Home › Off Topic › Forum Games

Please keep our forum rules in mind: we aspire to provide a safe environment for our users, so will not tolerate discriminatory, hateful, inflammatory or threatening posts. Thank you. http://mcgam.es/forum-rules

Please note that registration for the forum is age-restricted.

Michael Huang
Posts: **4,114** mod

Note: Archive thread will come soon enough if the number of questions reaches to some high number! It will come soon!

*For full view, click on the image portions. If you can't click on it, please drag the image next to the last browser tab or outside of the window for full view.*

*To be determined. Check out the problems later!*

__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.

Problems Wanted

Solved or Expired

I only update solutions for problems that are not easily solvable by certain answerers.Solutions to Previous, Tricky Problems

- Problem 5 - 57th IMO in Hong Kong
- Problem 12 - Rolling Around the Circle
- Problem 15 - Math in 8 Ball Pool III: Αυτά μου φαίνονται κινέζικα/It's So Greek to Me!
- Problem 17 - Cutting the Box
- Problem 19 - Math in 8 Ball Pool IV: Rolling Around Triquetra
- Problem 20b - Math in 8 Ball Pool VI: Remove the Whites!
- Problem 20c - Math in 8 Ball Pool VII: Wicked Witch of Odds
- Problem 27d - Trigonometry Chef
- Problem 27f - 1621 'N 2016 'N 2017
- Problem 27g - Pumpkin π's!

Statistics

@_raoul - 6

@M V Satya Mohan - 10

@omgsun - 1

Number of problems expired - 18

1 problem (6B) solved by @_raoul.

Post edited by Michael Huang on

Home › Off Topic › Forum Games

## Comments

4,114modI believe that since there aren't many solvers out there, I believe some easy problems should be the great start!

521. 11

2. 1:2

3. 63

2,239Pool Champion2. ratio of radii is 1:3, scaling 1d to 2d, we have one small semicircle = 1/9 of bigger, so 1:3

4,114modFirst answer- Incorrect. Seems like you forget to realize the extra 1's.Second- Correct, even though you didn't show proof.Third- Ditto.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.

Addendum:You don't need to solve ALL problems. Select one and work it out! Let me know which one you are working on!2,239Pool Championx in radians? pi as 180 degrees

52Problem 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:

(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:

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 = 1:2Problem 3:Proof: Somewhat similar method as OP's.

Problem 4:Answer =

6Proof:

Problem Image

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 = 62,239Pool Champion4,114modIn 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. ^.^

4,114modProblem No. 5 - 57th IMO in Hong KongTime Limit:1 weekProblem No. 6 - 2016 ExponentsTime Limit:2 weeks2,239Pool Champion6,864Pool Forum VIPToo bad it's not contagious...

2,239Pool Championher friend graduated from hs last year from 11th grade and went to harvard LOL

4,114modHarvard? Wow! I really want to go for, but I need lots of excellent admission factors.

Stanford is another great university - I heard.

52Problem 5:I could find

30solutions. If it is correct, I'll share the proof.4,114mod280Pool Prohttps://www.wolframalpha.com/input/?i=(+sum_(i%3D1)%5E2016+i%5E2016)+mod+2016

5,872Pool Forum VIPSeems like Harvard is an easy place to get in ...try Cambridge if you want a real academic seat of learning

4,114modCorrect! 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?52I 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: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 = 4or M+I =11, resulting inM = 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 = 7O = 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.

4,114modYou 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.