pak matematik 
Veshtroni me vemendje te gjitha veprimet e meposhtme. Te gjitha kane nje simetri qe te ben te mendosh.

Ja i pari:

3 shumezuar me 37:

3 x 37 = 111
6 x 37 = 222
9 x 37 = 333
12 x 37 = 444
15 x 37 = 555
18 x 37 = 666
21 x 37 = 777
24 x 37 = 888
27 x 37 = 999

Sa interesant:

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111=12345678987654321


Nje trapez tjeter:

1 x 9 +2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111




Ja dhe nje tjeter:

0 x 9 + 8 = 8
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
987654321 x 9 - 1 = 8888888888
9876543210 x 9 - 2 = 88888888888

I fundit:




1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

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Cilet numra mungojne ne serine e meposhtme?
0, __, 2, __, 9, __, 20, __, 26, __, 54, __, 77, __



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Te gjenden te gjitha treshet e numrave (a,b,c) te tilla qe:
- a=b^2+ c^2
- b= c
- pmp(b,c)=1
- bc/d eshte numur natyror per cdo d te thjeshte ku d= va

Shenime:- pmp ==> pjestuesi me i madh i perbashket
- ^ ==> ne fuqi



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Ne qoftese 0 < a < b < c < d atehere te vertetohet qe:

a/b + b/c + c/d + d/a > b/a + c/b + d/c + a/d