Hard Riddles and Math exercices

Ra8

I found some nice and challenging riddles and math exercises i'd like to share with you.
Don't answer if you already know the riddle/exercise and don't Google for the answer and post it here
1. Difficulty : 1/5

How many rectangles in a Sudoku grid? (Note that a square is also a rectangle)

2. Difficulty : 1/5

What's so special about this sequence?

8-5-4-9-1-7-6-10-3-2-0

3. Difficulty: 5/5

Note: the figure is not drawn to scale
ABC is isosceles on A and angle BAC = 20 degrees
D is on [AC] and angle CBD = 60 degrees
E is on [AB] and angle BCE = 50 degrees
Calculate with proofs angle BDE
->Don't draw the figure on the scale and measure the angle, this need demonstrations

In 4 and 5 you have to add as much as you want operations to get the result. You may arrange the numbers the way you want. You may not make 1 and 2 a 12 (twelve). You have to use all the numbers each one once.
(example 2 ,2 and 2 to make 6 => 2+2+2=6)

4. Difficulty 3/5

1, 3, 4, 6 to make 24 ; Operations allowed: + - * / ( )

5. Difficulty 2/5

Operations allowed: + - * / and all you can find on a scientific calculator.

1, 1, 1 to make 6
4, 4, 4 to make 6
8, 8, 8 to make 6
9, 9, 9 to make 6

MSD

1.
(1+2+....+n)^2

where n is the maximum number n a sudoku square, for example in a 9 x 9 sudoku square, the number of triangles would be (1+2+3+4+5+6+7+8+9)^2 => (n*(n+1)/2)^2 where n = 9 => ((9 * 10) / 2) ^ 2 => 2025 triangles

MSD

5.
(1+1+1)! = 6
(sqrt(4)+sqrt(4)+sqrt(4)) = 6
(qbrt(8)+qbrt(8)+qbrt(8)) = 6
sqrt(9*9) - sqrt(9) = 6

mahmoud-brk

2.
They are arranged alphabetically.

Eight Five Four Nine One Seven Six etc.

Ra8

Your answers are correct! here are some new:

6. Difficulty 1/5

7. Difficulty 1/5
Add one slash to complete this equation (example: 4 + 5 = 19 ,we add a slash before the 4 to be represented as 1 and the solution is 14 + 5 = 19)

5 + 5 + 5 =550

8. Difficulty 2/5
Complete this sequence: 1 - 7 - 28 - 84 - 210 - 462 - ???

mesa177

3. Solution:

ABC iscoceles at A => angle ABC = angle ACB = (180 - angle BAC)/2 = (180 - 20)/2 = 80
[sum of angles in triangle is 180]

angle ABD = angle ABC - angle DBC = 80 - 60 = 20 [adjacent angles]

angle BDC = 180 - (angle DBC + angle DCB) = 180 - (60 + 80) = 40
[sum of angles in triangle is 180]

=> angle BDA = 180 - angle BDC = 180 - 40 = 140 [supplementary angles]

angle BEC = 180 - (angle EBC + angle ECB) = 180 - (80 + 50) = 50

=> angle CEA = 180 - angle EBC = 180 - 50 = 130 [supplementary angles]
angle ECA = angle BCA - angle BCE = 80 - 50 = 30 [adjacent angles]

angle BDE = angle BDA - angle EDA = 140 - angle EDA
angle CED = angle CEA - angle DEA = 130 - angle DEA

=> angle BDE + angle CED = 140 - angle EDA + 130 - angle DEA
= 270 - (angle EDA + angle DEA)
= 270 - (180 - angle EAD)
= 90 + 20
= 110
=> angle BDE = 110 - angle CED(eq1)

Let I be intersection of lines (BD) and (CE) =>
angle CED + angle BED = 180 - angle EID = 180 - angle BIC [opposite angles are equal]
but angle BIC = 180 - (angle IBC + angle ICB) = 180 - (60 + 50) = 70

=> angle BED + angle CED = 180 - 70 = 110
=> angle BED = 110 - angle CED(eq2)

from eq1 and eq2 => angle BED = angle BDE = (180 - angle EBD)/2 = (180 - 20)/2 = 80

Ra8

Sorry mesa your answer is wrong... i'm trying to see where your mistake is:

mesa177 wroteLet I be intersection of lines (BD) and (CE) =>
angle CED + angle BED = 180 - angle EID = 180 - angle BIC [opposite angles are equal]

I think you meant: angle CED + angle BDE = 180- EID and this will lead you again to the same equation (eq1)

mesa177

Ra8 wroteSorry mesa your answer is wrong... i'm trying to see where your mistake is:

mesa177 wroteLet I be intersection of lines (BD) and (CE) =>
angle CED + angle BED = 180 - angle EID = 180 - angle BIC [opposite angles are equal]
I think you meant: angle CED + angle BDE = 180- EID and this will lead you again to the same equation (eq1)

Damn, yeah your right, trying again....

arithma

6. Pi/2

mahmoud-brk

8.

970.2

Please tell me I'm right.

EDIT: It's seems I'm wrong. I've made a mistake in the begining, but I'll give it a try later.

m0ei

arithma wrote6. Pi/2

I think it's pi/4 not /2.

m0ei

8. 924

mesa177

6. pi/4

still working on 3...

mahmoud-brk

m0ei, I was working on the following series 7 - 4 - 3 - 2.5 - 2.2, is it the same way you solved? we still don't know the right answer, but I liked to know how you tried to solve it.

mesa177

7. 5 + 5 + 5 ≠ 550

m0ei

@mesa; 3) is quite hard, you can't solve it by what you have, you need to add some points. I ain't going to spoil this because i already read about this same geometry problem, you can solve it with 8 different solutions... I would never knew the answer if i tried solving it, good luck.

Ra8

6. is correct pi/4
8. is 924 correct

mesa177 wrote7. 5 + 5 + 5 ≠ 550

not that :P ... think again

You still have 3, 4 and 7
I will add more later

MSD

I doubt that 4. is solvable with the given (unless there is some trick some where)

hj88

4. 6 /( 1 - (3/4))

for problem 3 i have spent 2 hours and i always go back to the same equation :s it's frustrating

Ra8

MSD wroteI doubt that 4. is solvable with the given (unless there is some trick some where)

I can assure you it can be solved with the given (only + - * / and parentheses)
My small brother solved it (he's 14)

Edit: hj88 gave the answer