We love math

arzafar

Junior Member
okay this is a fun thread for the wise guys (and gals) on tti.

Rules:
1. Questions must be related to basic math, simple logic or problem solving.
2. Only one problem may be posted at a time.
3. anyone can answer any number of times.
4. the questioner will then post the correct answer and its explanation!!!
5. the person who first posted the correct answer will be allowed to post the next question.
6. No google!
 

arzafar

Junior Member
Let start with a simple problem.

Count the number of separate squares in the figures give below

1.
4x4.bmp


2.
8x8.bmp
 

thariq2005

Praise be to Allah!
Salaamu 'alaikkum...

Ermmm first one is 21
second one is 85

Ahem if I get it wrong, I am going to pretend as although I havent come across this thread *smile*
 

arzafar

Junior Member
nice try bros.

Answers
1) 4^2 + 3^2 + 2^2 + 1^2 = 30
2) 8^2 + 7^2 + 6^2 + ....... + 1^2 = 204

explanation
no of small square tiles along each side, no of squares
4, 1^2=1
3, 2^2=4
2, 3^2=9
1, 4^2=16

ok ayman you give us another problem.
 

dimouf

New Member
Enigma ;)

Essala Alaykoum;
My enigma consists to find the next numbers of those following sequents:

I) 1, 3, 7, 15, 31, 63, ?
II) 6, 24, 60, 120, 210, 336, ?
III) 4, 9, 61, 52, 63, 94, ?

good luck
 

arzafar

Junior Member
46 :)

The pattern is to juxtapose the digits of the square of the next number in the series,

2x2 = 4 one number so unaffected
3x3 = 9 one number so unaffected
4x4 = 16, juxtaposed to 61
5x5 = 25, juxtaposed to 52
6x6 = 36, juxtaposed to 63
7x7 = 49, juxtaposed to 94,
8x8 = 64, juxtaposed to 46.

If I'm right, then the first number in the series should've been 1.

nice!
let's wait for dimouf to confirm the correct answer.
 

dimouf

New Member
Confirmation

nice!
let's wait for dimouf to confirm the correct answer.


I) ok
II) ok
III) ok

congratulation

the 3th is not obvious..the sequence starts with 2²=4 instead of 1²=1 ...in order to complicate it!!! ;)
for the next time ..i'll think to start at 9! :SMILY335:
 

dimouf

New Member
an other one

essalam alaykoum wa Rahmatou llah wa barakatouh!

for a,b and c three positive real numbers,we have to show this following inequality:



(a+b)(b+c)(c+a) >= 8abc (">=": superior or equal)

:SMILY335::SMILY335::SMILY335::SMILY335::SMILY335:
 

arzafar

Junior Member
(a+b)(b+c)(c+a)
= (ab + ac + bb + bc) (c+a)
= abc + acc + bbc + bcc + aab + aac + abb + abc
= 2abc + a(bb + cc) + b(aa + cc) + c(aa + bb)
= 2abc + 6abc - 6abc + a(bb + cc) + b(aa + cc) + c(aa + bbc)
= 8abc + a(bb - 2bc + cc) + b(aa - 2ac + cc) + c(aa - 2ab + bb)
= 8abc + [a(b-c)^2 + b(a-c)^2 + c(a-b)^2]

therefore (a+b)(b+c)(c+a) = 8abc + X
where X = [a(b-c)^2 + b(a-c)^2 + c(a-b)^2] >= 0

since 8abc + X >= 8abc,
(a+b)(b+c)(c+a) >= 8abc
 

dimouf

New Member
:)

(a+b)(b+c)(c+a)
= (ab + ac + bb + bc) (c+a)
= abc + acc + bbc + bcc + aab + aac + abb + abc
= 2abc + a(bb + cc) + b(aa + cc) + c(aa + bb)
= 2abc + 6abc - 6abc + a(bb + cc) + b(aa + cc) + c(aa + bbc)
= 8abc + a(bb - 2bc + cc) + b(aa - 2ac + cc) + c(aa - 2ab + bb)
= 8abc + [a(b-c)^2 + b(a-c)^2 + c(a-b)^2]

therefore (a+b)(b+c)(c+a) = 8abc + X
where X = [a(b-c)^2 + b(a-c)^2 + c(a-b)^2] >= 0

since 8abc + X >= 8abc,
(a+b)(b+c)(c+a) >= 8abc

A very nice Job!! the next level is level 1 :SMILY335::SMILY335::SMILY335:
 

arzafar

Junior Member
by definition
n! = n(n-1)(n-2)....1
=> (n-1)! = (n-1)(n-2)....1
=> (n-1)! = n!/n
substitute n=1, above
=> 0! = 1!/1
=> 0! = 1
 

arzafar

Junior Member
2 runners race toward each other, starting at 10 miles apart. The faster one runs at 12 mph, the other at 8 mph. At the same time, a bird - fascinated by the whole procedure - flashes back and forth between them until they meet. Assuming the bird flies at an average of 22mph, how far does it travel in all?
 
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