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papgust Community Manager

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Fri Jun 11, 2010 6:55 am

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Although this is not very important, it is fun to simply know this point.

If p is a prime number, then for any integer n,
(n^p - n) is ALWAYS divisible by p.

Example:
p = 3, n = 2

(2^3 - 2) is divisible by 3
OR
6 is divisible by 3

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papgust Community Manager

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																		 Fri Jun 11, 2010 8:13 pm
									 Quote

"Number Properties" flashcards are over by now. 

I will start posting flashcards in "Inequalities".

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GO GREEN..! GO VEG..!

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papgust Community Manager

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Fri Jun 11, 2010 8:15 pm

Quote

If x > y,
then x - y > 0 AND x = y + k (Where k > 0)

If x < y,
then x - y < 0 AND [x = y - k (Where k > 0) OR x = y + k (Where k < 0)]

papgust Community Manager

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Fri Jun 11, 2010 8:25 pm

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If x > y and 'k' is a real number,
then
x + k > y + k (Irrespective of the value of k)

papgust Community Manager

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Fri Jun 11, 2010 8:27 pm

Quote

I.
If x > y and k > 0,
then
k*x > k*y AND x/k > y/k

II.
If x > y and k < 0,
then
k*x < k*y AND x/k < y/k
[-- Inequality reverses when k < 0 --]

papgust Community Manager

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Fri Jun 11, 2010 8:32 pm

Quote

(IMPORTANT)

If x and y are both POSITIVE/NEGATIVE
AND x > y

THEN

1/x < 1/y

Examples:
1. x = 3, y = 2, x > y
So, 1/3 < 1/2

2. x = -2, y=-3, x > y
So, -1/2 < -1/3

papgust Community Manager

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Sat Jun 12, 2010 9:32 am

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If a1 > b1, a2 > b2, ... an > bn,
Then
(a1 + a2 + ... an) > (b1 + b2 + ... bn)

If a1 > b1, a2 > b2, ... an > bn,
Then
(a1 * a2 * ... an) > (b1 * b2 * ... bn) --- [ONLY When all values of a and b are POSITIVE]

papgust Community Manager

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																		 Sat Jun 12, 2010 9:35 am
									 Quote

If ax^2+bx+c > 0, where a > 0,

Then

x DO NOT LIE between xL and xU 

(xL and xU are lower and upper limits of x respectively)

If ax^2+bx+c < 0, where a > 0,

Then

x LIES between xL and xU 

(xL and xU are lower and upper limits of x respectively)

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GO GREEN..! GO VEG..!

Daily Quote:

“Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”

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papgust Community Manager

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Sat Jun 12, 2010 9:41 am

Quote

i. If | X1 | > | X2 |, Then X1 > X2 OR - X1 > X2

ii. If | X1 | < | X2 |, Then X1 < X2 OR - X1 < X2

jube Really wants to Beat The GMAT!

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Sat Jun 12, 2010 2:54 pm

Quote

papgust wrote:

If N is a perfect square, then the number of factors of N will ALWAYS be an ODD number.

If N is a NON-perfect square, then the number of factors of N will ALWAYS be an EVEN number.

Didn't know this - thanks a ton! I'm really loving your flashcard threads! thanks!

jube Really wants to Beat The GMAT!

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Sat Jun 12, 2010 3:02 pm

Quote

thanks a ton for all the stuff on remainders - it REALLY helps!

papgust Community Manager

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Sat Jun 12, 2010 8:20 pm

Quote

jube wrote:

thanks a ton for all the stuff on remainders - it REALLY helps!

Really glad to hear that my flashcards are helping you. Thank you very much! Stay tuned to this thread for more.

papgust Community Manager

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Sat Jun 12, 2010 8:23 pm

Quote

|x - a| < r, where a is a positive real number and a is a fixed real number,
then
a-r < x < a+r

|x - a| > r, where a is a positive real number and a is a fixed real number,
then
x < a-r (OR) x > a+r

papgust Community Manager

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Sat Jun 12, 2010 8:26 pm

Quote

Golden Rule:
When a NEGATIVE value is multiplied both sides in an INEQUALITY, then the INEQUALITY SIGN reverses/flips.

Example:
-1/x > -3/14

When -ve is multipled both sides,

1/x < 3/14 [-- Not only does the number sign changes, but also the inequality sign --]

papgust Community Manager

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																		 Sat Jun 12, 2010 8:34 pm
									 Quote

Always Remember to:

Plug in the solutions back to the absolute equations to see whether it satisfies, whenever you get multiple solutions while solving absolute equations (A very useful takeaway especially for solving DS questions)

Example:

Solve, |x| = 3x - 2.

i. x = 3x - 2

2x = 2

x = 1.

ii. -x = 3x - 2

4x = 2

x = 1/2.

You get 2 solutions 1 and 1/2. DO NOT CONCLUDE that there are 2 solutions and move on. Plug in both solutions back to the original equation (|x| = 3x - 2) and see whether the equation holds.

Sub. x = 1 in |x| = 3x - 2,

Equation SATISFIES!

Sub. x = 1/2 in |x| = 3x - 2,

Equation DOES NOT SATISFY!

Hence, x = 1 is the ONLY solution.

_________________
Download GMAT Math and CR questions with Solutions from Instructors and High-scorers:

http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html

-----------

GO GREEN..! GO VEG..!

Daily Quote:

“Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”

									Thanked by: sandi10017, tarunjagtap, eaakbari

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