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Papgust's GMAT MATH FLASHCARDS directory

Problem Solving — algebra and arithmetic
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by papgust » Fri Jun 11, 2010 5:55 am
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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by papgust » Fri Jun 11, 2010 7:13 pm
"Number Properties" flashcards are over by now.

I will start posting flashcards in "Inequalities".
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by papgust » Fri Jun 11, 2010 7:15 pm
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)]
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by papgust » Fri Jun 11, 2010 7:25 pm
If x > y and 'k' is a real number,
then
x + k > y + k (Irrespective of the value of k)
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by papgust » Fri Jun 11, 2010 7:27 pm
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 --]
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by papgust » Fri Jun 11, 2010 7:32 pm
(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
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by papgust » Sat Jun 12, 2010 8:32 am
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]
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by papgust » Sat Jun 12, 2010 8:35 am
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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by papgust » Sat Jun 12, 2010 8:41 am
i. If | X1 | > | X2 |, Then X1 > X2 OR - X1 > X2


ii. If | X1 | < | X2 |, Then X1 < X2 OR - X1 < X2
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by jube » Sat Jun 12, 2010 1:54 pm
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!
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by jube » Sat Jun 12, 2010 2:02 pm
thanks a ton for all the stuff on remainders - it REALLY helps!
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by papgust » Sat Jun 12, 2010 7:20 pm
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.
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by papgust » Sat Jun 12, 2010 7:23 pm
|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
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by papgust » Sat Jun 12, 2010 7:26 pm
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 --]
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by papgust » Sat Jun 12, 2010 7:34 pm
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.
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