
5 and 9 = 4/36
6 and 8 = 5/36
7 = 6/36

but we want 2-12 (all 11 sums) is to be equally likely for all 36 tosses. the probability for each then must be 1/11, right?

let x be the number on dice.
1/11 = P(sum 2) = P(sum 12)
= [P(x=1)]^2 = [P(x=6)]^2

P(x=1) = P(x=6) = √11/11

1/11 = P(sum 3) = P(sum 11)
= 2P(x=1)*P(x=2) = 2(P(x=6)*P(x=5)
= 2√11P(x=2)/11 = 2√11P(x=5)/11

P(x=2) = P(x=5) = √11/22

going on the same way/process will get you
P(x=3) = P(x=4) = 3√11/88

answer = (√11/11, √11/22, 3√11/88, 3√11/88, √11/22, √11/11)

or simplified to

(4/15, 2/15, 1/10, 1/10, 2/15, 4/15) instead of the conventional (1/6, 1/6, 1/6, 1/6, 1/6, 1/6)

q2 (just here as light reading, ok?)
24^3 = 13824 possible 3-sequences, one of them MAT. that's easy, right?

but if you toss 6 times, you actually have 4 3-sequences in it, as i think you already realized.
eg. in sequence of abcdef, you'll get
(abc) (bcd) (cde) (def) , so 4 in all. or n 3-sequences in (n+2) tossing.

if you tossed 3 times for one sequence, say you got (abc), and another 3 for the 2nd one, of (def), you didn't just have the 2 of them, but also created the other 2, (bcd) and (cde) inadvertently. so in your case of μ(M), α(A) and Θ(T) ;

"intentional" MAT happens in every 24^3,
"unintentional" MAT happens in
-MA followed by T-- or --M followed by AT-
number of ways for MAT
= 2(24P1)(576P1)
= 2(24^3)
number of ways possible
= (24^3)P2
= (24^3)(24^3 - 1)

P(MAT)
= P(intentional) + P(unintentional)
= 1/24^3 + n(MAT) / n(possible)
= 1/24^3 + 2/(24^3 - 1)
= [3(24^3) - 1] / [(24^3)(24^3 - 1)]
= 0.000217
= 0.0217%
= basically 1 in every 4608 3-sequences.

from n 3-sequences in (n+2) tosses, so i'm guessing you need 4610 tosses to get an MAT somewhere in the chain.

sorry, this is another stupid answer from me. do NOT believe it for your life and sanity.

Answer 4

This has some nice questions.

#8: T_n = n(n+1)/2 for n>=1. Then, 1/(T_n) = 2/((n)(n+1)). When computing the sum it helps to know that 1 / (n)(n+1) = 1/n - 1/(n+1). This telescopes so a bunch of terms cancel. I'll let you finish this one.

#33: I like this one especially. Before we answer this lets consider the expansion of (x+y)^n for positive integral n. If we can find a general expression for the sum of the squares of the coefficients of that we are done. Let's consider the coefficients

nC0, nC1, nC2, ..., nCn. We hope that their is some nice expression to represent

(1) (nC0)^2 + (nC1)^2 + (nC2)^2 + ... (nCn)^2.

Let's look at this from a counting perspective. Using the identity nCk = nC(n-k) we can rewrite (1):

(2): (nC0)(nCn) + (nC1)(nC(n-1)) + ... (nCk)(nC(n-k)) + ... + (nCn)(nC0).

Imagine we have two groups of n items (group A and B). In total there are 2n items between the groups. The first term in (2) tells us the number of way we can choose 0 things from group A and the number of ways we can choose n things from group B. Similarly, the second term tells us the number of ways to choose 1 item from A and n-1 items from B. We can continue this and say that the k+1 st term in (2) tells us the number of ways in which we can select k items from A and n-k items from B. Each of these terms seems to represent a unique way of selecting a total of n items from the two piles. This is very promising!

More precisely, the terms in (2) represent the total number of ways we can select n items from piles A and B which contain 2n items. Of course, there is a simpler way to express this, namely (2n)Cn. You can test the expansions of (x+y)^n for small values of n and verify that the sum of the squares of the coefficients does indeed equal (2n)Cn. To finish we recall that in the problem n = (2007!)! so our answer is (2((2007!)!))C((2007!)!).

As an aside, you can prove

(nC0)^2 + (nC1)^2 + ... (nCn)^2 = (2n)Cn using induction as well, but I like the counting approach!

#28: We are given that f(cosx) = cos(37x). As unsightly as f(cosx) may appear, it's not too hard to tweak. Naturally, we might think of the identity (sinx)^2 + (cosx)^2 = 1, but there is something nicer we can use.

Define g(x) = f(cosx). We ask ourselves is there some value of x we can select so that the f(cosx) transforms into f(sinx)? The answer is yes. Recalling the identity cos(x - pi/2) = sinx it's now natural to pick "x - pi/2" as our nice value for the argument. Then we get

g(x - pi/2) = f(sinx) = cos(37(x - pi/2)) = cos(37x - 37pi/2). Now our task is reduced to simplifying cos(37x - 37pi/2) which is much easier. See if you can finish this from here.

Answer 5

There is not any reply for this question. Because complete of five atypical numbers could be an atypical quantity and it are not able to be 20. Therefore, its not possible that the butcher slaught 20 cows in five days if he reduce the cows every day in atypical numbers.

Answer 6

I'll jump right in.
#4 evaluates to π
#8 sums up to 2

Answer 7

A WOOD CHUCKER

CHUCKS WOOD


& That's all I'm answering for you