ant0n Posted September 17, 2020 #401 Share Posted September 17, 2020 (edited) 46 minutes ago, ant0n said: Great Express this complex number z = (1 - i.sqrt(3))^11 in the form z = |z|.e^(arg(z)). That's the new maths problem to solve. Please explain how you get to the solution. Edited September 17, 2020 by ant0n Link to comment Share on other sites More sharing options...
Ozymandias Posted September 17, 2020 #402 Share Posted September 17, 2020 It is a bit onerous for one or two people to be providing solutions all the time and I will be busy tomorrow. Also, this type of problem might be too challenging for most and you might reconsider the difficulty. Myself and Dan used keep problems within the capability of people who had finished a general second-level maths programme. I'm going to let this go for a few days to see if anybody else wants to do it before posting a solution. Link to comment Share on other sites More sharing options...
ant0n Posted September 17, 2020 #403 Share Posted September 17, 2020 (edited) 36 minutes ago, Ozymandias said: It is a bit onerous for one or two people to be providing solutions all the time and I will be busy tomorrow. Also, this type of problem might be too challenging for most and you might reconsider the difficulty. Myself and Dan used keep problems within the capability of people who had finished a general second-level maths programme. I'm going to let this go for a few days to see if anybody else wants to do it before posting a solution. My exercise with the complex number is considered rather simple for an average scientific student in A-levels in France (whose age is 18, most of the time, when they get that degree). What age does 'general second-level maths programme' usually correspond to? z = (1 - i.sqrt(3))^11 |1 - i.sqrt(3)| = sqrt(1² + 3) = 2 (1 - i.sqrt(3))/2 = 1/2 - 1/2.sqrt(3) = cos(-pi/3) + i.sin(-pi/3) = e^(-i.pi/3) -> (1 - i.sqrt(3)) = 2.e^(-i.pi/3) z = (2.e^(-i.pi/3))^11 = 2^11.e^(-11i.pi/3) = 2^11.e^(i.pi/3) Please tell me what age usually corresponds to 'general second-level maths programme' and whoever feel free to post a new maths problem. Edited September 17, 2020 by ant0n Link to comment Share on other sites More sharing options...
ant0n Posted September 20, 2020 #404 Share Posted September 20, 2020 (edited) * New Maths Problem To Solve * Find the last two digits of 2^422 and please explain your reasoning here below. Edited September 20, 2020 by ant0n Link to comment Share on other sites More sharing options...
Ozymandias Posted September 23, 2020 #405 Share Posted September 23, 2020 On 9/20/2020 at 7:22 PM, ant0n said: * New Maths Problem To Solve Find the last two digits of 2^422 and please explain your reasoning here below. No takers, Anton, even though this is not that difficult. Link to comment Share on other sites More sharing options...
ant0n Posted September 23, 2020 #406 Share Posted September 23, 2020 52 minutes ago, Ozymandias said: No takers, Anton, even though this is not that difficult. I guess their calculator can't provide the answer they need ^^ Do you wish to display a reasoning that leads to the solution? Link to comment Share on other sites More sharing options...
danydandan Posted September 23, 2020 Author #407 Share Posted September 23, 2020 5 minutes ago, ant0n said: I guess their calculator can't provide the answer they need ^^ Do you wish to display a reasoning that leads to the solution? Sorry I've been meaning to drop a response but I've been stupidly busy in work.... Think a calculator would provide the answer, no? Just applying a simple prime factorisation of the exponent will yield the result. But 422 is a tricky number to do a prime factorisation on. I cannot work it out in my head unfortunately, so I'm out on this one. Link to comment Share on other sites More sharing options...
Ozymandias Posted September 23, 2020 #408 Share Posted September 23, 2020 1 hour ago, ant0n said: I guess their calculator can't provide the answer they need ^^ Do you wish to display a reasoning that leads to the solution? I know the solution (it's 04) but that is the easy bit. Right now I'm tired, it's too late and going to bed. I'll give the solution (which I expect will take some time to write out in a clear and logical manner) tomorrow. The prime factorisation of the exponent 422 is 211 x 2, which gets you nowhere. I can only compute the exact value of 2^X when X is no greater than 49. 2^49 =562,949,953,421,312. For powers higher than that the answer is an approximation given in scientific notation form. I'll give the solution tomorrow. Link to comment Share on other sites More sharing options...
Golden Duck Posted September 23, 2020 #409 Share Posted September 23, 2020 (edited) 20 minutes ago, Ozymandias said: I know the solution (it's 04) but that is the easy bit. Right now I'm tired, it's too late and going to bed. I'll give the solution (which I expect will take some time to write out in a clear and logical manner) tomorrow. The prime factorisation of the exponent 422 is 211 x 2, which gets you nowhere. I can only compute the exact value of 2^X when X is no greater than 49. 2^49 =562,949,953,421,312. For powers higher than that the answer is an approximation given in scientific notation form. I'll give the solution tomorrow. Last digit of a power if 2 is one of 2, 4, 8, or 6. 422 mod 4 = 2. So, the last digit is 4. (10x + 6) × 6 will always end in 6. Also: 2^4 = 16. (10x + 6) × 4 will always equal 4 mod 10. I can observe a pattern in the 10^1 digit but I was trying to find a paper method of proving it. This a luck method. 2^20 = 76 mod 100. (100x + 76) × 76 = 76 mod 100 420 = 20 × 21. 4×76 = 304 = 04 mod 100 There must be a better (more elegant) number theory proof. Edited September 23, 2020 by Golden Duck 1 Link to comment Share on other sites More sharing options...
danydandan Posted September 24, 2020 Author #410 Share Posted September 24, 2020 (edited) 7 hours ago, Golden Duck said: Last digit of a power if 2 is one of 2, 4, 8, or 6. 422 mod 4 = 2. So, the last digit is 4. (10x + 6) × 6 will always end in 6. Also: 2^4 = 16. (10x + 6) × 4 will always equal 4 mod 10. I can observe a pattern in the 10^1 digit but I was trying to find a paper method of proving it. This a luck method. 2^20 = 76 mod 100. (100x + 76) × 76 = 76 mod 100 420 = 20 × 21. 4×76 = 304 = 04 mod 100 There must be a better (more elegant) number theory proof. In my head I had. ((2^2)^211). Gets you the result but I can't do it in my head. The last two digits are 92 I think but that's a guess. Edit: I think the last two digits are 00 now that I think about it. Edited September 24, 2020 by danydandan 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 24, 2020 #411 Share Posted September 24, 2020 (edited) As I stated last night, the last two digits of the result when you calculate 2^422 is 04; that is, the result has zero 10s and 4 units. I approached this as a geometric sequence in which four-hundred-and-forty-two 2s multiply each other sequentially and, each time, a new term in the sequence is generated. Obviously, these terms cannot have an odd number in their unit place-holder, so no 1, 3, 5, 7 or 9. If you are repeatedly multiplying by 2 then 0 in the unit place-holder is equally impossible because to get it you would need to multiply 5 by 2 and 5 is an odd number. The only possible numbers allowable in the unit-place are 2, 4, 6 and 8. If you generate the aforesaid sequence you get those digits repeating themselves in the unit place in a cyclic pattern 4 - 8 - 6 - 2 ad infinitum. I think of a mechanical counting device with the 'unit' wheel set at 2. Now, as we begin doubling the numbers shown on the counter, the 'unit' wheel will always rotate through the cycle 4 - 8 - 6 - 2. Staying with this analogy, I have written down the first 23 terms of this sequence. Because we are interested in the last two digits of 2^422 I have enlarged and bolded the digits that would appear on the 'unit' and 'tens' wheels of the counter, starting with the first term T_{1}= 2. This would show as a zero on the 'tens' wheel and a two on the 'unit' wheel, etc. T_{1} = 02 T_{2} = 2 x 2 = 04, T_{3} = 4 x 2 = 08, T_{4} = 8 x 2 = 16, T_{5} = 16 x 2 = 32, T_{6} = 32 x 2 = 64, T_{7} = 64 x 2 = 128, T_{8} =128 x 2 = 256, T_{9} = 256 x 2 = 512, T_{10} = 512 x 2 = 1024, T_{11} = 1024 x 2 = 2048, T_{12} = 2048 x 2 = 4096, T_{13} = 4096 x 2 = 8192, T_{14} = 8192 x 2 = 16384, T_{15} = 16384 x 2 = 32768, T_{16} = 32768 x 2 = 65536, T_{17} = 65536 x 2 = 131072, T_{18} = 131072 x 2 = 262144, T_{19} = 262144 x 2 = 524288, T_{20} = 524288 x 2 = 1048576, T_{21} = 1048576 x 2 = 2097152, T_{22} = 2097152 x 2 = 4194304, T_{23} = 4194304 x 2 = 8388608, T_{24} = 8388608 x 2 = 16777216, T_{25} = 16777216 x 2 = 33554432, T_{26} = 33554432 x 2 = 67108864, ............. The appearance of a cyclic pattern on the 'units' wheel begs the question: do the combined 'tens' and 'units' wheel exhibit the same kind of cyclic behaviour? If so, what is it? It turns out that the double digits of the tens and units (bolded) start to repeat themselves at T_{22} where 04 reappears, likewise T_{23} has 08, and so on it goes, .... So beginning with T_{2}, the last two digits of all the terms generated sequentially by calculating 2^422 repeat every 20 terms. To verify just ask a calculator to check that 2^22 ends in 04 and 2^42 ends in 04. They both do. 2^22 = 4194304 (as above) and 2^42 = 4,398,046,511,104. So 2^422 = (2^2)(2^420), also ends in 04 even though I can't right the answer down. Summarising, 2^(20n + 2) always ends in 04 when n = 0, 1, 2, 3, ..... Edited September 24, 2020 by Ozymandias 3 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 25, 2020 #412 Share Posted September 25, 2020 (edited) Problem: Show that if 7 marbles are randomly scattered inside a regular hexagon of side 1 metre, then the centre of at least two of them must lie within 1 metre of each other. Edited September 25, 2020 by Ozymandias Link to comment Share on other sites More sharing options...
Ozymandias Posted September 25, 2020 #413 Share Posted September 25, 2020 (edited) 10 hours ago, Ozymandias said: Problem: Show that if 7 marbles are randomly scattered inside a regular hexagon of side 1 metre, then the centre of at least two of them must lie within 1 metre of each other. Please be advised that in my rush to provide the above new teaser I did not think carefully enough about the problem that I posted and have not worded or framed it correctly. Apologies for that. These problems are created by myself and in this case it showed. The new problem would have been better presented as follows: Problem: Show that if 8 points are randomly selected on or inside a regular hexagon of side 1 metre, then at least two of these points must lie within 1 metre of each other. Edited September 25, 2020 by Ozymandias 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 26, 2020 #414 Share Posted September 26, 2020 (edited) Here's a regular hexagon which side is 1 metre. I can randomly get these 8 red dots ( = there's a non-null probability to get them this way). However none of them is distant of at least one metre from the other dots. What do you exactly mean by "randomly"? Edited September 26, 2020 by ant0n Link to comment Share on other sites More sharing options...
Harte Posted September 26, 2020 #415 Share Posted September 26, 2020 Not rigorous. Given: The radii of every regular hexagon are congruent to any side. Therefore the maximum distance from the center in this case is one meter - if one point is at the center and the other points are on the hexagon's vertices. Therefore seven points can be arranged in such a way that they are all one meter from the nearest points. 1. Any point not on a side of the hexagon and not at the center is less than one meter from the center. 2. Any point not on the side of the hexagon and not at the center is less than one meter from the nearest vertex. 3. A point at the center is one meter from every vertex. 4. Points on the vertices are all 1 meter from the two nearest other points on the hexagon's vertices (they are endpoints of consecutive congruent segments.). 5 There are six vertices, which leaves two "extra" points. 6. If these two points are on any sides of the figure, then they are within one meter of the two points at the nearest two vertices.(#4 above) 7. One point may be placed at the center, which is at the maximum separation from the points at the vertices (one meter.) (#3 above) 8. If no point is at the center, then both "extra" points are within one meter of the nearest vertex.(#2 above) 9. If a point is placed at the center, then the second extra point must be placed on the interior or on a side, but not at the center. This second "extra" point is less than one meter from the point at the nearest vertex AND the point at the center (if any.) (#2 and #3 above) Harte Link to comment Share on other sites More sharing options...
Harte Posted September 26, 2020 #416 Share Posted September 26, 2020 Two circles of different radii are concentric. A tangent drawn to the inner circle creates a chord of length 20 in the outer circle. Find the area of the region outside the smaller circle and inside the larger circle. Harte Link to comment Share on other sites More sharing options...
Ozymandias Posted September 26, 2020 #417 Share Posted September 26, 2020 (edited) @ant0n - randomly means chosen without deliberation where each selection is as likely as any other. The number of random selections of 8 points is infinite and your selection has all 8 points less than 1 metre apart. But the problem asks you to show that for all those infinite selections possible there will always be at least one pair of points that will be less than 1 metre apart. @Harte you are correct. Like you, this is how I visualised the proof that there must always be two points closer to each other than 1 metre. My hexagon is not accurately drawn but you get the idea. A regular hexagon of side 1m comprises six equilateral triangle of side 1m. Therefore six of the eight points can be arranged on the hexagonal vertices and a seventh at the centre. This leaves the eight [red] point which must be in a position that leaves it less than 1m from one of the other seven points. Edited September 26, 2020 by Ozymandias 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted September 27, 2020 #418 Share Posted September 27, 2020 On 9/26/2020 at 3:55 PM, Harte said: Two circles of different radii are concentric. A tangent drawn to the inner circle creates a chord of length 20 in the outer circle. Find the area of the region outside the smaller circle and inside the larger circle. Harte I am surprised there are no takers on this? It is a classic annulus area problem. If ever a solution to a puzzle involving algebra and Pythagoras could be classified as a party trick, this is it!! Link to comment Share on other sites More sharing options...
Harte Posted September 28, 2020 #419 Share Posted September 28, 2020 It's actually very easy. I hesitated to post it. It would be done by now if I had posted a figure of it I bet. Harte Link to comment Share on other sites More sharing options...
ant0n Posted September 28, 2020 #420 Share Posted September 28, 2020 (edited) R = radius of the big circle r = radius of the small circle Pythagore: R² = r² + (20 / 2)² = r² + 100 therefore R² - r² = 100 Area between the small disk and the big circle = pi.R² - pi.r² = pi.(R² - r²) = 100.pi square units. Edited September 28, 2020 by ant0n Link to comment Share on other sites More sharing options...
ant0n Posted September 28, 2020 #421 Share Posted September 28, 2020 (edited) Your comments leave me speechless by the way. I read about that problem to solve a few hours ago. I should have prevented myself from playing and finding that solution, just to make you feel even more convinced no reader was able to provide the solution. Edited September 28, 2020 by ant0n Link to comment Share on other sites More sharing options...
Harte Posted September 28, 2020 #422 Share Posted September 28, 2020 I wasn't convinced no reader would supply an answer. I said I was surprised. Though, admittedly, there's probably not much traffic in this thread. You can see why a diagram would have resulted in a quick solution. Harte 1 Link to comment Share on other sites More sharing options...
ant0n Posted September 29, 2020 #423 Share Posted September 29, 2020 Anyone feel free to publish another maths problem to solve. The show must go on ♫♪ Link to comment Share on other sites More sharing options...
Harte Posted September 29, 2020 #424 Share Posted September 29, 2020 I believe the rule is he who solved the last one must post the next one. Harte 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted October 1, 2020 #425 Share Posted October 1, 2020 Since nobody is bothering to upload another teaser and I have one of my own to hand .............. Problem: a triangle ABC has sides 7m, 8m and 9m in length. Its area is 26.833m^2. What is the radius of its incircle? Link to comment Share on other sites More sharing options...
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