Software Applications Programming Errors and solutions.

The sum of two numbers is a square. The sum of their squares is also a square. Find a method to get all such numbers. Answer let a + b = x^2 and a^2 + b^2 = y^2. the second equation is the characteristic Pythagorean triplet. so we can write a and b as p^2-1 and 2p where p is a positive integer > 1 . therefore, (p^2-1) + (2p) has to be a square number. [which is not true except for p=1/2] thus there is no base Pythagorean triplet which satisfies this relation, but multiples of them may easily do so. like: 3,4,5. 3+4 = 7 not a square. but for 21,28,35. 21 + 28 =7^2. thus (p^2+2p-1)(p^2-1) and (p^2+2p-1)(2p) or better still, q^2*(p^2+2p-1)(p^2-1) and q^2*(p^2+2p-1)(2p) satisfy the given condition, where p,q are both positive integers.

Properties of Irrational numbers If x be an irrational number then there are infinitely many relatively prime integers p and q such that | p/q - x | < 1/q^2 If x is a rational number then there are finitely many solutions to the above equation. This Theorem was first discovered by Dirichlet Further let {y} be the fractional part of y. x is an irrational number then the sequence {nx} n = 1,2,3, ... is not periodic. If x is a rational then this sequence is periodic. A rational number can be represented only by a a finite continued fraction where as an irrational number can be represented by an infinite continued fraction. Each convergent of the continued fraction representation of a irrational number is a better rational approximation of the given irrational number, than any previous convergent. Eg: The continued fraction representation of Pi is [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ..] hence the successive rational approximation of Pi are 3, 2

Fibonacci numbers in pascal's triangle? The Proof follows from the following identity: Let F(r) be the r-Th Fibonacci number, C(n,r) be the number of combination of n things taken are at a time and [x] be the greatest integer function then Summation C(n-k, k) = F(n+1) where k goes form k=0 to k = [n/2] Now use induction on the variable n.

Improve your reasoning power Question - When a no. is divided by 2, rem. is 1 when divided by 3, rem. is 2. and so on...till it is divided by 10, remainder is 9.What's the number? Answer- if the no. be n, (n+1) is divisible by all of {2,3,4,..,10}. hence (n+1) is divisible by 2520. [ 2 |(n+1) & 3 |(n+1) => 6 |(n+1) & 4 | (n+1) => 12 |(n+1) (12=l.c.m(6,4)) & 5 |(n+1) => 60 |(n+1). continuing in this way 2520 |(n+1)] therefore any no. of the form n=(2520p-1) will suffice where p is a natural no. of course the smallest such natural no. is 2519. Mathematics equation problem X^3-Y^2=2 FIND ALL INTEGRAL SOLUTION OF X,Y Answer: case 1 x-3=y-5 y=x+2 original eq gives x^3-x^2-4x-2=0 no integral solution for this. case 2 x-3=y+5 y=x-8 original eq gives x^3-x^2+16x-66=0 (x-3)(x^2+2x+22)=0 x=3 y=-5 as eq invoves y^2 hence y as wellas -y both satisfies the pairs are (3,5) and (3,-5)

Find the remainder when the digits 1 to 99 written side by side is divided by 11 i.e: (123456789101112...99)mod 11=? 123456789101112.....979899 counting from right, sum of digits at odd place = 9(9+8+...+1+0)+(9+7+5+3+1) sum of digits at even place = 10(9+8+...+1)+(8+6+4+2) sum of odd - sum of even = -40 = 4 (mod 11)

Code to send email using PLSQL You can use UTL_SMTP and UTL_TCP to send email attachment from Oracle Version 8i onwards. Sample Code: DECLARE v_From VARCHAR2(80) := 'oracle@mycompany.com'; v_Recipient VARCHAR2(80) := 'test@mycompany.com'; v_Subject VARCHAR2(80) := 'test subject'; v_Mail_Host VARCHAR2(30) := 'mail.mycompany.com'; v_Mail_Conn utl_smtp.Connection; crlf VARCHAR2(2) := chr(13)||chr(10); BEGIN v_Mail_Conn := utl_smtp.Open_Connection(v_Mail_Host, 25); utl_smtp.Helo(v_Mail_Conn, v_Mail_Host); utl_smtp.Mail(v_Mail_Conn, v_From); utl_smtp.Rcpt(v_Mail_Conn, v_Recipient); utl_smtp.Data(v_Mail_Conn, 'Date: ' || to_char(sysdate, 'Dy, DD Mon YYYY hh24:mi:ss') || crlf || 'From: ' || v_From || crlf || 'Subject: '|| v_Subject || crlf || 'To: ' || v_Recipient || crlf || 'MIME-Version: 1.0'|| crlf || -- Use MIME mail standard 'Content-Type: multi

SET NOCOUNT ON DBCC UPDATEUSAGE(0) -- Database size. Following code you can write in stored procedure or you can directly run on query browser to get all table size in selected database. EXEC sp_spaceused-- Table row counts and sizes. CREATE TABLE #test ( [name] NVARCHAR(128), [rows] CHAR(11), reserved VARCHAR(18), data VARCHAR(18), index_size VARCHAR(18), unused VARCHAR(18)) INSERT #t EXEC sp_msForEachTable 'EXEC sp_spaceused ''?''' SELECT *FROM #test-- # of rows. SELECT SUM(CAST([rows] AS int)) AS [rows]FROM #test DROP TABLE #test I hope this code will help you a lot on database administration.