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#1651 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-10 08:41:18
For k+55 problem.
I can give you smaller set than yours. Let P(x) is the product of all prime numbers less than x. Then all the numbers of the set
P(1001)+2
P(1001)+3
...
P(1001)+1000
are non-prime.
Thus the first number of your set is 999!+2 =
4023872600770937735437024339230039857
1937486421071463254379991042993851239
8629020592044208486969404800479988610
1971960586316668729948085589013238296
6994459099742450408707375991882362772
7188732519779505950995276120874975462
4970436014182780946464962910563938874
3788648733711918104582578364784997701
2476632889835955735432513185323958463
0755574091142624174743493475534286465
7661166779739666882029120737914385371
9588249808126867838374559731746136085
3795345242215865932019280908782973084
3139284440328123155861103697680135730
4216168747609675871348312025478589320
7671691324484262361314125087802080002
6168315102734182797770478463586817016
4365024153691398281264810213092761244
8963599287051149649754199093422215668
3257208082133318611681155361583654698
4046708975602900950537616475847728421
8896796462449451607653534081989013854
4248798495995331910172335555660213945
0399736280750137837615307127761926849
0343526252000158885351473316117021039
6817592151090778801939317811419454525
7223865541461062892187960223838971476
0885062768629671466746975629112340824
3920816015378088989396451826324367161
6762179168909779911903754031274622289
9880051954444142820121873617459926429
5658174662830295557029902432415318161
7210465832036786906117260158783520751
5162842255402651704833042261439742869
3306169089796848259012545832716822645
8066526769958652682272807075781391858
1788896522081643483448259932660433676
6017699961283186078838615027946595513
1156552036093988180612138558600301435
6945272242063446317974605946825731037
9008402443243846565724501440282188525
2470935190620929023136493273497565513
9587205596542287497740114133469627154
2284586237738753823048386568897646192
7383814900140767310446640259899490222
2217659043399018860185665264850617997
0235619389701786004081188972991831102
1171229845901641921068884387121855646
1249607987229085192968193723886426148
3965738229112312502418664935314397013
7428531926649875337218940694281434118
5201580141233448280150513996942901534
8307764456909907315243327828826986460
2789864321139083506217095002597389863
5542771967428222487575867657523442202
0757363056949882508796892816275384886
3396909959826280956121450994871701244
5164612603790293091208890869420285106
4018215439945715680594187274899809425
4742173582401063677404595741785160829
2301353580818400969963725242305608559
0370062427124341690900415369010593398
3835777939410970027753472000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
000000000002.
And the first number in my list is:
1959034064499908343126250819820638104
6123972390589368223882605328968666316
3798706618519516487894823215962295591
1543601914918952972521526672829228299
0852649023362731392404017939142010958
2613936349594714837571967216722434100
6711851622766113313519248884898991489
2157188308679896875137439519338903968
0949055497503864071060338365866606835
3920101163591790003990449506520329974
9542985993134669814805318474080581207
891125912.
#1652 Re: Introductions » Здрасти by krassi_holmz » 2005-12-10 06:23:00
Yes, it's something like Russian.
And for the phonetic system Russian is better than English, but it's not so good as Bulgarian. 
#1653 Introductions » Здрасти by krassi_holmz » 2005-12-04 19:40:14
- krassi_holmz
- Replies: 13
Hello.
I'm Krassi. I'm from Bulgaria (you don't know it, I guess). I'm fourteen. I know some maths and I come here to have fun, because...
MATHS IS FUN, RIGHT?
#1654 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-04 19:25:43
2/1001!+2
3/1001!+3
...
1000/1001!+1000
1001/1001!+1001
#1655 Re: Help Me ! » Need Help Fast » 2005-12-04 18:20:46
2sin3xcos3x=2(sin6x)/2=sin6x, so
sin6x=1
x = (1/6)(Pi/2)=Pi/12
#1656 Re: Help Me ! » geometry » 2005-12-04 18:09:52
Ganesh,
I think c) proof isn't very right. Because you have "let" here. Here's a proof without "let":
(b means black, w means white)
Let...
b1+w1=b2
b2+w2=b3
so
b1+w1+w2=b3; b1+(w1+w2)=b3,
but
b1+w3=b3, so w1+w2=w3.
#1657 Re: Help Me ! » Integer system » 2005-12-04 17:29:08
More colouricaly, we search for 4 squares such the sum of first and second is equal to third and the sum of second and third is equal to fourth. So we can make a generalized question. Does the system:
|a1²+a2²=a3²
|a2²+a3²=a4²
|...
|a{N-2}²+a{N-1}²=aN²
have integer solutions?
I'm sure the upper system hasn't general solution.
Why?
#1658 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-04 17:16:53
#k+54
Because N[abcdabcd]=N[abcd]+10000.N[abcd]=10001.N[abcd]=73.137.N[abcd], where N[abcd] means number abcd.
#1659 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-04 08:16:42
About the problem k+43:
x^2+1/x^2=22
x^2+2+(1/x)^2=25
(x+1/x)^2=25
Let x > 0
x+1/x=A=5
A^3=125=(x^3+1/x^3)+3*5
x^3+1/x^3=125-15=110
A^5=3125=(x^5+1/x^5)+5*110+10*5
x^5+1/x^5=3125-5*110-10*5=2525
A^7=78125=(x^7+1/x^7)+7*2525+21*110+35*5
x^7+1/x^7=78125-(7*2525+21*110+35*5) =57965
A^9=1953125=(x^9+1/x^9)+9*57965+36*2525+84*110+126*5
x^9+1/x^9=1953125-(9*57965+36*2525+84*110+126*5)=1330670
Is this true? Check it. I'm sure there's an easier way.
#1660 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-12-04 07:15:24
The sum of all cells in the magic square is ∑=n²(n²+1)/2. We have n rows with same sum in each and the sum of the rows is equal to the sum of the cells, so the sum in each row is equal to the sum of the cells divided by the number of the rows, e.a. A=n²(n²+1)/2n=(n³+n)/2
sorry for the syntax, I don't know English very well.
#1661 Re: Help Me ! » equation » 2005-12-04 05:58:52
Sorry Ricky. But Sarah can still answer the question and can verify it with mine.
#1662 Re: Help Me ! » equation » 2005-12-04 03:42:05
y-2x=4 ⇒y=2x+4 ⇒
mx+y=2 <=>mx+2x+4=2
(m+2)x+4=2
(m+2)x=-2
1. If m+2=0;m=2
0x=-2
No solution
2. If m+2≠0;m≠2
x=-2/(m+2)
So:
No solution-
m=2
One solution-
m≠2
Infinite many solutions-
m∈{}
sorry for the syntax, I don't know English well.
#1663 Re: Help Me ! » Integer system » 2005-12-04 03:19:54
Let write the system as:
|x^2+y^2=a^2
|b^2+y^2=x^2
Thus we get 2 pythagorean triples (Sorry if the syntax is incorrect. I don't know English well)
x=u^2-v^2
y=2uv
a=u^2+v^2
and
b=w^2-z^2
y=2wz
x=w^2+z^2.
To solve the system is enough to solve:
|2uv=2wz
|u^2-v^2=w^2+z^2
<=>
|uv=wz
|u^2=w^2+z^2+v^2
what to do further?
#1664 Help Me ! » Integer system » 2005-12-04 00:22:31
- krassi_holmz
- Replies: 6
Can anybody solve the system
|x^2+y^2=a^2
|x^2-y^2=b^2,
where x, y, a and b are integers?