BMAD
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Posts posted by BMAD
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In retrospect, when I thought up the question, I realized that we shouldn't assume that we have a ninety degree angle. In geometric terms, if you've measured so all the sides are the correct length (opposite sides are the same length), then you have a parallelogram, and what I want to know is how far the angles are from right angles, based on the difference in the diagonals.
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Here is an algorithmic approach to solving this problem that may help with finding the equation. Assume that between L and W is a right angle, use a string to find the length of the hypotenuse. Use that string to arc out the possible endpoints of the other diagonal using the length from the string. Use the same rope to find the length of L or W. Go to the end point of either L or W away from the given ninety degree angle. Arc the line again and where it crosses with the other arc is where you need to put your fourth corner. Done!
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19 hours ago, gavinksong said:
I think I see the intuition behind that guess, but alas that is not it. Counterexample in the spoiler.
To rephrase your response, the maximum number of coins is 2^(n-1). So with three uses of the balance scale, you should only be able to pick out the fake from at most 4 coins.
However, in the original problem, you have 12 coins, and you are allowed three uses of the balance scale. (Hint: do not let this example mislead you. The original problem is possible with more than 12 coins.)
I don't see how that's possible if we don't know whether the fake is heavier or lighter in advance.
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Let's say that there are three congruent circles that are tangent to each other. Draw a fourth congruent circle with its center at the center of the gap formed between the three circles. If all four circles are unit circles, what is the area of the overlap?
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You understood it correctly. You have one of the numbers but there are actually more that meet the requirements. I just want the lowest answers.
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I do see that, which is why we must keep L and W arbitrary
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Everyday I make the same number of cookies. On Monday, five people came over for cookies. I gave each of them the same amount and kept one for me. The next day, I baked the same amount but this time eleven people came over. I gave them all an equal share and kept one for me. On the third day, I baked the same amount of cookies as the other days, gave thirty-five people the same amount, and of course kept one for me. What are the three possible lowest amounts of batches I made each day?
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Not anywhere otherwise there is no way someone would confuse the shape for a rectangle as it would lose its fourth side and by symmetry much of that circle is immaterial.
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If the bag contains coins that equal 2^m then it would take m+1 try to find the fake coin.
The spoiler option seems to have disappeared from the mobile app.
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On 8/5/2017 at 11:40 AM, Indra said:
For the shelf with 11 DVD's, the most moves you would need are 9 moves.
For two selves, the most moves you may have to make are 19 moves.
I like the solution of 7, could you explain why it isn't?
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Four men (A, B, C, and D) are kept in a room a punishment for their crimes. Each day, they retreat to a corner where they are fed an amount of a certain type of food. In total, they receive thirteen pieces of the assigned food. On this day, they all retreated to their corners to receive beef jerky. After enjoying their food, they came back to the center. In front of the group, A asked B, "Did you get more than me?" To which B replied, I don't know. Curious then, B turned to C and said, "Did you get more than me?" To which C replied, I don't know. Even more curious, C turned to D and said, "Did you get more than me?" To which D replied, "Yes and I know how much each of you got!"
How much food did A and D get?
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On 8/8/2017 at 8:44 AM, gavinksong said:
That's actually a pretty interesting point.
You're saying that even if A knows for sure that he got more, it's possible he is trying to determine whether B only got one piece or not (unless A got the maximum number of pieces).
Perhaps this is reading too much into the problem, but if that's the case, why wouldn't A simply ask how many pieces B got?
Because then we wouldn't have a puzzle.
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On 8/6/2017 at 4:09 AM, araver said:
Question:
1. Do we have anything else other than a (marked) ruler (as implied by the 6 cm measurement)?
Second question in a spoiler
After reading the question, I think "moving one of the corners" must mean that already 3 corners form a rectangular triangular, otherwise you wouldn't be able to create a rectangle by moving only one corner.
As such, drawing point A as the point that needs to be moved I understand that one of the diagonals is greater (or smaller) than the other by 6 cm. Please let me know if I understood "off by" correctly.
If so, I understand the problem is to "move" point A to point B (an example in the next figure).
However if one can measure L and W and has managed to draw a 90 degree angle between them, this means perpendiculars can be drawn to create the rectangle safely. How would an equation help?
You understand the problem well and yes you only have a ruler. As for the equation, moving the corner of one of the rectangular corners specifically the one opposite of the 90 angle, would adjust the angles adjacent to the 90 degree angle relative to L, W L2, W2. There is an equation that we can create so that we can check to ensure that l=l2 and w=W2 with the assumption that there is a 90 degree angle between L and W.
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Let's say I attempted to draw a rectangle with dimensions of L x W. When I compared the diagonals, I noticed that they were off by 6 cm. Create an equation so that I can determine how far and where I should move one of the corners to fix my rectangle.
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Four men (A, B, C, and D) are kept in a room a punishment for their crimes. Each day, they retreat to a corner where they are fed an amount of a certain type of food. In total, they receive thirteen pieces of the assigned food. On this day, they all retreated to their corners to receive beef jerky. After enjoying their food, they came back to the center. In front of the group, A asked B, "Did you get more than me?" To which B replied, No. Curious then, B turned to C and said, "Did you get more than me?" To which C replied, No. Even more curious, C turned to D and said, "Did you get more than me?" To which D replied, "Yes and I know how much each of you got!"
How much food did B and C get?
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On 8/2/2017 at 5:46 AM, gavinksong said:
Haha, this is wonderful. BMAD, you must mean 0.7 waist-to-hip ratio and not the other way around.
Which ever one gives me a bigger butt
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On 7/21/2017 at 9:59 AM, sushma said:
int no_of_happy_bug(int gen) {
int happy = 1;
int sad = 0;
int neutral = 0;
int generation = 1;while (generation <= gen) {
int h = 0;
int s = 0;
int n = 0;
if (happy >= 1) {
s = s + happy;
n = n + happy;
}if (sad >= 1) {
h = h + (2 * sad);}
if (neutral >= 1) {
h = h + neutral;
s = s + neutral;
}happy = h;
sad = s;
neutral = n;
generation++;
}
return happy;
}by explicit, I was referring to a function that for any mth generation you can generate the h, s, and n without the need to recursively derive the list.
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The second half of the solution seems to be inefficient though. Once we stop at hole 4 and conclude that the fox is in fact at an even hole the only hole left is hole number 2 which means that we don't need to start all over. Simply inspecting three and then two would be enough to find the fox.
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not without assuming something about the population.
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For a particular bug population, a bug is either classified as happy, sad, or neutral.
For each generation,
all happy bugs birth 1 sad and 1 neutral bug.
all sad bugs birth 2 happy bugs
all neutral bugs birth one happy bug and one sad bug
all birthing bugs die after birthing. No bug lives more than one generation and birth at the same time. These bugs do not require mating to reproduce.
The initial generation only consists of one happy bug.
Write an explicit function to determine for the nth generation how many happy bugs there are.
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Player A has one more coin than player B. Both players throw all of their coins simultaneously and observe the number that come up heads. Assuming all the coins are fair, what is the probability that A obtains more heads than B?
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1
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On 1/21/2017 at 11:05 AM, bonanova said:
This question appears similar to the one that asks for a proof that 0.9999... = 1 which is paradoxical until we establish that the notation "..." stands for infinite repetition. That is, the equality holds only if there is no "last" nine: they go on forever. But then the expression given in the OP has no clear meaning. It implicitly claims that after an infinite number of zeros there can be (a) five more zeros and then (b) a final unity. There can be neither. The infinity represented by "..." has the lowest cardinality of all the infinities, but it is infinity nonetheless. Nothing follows it.
If we change the meaning of "..." we might say that 0.000...000001 reads: limit (n->inf) [0.000 followed by n zeros followed by five zeros followed by 1]. In that case, the expression evaluates to zero and the proof is simple. The expression is positive, decreasing and bounded below by zero. For any positive epsilon, no matter how small, there is an n for which the expression is (positive and) smaller than epsilon.
One of my students used this as their proof, I am curious about what you think of it bonanova,
Assume that 0.000....1 exists then 1 - 0.000....1 = 0.999....
Since we can show 0.99999 = 1 and 0.000...1 + 0.999... = 1 then 0.000...1 must equal zero as 0+1=1.
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Prove or disprove that 0.000....000001 equals 0
I bake
in New Logic/Math Puzzles
Posted · Edited by BMAD
Yes it follows this pattern:
1+5x7x11xN
Where N=0,1,2,3,4,5,...
This will ensure the remainder is always one.