**Numbers**

-> For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

-> For any positive integer n

2<= (1+1/n)^n <=3

-> The greatest common divisor is found by looking at the prime factorizations or using the Euclidean algorithm.

-> The least common multiple of a and b is found by looking at the prime factorizations or (ab)/gcd(a,b).

-> Two numbers are said to be relatively prime in the greatest common factor is 1.

-> If gcd(a, b)=d, then there exist integers x and y so that ax+by=d.

-> If d divides both a and b, then d divides a+b and d divides a-b.a=b mod m iff m divides a-b iff a and b both have the same remainder when divided by m.

-> ap-1 = 1 mod p (a is not a multiple of p)af(m)=1 mod m ( gcd(a, m) =1)

-> If a probability experiment is repeated n times and the probability of success in one trial is p, then the probability of exactly r successes in the n trials is nCr (p)r(1-p)(n-r).

-> The number of zeros at the end of n! is determined by the number of 5’s. To find this you do the following process: n/5 = n1 and some remainder. Drop the remainder and compute n1/5 = n2 plus some remainder. Drop the remainder and compute n2/5 = n3 plus some remainder, etc. The number of zeros is n1+n2+n3+n4...

-> The sum of any consecutive integers k through n, with n being the larger,simply use this equation:(n+k)(n-k+1)------------ 2

-> a^2+b^2+c^2 >= ab+bc+ca If a=b=c , then the equality holds in the above. -> a^4+b^4+c^4+d^4 >=4abcd

-> (n!)^2 > n^n (! for factorial)

-> If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s .

-> Consider the two equations a1x+b1y=c1 a2x+b2y=c2

-> (m+n)! is divisible by m! * n! .Then , If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations. If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to ) If a1/a2 <> b1/b2 , then we have a unique solutions for these equations..

-> When a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .

-> Let 'x' be certain base in which the representation of a number is 'abcd' , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d