Clipmarks | DWarrior's 'gre-gmat' clips

Tougher Combinatorics Formulas

Sat, 30 Aug 2008 15:23:51 GMT

clipped by: DWarrior

This is a well known problem which states that there are C(n+r -1, r-1) distinct number of nonnegative integer valued vectors (x1,x2,...,xr) satisfying x1 + x2 + ... + xr = n with "disregard to order". The formula where order matters (your example) is n!*C(n+r-1,r-1).

Tags: gre-gmat

Three Modulus Identities

Thu, 29 Nov 2007 22:56:44 GMT

clipped by: DWarrior

Clip Source: www.urch.com

(a+b) mod c = [(a mod c) + (b mod c)] mod c
(a*b) mod c = [(a mod c) * (b mod c)] mod c

(a^b) mod c = [(a mod c)^b] mod c

Tags: gre-gmat

Net Percent Increase

Thu, 29 Nov 2007 17:13:18 GMT

clipped by: DWarrior

Clip Source: www.urch.com

if price is increase by x% then by y% the resulting net increase after the successive increments is always (X + Y + XY/100)

Tags: gre-gmat

Terminating Decimal

Thu, 29 Nov 2007 10:00:13 GMT

clipped by: DWarrior

Clip Source: www.urch.com

For any quotient to have a terminating decimal, the denominator of the equation must have 2 and/or 5 as its only prime factor(s).

Tags: gre-gmat

Solving remainder problems

Sat, 24 Nov 2007 07:51:00 GMT

clipped by: DWarrior

Clip Source: www.urch.com

Set up the original remainder equality. p will always be a positive integer. Look for any restrictions, like it having to be odd/even, or anything else. Then plug a value for p that meets all restrictions to get a sample x, plug the x back into the new number and find the remainder.

Clip Source: www.urch.com

1. Twice the sum of 3 integers x ,y and z when divided by 7 gives remainder 1

what is the remainder when x+y+z is divided by 7?

Clip Source: www.urch.com

realize that we'll always be working with (x+y+z) and never with the individual variables, so you should simplify the problem by substitution (S for x+y+z). That way, there's a greater chance that you'll "see" the solution.

set up the original problem:

2S=7p+1, where p is some multiple of 7 divided by 7. We know 2S must be even, and therefore p must be odd (do you see why?).

So now just plug in any odd number for p and you'll get a valid S. For example, use p=3:
2S=21+1=22
S=11

11/7 leaves remainder of 4.

Tags: gre-gmat

Distance between point and a line

Wed, 07 Nov 2007 01:19:38 GMT

clipped by: DWarrior

Clip Source: www.mathbin.net

Distance between the line ax+by+c=0 and point (X,Y) is:

Tags: gre-gmat

Divisibility by 11 (easiest method)

Tue, 06 Nov 2007 07:54:33 GMT

clipped by: DWarrior

Clip Source: en.wikipedia.org

Subtract the last digit from the rest. 627: 62 - 7 = 55.

Tags: gre-gmat

Inflection Points on Normal Curve

Mon, 05 Nov 2007 19:44:28 GMT

clipped by: DWarrior

Clip Source: www.urch.com

inflection points of a normal curve occur at +/- 1 SD away from the mean.

Clip Source: www.urch.com

If you apply the 68/95/99.5 rule, we know the inflection point occurs at 34%, which is after the 30% mark, meaning the curve is bulging upwards at 30%.

450 is half-way between the two y-values, while 30% is half-way between the two x-values. The curvature is exaggerated for emphasis:

k<30, so B

Tags: gre-gmat

Interesting Problem

Mon, 05 Nov 2007 19:35:29 GMT

clipped by: DWarrior

Clip Source: www.urch.com

If 72.42= k*(24+n/100) , where k and n are positive integers and
n>100, then what is the value of k+n?
a)17
b)16
c)15
d)14
e)13

Clip Source: www.urch.com

L.H.S = 72.42 = 72 + 42/100

R.H.S = 24k + k*n/100

24k = 72
k=3
n=14

k+n = 17
Answer : A

Tags: gre-gmat

Power of a number in a factorial - Problem/Solution

Mon, 05 Nov 2007 06:37:59 GMT

clipped by: DWarrior

Clip Source: www.urch.com

What is the number of Odd divisors of 20! ?

Clip Source: www.urch.com

We have to find out Number of Odd factors of 20!. So Let us start by Expressing 20! in terms of Prime factors. Since

N = 20 X 19 X18...................1 = 2^m + 3^n + 5^p + 7^q. Our First goal is to find the Highest Power of 2, 3, 5, 7, 11, 17, 19 which divide 20!.

Highest Power of 2 Which will divide 20! is = 20/2 + 20/4 + 20/8 + 20/16

= 10 + 5 + 2 + 1 = 18. Remember only Quotients need to be considered

the highest Power of 3, 5, 7,11,13,17,19 is 8 ,4,2,1,1,1,1 respectively.

express 20! as

20! = 2^18 + 3^8 + 5^4 + 7^2 + 11^1 + 13^1 + 17^1 +19^1.

Total no of odd Divisors is (8+1) x ( 4+1) x ( 2+1) X (1+1) x(1+1) x(1+1)x(1+1) = 9 x 5x3x16

Clip Source: www.urch.com

why the + 1 when multiplying:

Because the 3 ^ 0 should also be counted in determining the number of factors. Similarly, the powers for other primes have a +1.

why they are multiplied:

To determine the total no. of distinct factors (ways they can be combined) that can be formed from the given set of prime factors

Tags: gre-gmat

Equilateral Triangle properties

Mon, 05 Nov 2007 06:27:05 GMT

clipped by: DWarrior

Clip Source: www.urch.com

Here are some Equilateral Triangle properties :

For an Equilateral Triangle with side 'a':

Circumscribed Circle Radius

Inscribed Circle Radius

Area

Perimeter

Altitude

Median

Tags: gre-gmat

Square and Cube Identities

Mon, 05 Nov 2007 06:25:46 GMT

clipped by: DWarrior

Clip Source: www.urch.com

Here are some formulae which you may find useful for solving some problems.

Tags: gre-gmat

Mixtures Formula

Mon, 05 Nov 2007 06:19:58 GMT

clipped by: DWarrior

Clip Source: www.urch.com

This is a standard problem applicable to mixtures as well as some problems on averages.

Code:

A                 :           B
a                             b
                        
                  c
    
(b~c)          :        (a~c)

What I've shown above is that if A and B are the items whose certain parameter is given as 'a' and 'b' respectively and same parameter of resultant mixture C is given to be 'c',
Then the ratio in which A and B have been mixed is (c~b) : (c~a)
['~' sign stands for difference not subtraction]

Tags: gre-gmat

Identify power of a prime number in a factorial

Mon, 05 Nov 2007 06:07:02 GMT

clipped by: DWarrior

Clip Source: www.urch.com

What is the highest power of 7 in 5000! (five thousand factorial)?
Ans: 832

Clip Source: www.urch.com

5000/7 = 714
714/7 = 102
102/7 = 14
14 / 7 = 2

2 / 7 = 0

714+102+14+2+0 = 832

Don't ask me how. I just know that to identify power of a number in a factorial, u need to sum the successive quotients.

Tags: gre-gmat