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dimis
Responsible
Supreme Hero
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posted April 30, 2010 08:06 PM |
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Ok, give us these two numbers.
Btw, your explanation justifies my comment. If we write down a (=one) number, then we can not be certain that this was the number the other guy wrote down, since there are two solutions.
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ihor
Supreme Hero
Accidental Hero
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posted April 30, 2010 08:09 PM |
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dimis
Responsible
Supreme Hero
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posted April 30, 2010 08:19 PM |
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Q3. Comparing Ages
Alright!
Q3. Now I'm four times older than my sister was when she was half as young as I was. In 15 years our combined age will be 100. How old are we now ?
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ihor
Supreme Hero
Accidental Hero
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posted April 30, 2010 08:31 PM |
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Edited by ihor at 20:45, 30 Apr 2010.
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Your age is 40, your sister's is 30.
Edit: solution missing to give chance to solve it for others.
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dimis
Responsible
Supreme Hero
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posted April 30, 2010 08:41 PM |
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Q4. A Mean Shot
You are too fast for me ihor! Alright, let's move to another one!
Q4. Thirty people took part in a shooting match. The first participant scored 80 points, the second scored 60 points, the third scored the arithmetic mean of the number of points scored by the first two, and each subsequent competitor scored the arithmetic mean of the number of points scored by all the previous ones. How many points did the last competitor score ?
(N. Antonovich)
Clarification by me: The arithmetic mean of some numbers is the average of those numbers.
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ihor
Supreme Hero
Accidental Hero
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posted April 30, 2010 08:51 PM |
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I think enough for today. I'll think on this problem tomorrow.
A bit offtopic question dimis:
Is this some kind of collection of slavic problems? All of the names of authors you stated have slavic origin.
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dimis
Responsible
Supreme Hero
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posted April 30, 2010 08:53 PM |
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I will reveal my source in the end. 
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Ecoris
Promising
Supreme Hero
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posted April 30, 2010 09:50 PM |
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Edited by Ecoris at 21:52, 30 Apr 2010.
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It took me 1 minute to guess the number(s). Are to prove that there are only two solutions?
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dimis
Responsible
Supreme Hero
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posted April 30, 2010 10:02 PM |
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Can we prove it ?
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 30, 2010 10:31 PM |
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Edited by Corribus at 22:33, 30 Apr 2010.
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Q4 seems to be a little bit of a trick question, if I'm reading it correctly. Is the answer 70? Each shooter after the first two score 70 points; the mean never changes.
EDIT: My guess is you can prove it. For instance, it is easy to prove that 1 cannot appear at either end of the number, in order to satisfy the rest of the rules.
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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dimis
Responsible
Supreme Hero
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posted April 30, 2010 10:34 PM |
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Edited by dimis at 22:35, 30 Apr 2010.
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That's the moral of the story which is obvious to you now Corribus.
Adding the mean to a set of numbers, does not change its mean.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted April 30, 2010 10:40 PM |
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Edited by dimis at 23:03, 30 Apr 2010.
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Q5. Billions and Billions
Alright, let's move on to another one.
Q5. The product of a billion natural numbers is equal to a billion. What's the greatest value the sum of these numbers can have ?
(G. Galperin)
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dimis
Responsible
Supreme Hero
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posted April 30, 2010 10:49 PM |
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Shortest Proof for Q2
Unlike in other situations, now we want to admire the one who has the shortest. PROOF that is ...
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted April 30, 2010 11:10 PM |
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Edited by Corribus at 23:18, 30 Apr 2010.
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Quote:
Unlike in other situations, now we want to admire the one who has the shortest. PROOF that is ...
Ok, here's my proof.
(1) A four can only be in the first, second or third position (likewise: 6th, 7th and 8th), as there must be four numerals between them. (Also, if a 4 is in the third position, the other 4 is in the 8th position, which by mirror symmetry is the same as having a 4 in the 1st position, so we'll say that there must be a 4 in the 1st or 2nd position.)
(2) If the digit "1" is on the end (first position), then a 1 must also be on the third position, and therefore 4 must be in the second position (see #1) and 7th position. This leaves no satisfactory arrangement of the 2's and 3's (141---4-). Therefore, a 1 cannot be in the first position.
(3) Let's consider whether the 4 can be in the 2nd position in any circumstance. We now know that in this case, the 1s would have to be in positions 3 and 5 (or 4 and 6, but these two cases are identical when 4s are at 2 and 7). However, in this case we are again left with no satisfactory arrangement of the 2's and 3's. Therefore, 4 must be in positions 1/6 (or 3/8).
(4) If 4 is in positions 1/6 then 1 must be positions 2/4, 3/5 or 5/7. If 1 is in position 3/5 or 5/7 then the 3s would have to be in positions 4/8 in both cases, which only leaves either 2/7 or 2/3 open, both of which are unsatisfactory for the 2s. Therefore, the 1s must be in positions 2/4.
(5) This gives us 41-1-4--. Clearly, position 3 cannot be occupied by a 2, so it must be a 3. Which means position 5 must be a 2, as must position 8, and position 7 must be a 3. Therefore, the only solution can be 41312432 or its mirror image, 23421314.
Yeah, not very elegant. 
EDIT: For Q5, deleted. I have a guess but I'll write it later.
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I'm sick of following my dreams. I'm just going to ask them where they're goin', and hook up with them later. -Mitch Hedberg
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted May 01, 2010 03:04 AM |
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Edited by dimis at 03:18, 01 May 2010.
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Q6. Fashion Statement
What's up with the rest of you guys ? Do you want some ladies in the picture in order to participate ?
Q6. Three girls went out in white, green, and blue dresses. Their shoes were also one of these three colors. (Each wore a matched pair of shoes!) Only Anna had the same color dress as shoes. Neither Betty's dress nor her shoes were white, and Katie's shoes were green. What was the color of each girl's dress ?
So, now we have Q5 and Q6 which are open.
Alright,
Later!
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winterfate
Supreme Hero
Water-marked Champion!
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posted May 01, 2010 03:36 AM |
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Katie's shoes are green. Since only Anna had the same color dress as shoes, Katie cannot have had a green dress. Since Betty cannot have a white dress (or shoes), she must've had a green dress or a blue dress.
By process of elimination:
You could say:
Anna had a white dress.
Katie had a blue dress.
Betty had a green dress.
Because:
Betty could have blue shoes
Katie has green shoes
Anna could have white shoes.
<3 logic problems. Beats math by a bunch!!
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If you supposedly care about someone, then don't push them out of your life. Acting like you're not doing it doesn't exempt you from what I just said. - Winterfate
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ihor
Supreme Hero
Accidental Hero
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posted May 01, 2010 09:50 AM |
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I'm too lazy to post my solution of Q2.
It is similar by beauty to Corribus's one.
I'll just say that I investigated what digit could be at the first position and proved that only 2 or 4 possible. Then investigated possible digits on other places.
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Ecoris
Promising
Supreme Hero
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posted May 01, 2010 11:35 AM |
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Edited by Ecoris at 11:37, 01 May 2010.
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Q5: Let B = 10^9 (a billion) and let a_1,a_2,...,a_B be a sequence of one billion natural numbers whose product is B. I claim that if two of the numbers are strictly greater than 1, then the sequence does not yield the greatest possible sum.
Proof: Assume the sequence has two numbers that are greater than 1; we may assume that a_1 and a_2 are two such numbers. The product of the elements of the sequence a_1*a_2,1,a_3,a_4,...,a_B (i.e. the sequence where a_1 is replaced by a_1*a_2 and a_2 by 1) is still B but the sum is strictly greater since a_1 > 1 and a_2 > 1 implies that (a_1 - 1)*a_2 > a_1 - 1, hence a_1*a_2 + 1 > a_1 + a_2.
As such, an "optimal" sequence can at most have one element that is not equal to 1. That element must then equal B, therefore the solution (up to ordering) is unique and its sum is 1,999,999,999 (= 2B - 1).
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Ecoris
Promising
Supreme Hero
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posted May 01, 2010 12:09 PM |
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Edited by Ecoris at 12:13, 01 May 2010.
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Proof of Q2:
The problem is equivalent to the following: Find natural numbers a,b,c,d such that {a,b,c,d,a+2,b+3,c+4,d+5} = {1,2,3,4,5,6,7,8}.
Add the elements of both sets:
2(a+b+c+d) + 14 = 36, i.e. a+b+c+d = 11
Since 1,2 have to be found among a,b,c,d, the remanining two terms in the sum a+b+c+d must be 3 and 5, since all the numbers a,b,c,d have to be distinct. (1+7, 2+6, 3+5 and 4+4 are the only ways to write 8 as a sum of two natural numbers and we can only use 3+5). So {a,b,c,d} = {1,2,3,5} and then {a+2,b+3,c+4,d+5} = {4,6,7,8}.
Clealy we have that:
a=5 or b=5
and
a+2 = 4 or b+3 = 4
This gives us two cases:
1) a = 5 and b = 1
and
2) b = 5 and a = 2
In case 1), {c,d} = {2,3} but c = 3 and d = 2 is not possible since then c+4 = d + 5. Therefore we must have c = 2, d = 3 which does indeed give a solution. (Which corresponds to 23241314).
In case 2), {c,d} = {1,3}, but c = 1 is clearly not possible since then c + 4 = 5 = b. Therefore we must have c = 3 and d = 1 which gives another solution. (Which corresponds to 41312432).
Edit: Corribus' proof has 50% more characters, so mine is shorter .
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted May 01, 2010 03:56 PM |
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Quote:
Edit: Corribus' proof has 50% more characters, so mine is shorter .
I'm just wordy.
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