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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Tue Jun 08, 2010 5:31 pm

Playing with Multiples of N:

(I)
If you add/subtract multiples of number 'N', the result is also a multiple of 'N'.

Examples:
35+21 = 56 [Multiple of 7]
20-15 = 5 [Multiple of 5]

TAKEAWAY:
In general, if N is a divisor of both x and y, then N is a divisor of both x+y and x-y.

(II)
If you add/subtract a multiple of N to/from a non-multiple of N, the result is a non-multiple of N.

Example:
9-5 = 4 [(Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)]

(III)
If you add/subtract 2 non-multiples of N, the result could either be a multiple or a non-multiple of N.

Examples:
19+13 = 32 [(Non-Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)]
19+14 = 33 [(Non-Multiple of 3) - (Non-Multiple of 3) = (Multiple of 3)]

EXCEPTION:
When N = 2, two odds always sum to an even number (Multiple of 2).

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Tue Jun 08, 2010 5:39 pm

GCF Facts:

1. GCF of integers 'm' and 'n' CANNOT be larger than the difference between 'm' and 'n'.

Assume that GCF of m and n is 12. m and n are both multiples of 12. Consecutive multiples of 12 are 12 units apart from each other on the number line. Therefore, m and n CANNOT be less than 12 units apart.

2. Consecutive multiples of n have a GCF of n.

4 and 8 are multiples of 4. Thus 4 is a common factor of both the numbers. 4 and 8 are exactly 4 units apart from each other on the number line. Thus, 4 is the greatest common factor (GCF) of 4 and 8. That is why GCF of any 2 consecutive numbers is ALWAYS 1 as both are multiples of 1.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Tue Jun 08, 2010 5:47 pm

Consecutive Integers:

1. In any set of 3 consecutive integers, one of the integers is ALWAYS divisible by 3.

2. (x-1) * x * (x+1) is divisible by 8, if x is odd.

(x-1) and (x+1) are even. (x-1) is atleast divisible by 2 (Since it always has a 2). As (x-1) and (x+1) are consecutive multiples of 2, (x+1) will have an additional 2 units apart from another 2 (i.e. 4). So, (x+1) is divisible by 4.
Thus, (x-1) * x * (x+1) is divisible by 8

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Wed Jun 09, 2010 6:43 pm

Factor Foundation Rule:

"If x is a factor of y and y is a factor of z, then x is a factor of z".

In other words, any integer is divisible by all of its factors and it is also divisible by all of the factors of its factors.

Example:
Consider 36.

36 can be broken down into 12 * 3

12 can be broken down into 2^2 * 3

So, 2^2 is also a factor of 36

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

Papgust's GMAT MATH FLASHCARDS directory

by papgust » Wed Jun 09, 2010 6:50 pm

Divisibility Rules - ODDs and EVENs:

In general,

Odd integer divided by any other integer CANNOT produce an even integer.
Odd integer divided by an even integer CANNOT produce and integer.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Wed Jun 09, 2010 6:55 pm

Exponent Rules:

1. x^a * x^b = x^(a+b)

2. (a^x)^y = a^xy = (a^y)^x

3. a^x * b^x = (ab)^x

4. x^(a/b) = b root (x^a) = (b root (x))^a

5. x^a / x^b = x^(a-b)

6. (a/b)^x = a^x / b^x

7. x^-a = 1 / x^a

8. a^x + a^x + a^x = 3 * a^x

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Wed Jun 09, 2010 6:57 pm

Properties of Roots:

1. n root(x) / n root(y) = n root(x/y)

2. n root(x) * n root(y) = n * root(xy)

3. b root(x^a) = (b root(x)) ^ a = x^(a/b)

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Wed Jun 09, 2010 6:59 pm

Evenly Spaced Sets:

i. Mean and Median are equal
ii. Mean and Median of the set = Average (First + Last terms)
iii. Sum of the elements = Mean of the set * Number of elements in the set

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Wed Jun 09, 2010 7:01 pm

Counting Integers:

For Consecutive integers:

(Last term - First term + 1)

For Consecutive multiples:

[(Last term - First term) / increment] + 1, where increment is the difference between each consecutive term in the set.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Jun 10, 2010 6:18 pm

Average of an ODD number of consecutive integers will ALWAYS be an integer.

E.g: Average of 5 numbers (1,2,3,4,5) is 3 (an integer).

Average of an EVEN number of consecutive integers will NEVER be an integer.

E.g: Average of 5 numbers (1,2,3,4,5,6) is 3.5 (NOT an integer).

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Jun 10, 2010 6:22 pm

The product of "k" consecutive integers is ALWAYS divisible by K!

Example:
Consider numbers 5,6,7,8,9,10 (6 consecutive numbers). Their PRODUCT is divisible by 6!

5*6*7*8*9*10 / 6*5*4*3*2*1 = 210.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Jun 10, 2010 6:27 pm

For any set of consecutive integers with an ODD number of items, the sum of all integers is ALWAYS a multiple of the number of items.

Example:
Consider 1,2,3,4,5. Number of items is 5.

1+2+3+4+5 = 15. 15 is a multiple of 5 (the number of items).

For any set of consecutive integers with an EVEN number of items, the sum of all integers is NEVER a multiple of the number of items.

Example:
Consider 1,2,3,4,5,6. Number of items is 6.

1+2+3+4+5+6 = 21. 21 is NOT a multiple of 6 (the number of items).

--[You can try out any set of numbers to see whether the rule holds]--

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Jun 10, 2010 6:32 pm

Another GCF:

If "n" is NOT divisible by "k" and GCF (n,k) = "z"
THEN
the remainder (When "n" is divided by "k") will be "z"

Example:

Assume n=20, k=15. 20 is NOT divisible by 15.

GCF (20,15) = 5.

Remainder when 20 is divided by 15 is 5 (Which is the GCF(20,15)).

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Jun 10, 2010 6:35 pm

Terminating/Non-Terminating decimals: (VERY IMPORTANT)

If the fraction is completely simplified and the denominator only have 2's and 5's (or) only 2's or only 5's, then it is a terminating decimal.

If the fraction is completely simplified and the denominator of the fraction has prime factors other than 2's and/or 5's , then it is a repeating/non-terminating decimal.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Fri Jun 11, 2010 5:50 am

Ruthless ROOTS:

If n is a POSITIVE integer, then the nth root of any number (> 1) will ALWAYS be > 1.

Example: (Official Problem)

What is root (4) + 3root (4) + 4root (4) approximately?

root (4) = 2
3root (4) > 1
4root (4) > 1

So, root (4) + 3root (4) + 4root (4) is approximately > 4.

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