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General Puzzles, Problems & Riddles! (View 1st Post BEFORE You Answer or Ask!)

Michael HuangMichael Huang Posts: 4,114 mod
edited January 2017 in Forum Games
Note: Archive thread will come soon enough if the number of questions reaches to some high number! It will come soon!

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u2e2ncp3lsxx.png
ex9ogv3i3xib.png
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Problems Wanted
Solved or Expired

To be determined. Check out the problems later!
Solutions to Previous, Tricky Problems
I only update solutions for problems that are not easily solvable by certain answerers.
Statistics
Solved by Members
@_raoul - 6
@M V Satya Mohan - 10
@omgsun - 1
Number of problems expired - 18

Bonus Problems Solved
1 problem (6B) solved by @_raoul.
Post edited by Michael Huang on
«13456710

Comments

  • Michael HuangMichael Huang Posts: 4,114 mod
    ** New Problems Updated **

    I believe that since there aren't many solvers out there, I believe some easy problems should be the great start!
  • I'm not providing proof so that others could have a go. Answers:

    1. 11
    2. 1:2
    3. 63
  • omgsunomgsun Posts: 2,239 Pool Champion
    edited October 2016
    on number 3, just use wut u have to find congruent squares(by shared sides), sum side lengths, then just crunch some numbers

    2. ratio of radii is 1:3, scaling 1d to 2d, we have one small semicircle = 1/9 of bigger, so 1:3
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    I'm not providing proof so that others could have a go. Answers:

    1. 11
    2. 1:2
    3. 63

    First answer - Incorrect. Seems like you forget to realize the extra 1's. ;)
    Second - Correct, even though you didn't show proof.
    Third - Ditto.
    omgsun wrote: »
    on number 3, just use wut u have to find congruent squares(by shared sides), sum side lengths, then just crunch some numbers

    The problem is simply solved by looking at the lengths shown in the diagram. There is no need to crunch in numbers if you can start with enough information.

    Firstly, we have 3-by-3, 4-by-4 and 5-by-5 squares. Then, (3 + 4) * (4 + 5) = 63.

    ** New Problem **
    mabqwmhmxeuf.jpg

    If the area of the shaded region equals wriovg6r0kfx.gif, determine the value of...
    0n0ar4af87og.gif

    Note: Assume that the white semicircles inside the large circle have the radius of 1.

    Addendum: You don't need to solve ALL problems. Select one and work it out! Let me know which one you are working on!
  • omgsunomgsun Posts: 2,239 Pool Champion
    edited October 2016
    i thought the semicircles didn't overlap before...

    x in radians? pi as 180 degrees
  • M V Satya MohanM V Satya Mohan Posts: 52
    edited October 2016
    Problem 1:

    Somehow I missed the fact that k is starting from 0 in the numerator, but included it in the denominator. So this will result in one extra 1 in numerator... So the answer is 111.

    Proof:
    From Wilson's remainder theorem, we have
    (p-1)! mod p = p-1
    Extending, we get
    (p-2)! mod p = 1, where p is a prime number.

    After expanding denominator, we have 1111.... (1008 + 1 times)
    note that 1009 is a prime number, so we can apply above theorem.
    And in the numerator we have 1111.... (1007! + 1 + 1), which can be written as ((1009-2)! + 1 + 1).
    So (1009-2)! mod 1009 = 1. and using remaining two 1's, we have 111 as remainder.

    Problem 2:

    Proof:
    9icxqj8ofd4l.png

    In the figure, R = 3*r
    So,
    red area = 3* area of small semi circle = 3* (pi*r^2)/2
    now blue area = area of bigger semi circle - red area = (pi*R^2)/2 - 3*(pi*r^2)/2
    substituting R = 3*r, we get
    blue area = (pi*9*r^2)/2 - 3*(pi*r^2)/2
    = (3*pi*r^2)/2*(3-1)
    = 3*pi*r^2
    red area : blue area = 3* (pi*r^2)/2 : 3*pi*r^2

    Solving, we get, red area : blue area = 1:2

    Problem 3:
    Proof: Somewhat similar method as OP's.

    Problem 4:

    Answer = 6

    Proof:
    vmdljbx6nshf.jpg
    Problem Image

    det2z04ztbwl.png
    My construction (s1 and s2 are semi circles, t1 and t2 are triangles)

    Area of shaded region = area of bigger semi circle - area of unshaded region ........(eqn 1)
    Area of unshaded region = area of s1 + area of s2 + area of green region ................(eqn 2)
    Area of green region = area of sector - area of two segments ......................................(eqn 3)

    Now,
    area of segment = area of sector - area of t1

    note that t1 is equilateral, therefore

    area of segment = 60/360* pi*r^2 - squareroot(3)/4*side^2

    here r = side = 1, hence
    area of segment = pi/6 - squareroot(3)/4

    using this in eqn 3, we get

    area of green region = 60/360*pi*r^2 - 2*(60/360* pi*r^2 - squareroot(3)/4*side^2)
    = - pi/6 + squareroot(3)/2

    using this in eqn 2, we get

    area of unshaded region = pi*r^2/2 + pi*r^2/2 + (-pi/6 + squareroot(3)/2)
    = 5*pi/6 + squareroot(3)/2

    using this in eqn 1, we get

    Area of shaded region = pi*R^2/2 - 5*pi/6 + squareroot(3)/2, here R = 2, so substituting
    = 2 * pi - 5*pi/6 + squareroot(3)/2
    = 7*pi/6 - squareroot(3)/2 ..........................................................(eqn 4)

    Acc to question, area of shaded region = 7x - cos(x) ...........................................................(eqn 5)

    comparing eqn 4 and eqn 5, we get

    x = pi/6, therefore
    pi/x = 6
  • omgsunomgsun Posts: 2,239 Pool Champion
    well i guess i didn't get there first... oh well
  • Michael HuangMichael Huang Posts: 4,114 mod
    Perfect, @M V Satya Mohan ! You have worked out the problem easily!
    omgsun wrote: »
    i thought the semicircles didn't overlap before...

    x in radians? pi as 180 degrees

    In fact, they do! I posed the question with the note that the overlapped shapes are semicircles with radius of 1 unit.

    O.K. Seems like you like judging problems. ^.^
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    ** New Problems **
    Decided to find bit different problems instead of relying on geometry! Let's see if anyone can work out the following puzzle or math problem. ^.^

    Problem No. 5 - 57th IMO in Hong Kong
    Time Limit: 1 week
    yvn0dvry1p41.gif
    The above depicts the cryptogram, the mathematical puzzle representing digits that makes the system true. Few conditions to consider are:
    1. Each letter must represent the different integer.
    2. Each leading digit is non-zero.
    Find the total number of possible solutions to this cryptogram.

    Problem No. 6 - 2016 Exponents
    Time Limit: 2 weeks
    xqsbjt9agwqa.gif
    What is the remainder when the above is divided by 2016?
  • omgsunomgsun Posts: 2,239 Pool Champion
    edited October 2016
    jesus im not a superhero LOL, although my friend's older sister made MOSP in 10th grade last year, got early admission to Harvard but turned down for an extra year of high school
  • *Aurora**Aurora* Sorry, I'm a lady...Posts: 6,864 Pool Forum VIP
    omgsun wrote: »
    jesus im not a superhero LOL, although my friend's older sister made MOSP in 10th grade last year, got early admission to Harvard but turned down for an extra year of high school

    Too bad it's not contagious...
  • omgsunomgsun Posts: 2,239 Pool Champion
    edited October 2016
    lmao

    her friend graduated from hs last year from 11th grade and went to harvard LOL
  • Michael HuangMichael Huang Posts: 4,114 mod
    omgsun wrote: »
    lmao

    her friend graduated from hs last year from 11th grade and went to harvard LOL

    Harvard? Wow! I really want to go for, but I need lots of excellent admission factors.

    Stanford is another great university - I heard.
  • M V Satya MohanM V Satya Mohan Posts: 52
    edited October 2016
    Problem 5:

    I could find 30 solutions. If it is correct, I'll share the proof.
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    Not even close! Try again! Actually, there is a mathematical + logical approach for the Hong Kong cryptogram problem. If anyone gets the interesting problem without any assistance, that would even be great!
  • _raoul_raoul Posts: 280 Pool Pro
    Problem 6 is 48.

    https://www.wolframalpha.com/input/?i=(+sum_(i%3D1)%5E2016+i%5E2016)+mod+2016
  • Paulb228Paulb228 Posts: 5,872 Pool Forum VIP
    omgsun wrote: »
    lmao

    her friend graduated from hs last year from 11th grade and went to harvard LOL

    Seems like Harvard is an easy place to get in ...try Cambridge if you want a real academic seat of learning
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    _raoul wrote: »
    Problem 6 is 48.

    https://www.wolframalpha.com/input/?i=(+sum_(i%3D1)%5E2016+i%5E2016)+mod+2016

    :tongue:

    Correct! Next time, no computational devices UNLESS needed!

    BONUS (3 days left!): Can you actually find the answer WITHOUT using computational devices and websites? What is the remainder when you divide the following by 2016?

    qwrcaukyzeli.gif

  • M V Satya MohanM V Satya Mohan Posts: 52
    edited October 2016
    Problem 5:

    I could find 30 solutions. If it is correct, I'll share the proof.

    I found that I included solutions which violated rule no. 2 and I missed some other solutions. So I correct my answer to 50. I did try to apply some logic into it which is as follows:
    1. Divided the problem in columns as below:yr6hgyqie4mp.png

    2. Noted that, the carry from C3 to C2 would never exceed "2". So if carry is "2", to satisfy C2, I must be "8". If the carry is "1", I must be "9", which is not possible , because then the carry would exceed "1". And a carry of "0" is not possible. So we fix I = 8.

    3. Now in C3, any value N, which satisfies the conditions, would lead to the eqn:
    M + I = 2 or (M + I = 12(when carry from C4 to C3 is 2) and M + I = 11(when carry from C4 to C3 is 3).
    M + I = 2 is not possible as I = 8, So we fix M + I = 12, resulting in M = 4 or M+I =11, resulting in M = 3.

    4. Now in C4, any value of G, which satisfies the conditions. would lead to eqn:
    O + N = 7 or O + N = 17.
    Now if O + N = 17, then O = 8, N = 9 or O = 9, N = 8, which are not possible as we fixed I = 8.
    So O + N = 7. This leads to a carry of "2" from C4 to C3, therefore we revisit point 3 and conclude that M can't be 3, Therefore we fix M = 4.

    5. Now since O + N = 7, O and N can take following values:
    O = 0, N = 7
    O = 1, N = 6
    O = 2, N = 5

    O = 5, N = 2
    O = 6, N = 1
    O = 7, N = 0

    O and N can niether be 3 or 4, as M = 4.

    6. Now G can take any value from 0-9 except 4 and 8.

    7. Now in C1, carry would always be "1". So 3 + H = K, where H != 0.

    Now what I did was for every value of G, iterate 3 times, for values of O and N (bolded out in point no. 5). Now since G, I, M, N and O are fixed H and K can be checked to satisfy the eqn in point no. 7.

    Now after all the values of G are covered, double the number of soultions obtained to compensate for the remaining 3 values of O and N (not bolded in point no. 5).

    Still my method is tedious, because one has to write down solutions in order to count it. It'd be great to know a way which can direclty give the no. of possible solutions.
  • Michael HuangMichael Huang Posts: 4,114 mod
    Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.
«13456710
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