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# Math: Number Theory

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10 Jun 2010, 15:08
Thanks a lot of the detailed explanation !!!!
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12 Jun 2010, 19:11
thanks for sharing!!
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23 Jun 2010, 11:21
My only problem is that I was not able to spot such a good site earlier.
Great effort by Bunuel. Thanks a lot.
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21 Jul 2010, 18:38
Thank you so much for compiling this! I have a few questions on the reasoning behind the rules... Sorry for the very long post.

Why is:

Perfect Square
• The sum of distinct factors of a perfect square is ALWAYS ODD.
• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
• Perfect square always has even number of powers of prime factors.

Consecutive Integers
• The product of n consecutive integers is always divisible by n!.
Given n=4 consecutive integers: \{3,4,5,6\}. The product of 3*4*5*6 is 360, which is divisible by 4!=24.

Evenly Spaced
• If the first term is a_1 and the common difference of successive members is d, then the n_{th} term of the sequence is given by:
a_ n=a_1+d(n-1)

Terminating Decimal
Why must the denominator be 2^n5^m?

Exponents
Why are:
a^n-b^n is ALWAYS divisible by a-b.
a^n + b^n is divisible by a+b if n is odd, and not divisible by a+b if n is even.
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21 Jul 2010, 20:33
I'm so sorry to make you explain them, but I have a really hard time memorizing rules that I don't understand... thanks again!
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23 Jul 2010, 06:01
Thanks for this post.
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25 Jul 2010, 21:43
Hi Bunuel,

Thanks for this wonderful post on number properties.

I was just wondering if we can add topics related question numbers from official book as we have done for Triangles.

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21 Aug 2010, 23:03
Great Collection.. kudos..
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07 Sep 2010, 21:20
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09 Sep 2010, 12:16
great compilation.......
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14 Sep 2010, 05:33
Bunuel wrote:
NUMBER THEORY
Consecutive Integers

Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers.

• Sum of $$n$$ consecutive integers equals the mean multiplied by the number of terms, $$n$$. Given consecutive integers $$\{-3, -2, -1, 0, 1,2\}$$, $$mean=\frac{-3+2}{2}=-\frac{1}{2}$$, (mean equals to the average of the first and last terms), so the sum equals to $$-\frac{1}{2}*6=-3$$.

• If n is odd, the sum of consecutive integers is always divisible by n. Given $$\{9,10,11\}$$, we have $$n=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If n is even, the sum of consecutive integers is never divisible by n. Given $$\{9,10,11,12\}$$, we have $$n=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of $$n$$ consecutive integers is always divisible by $$n!$$.
Given $$n=4$$ consecutive integers: $$\{3,4,5,6\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.

Evenly Spaced Set

Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers $$\{9,13,17,21\}$$ is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set.

• If the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by:
$$a_ n=a_1+d(n-1)$$

• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\{7,11,15,19\}$$, $$mean=median=\frac{7+19}{2}=13$$.

• The sum of the elements in any evenly spaced set is given by:
$$Sum=\frac{a_1+a_n}{2}*n$$, the mean multiplied by the number of terms. OR, $$Sum=\frac{2a_1+d(n-1)}{2}*n$$

• Special cases:
Sum of n first integers: $$1+2+...+n=\frac{1+n}{2}*n$$

Sum of n first odd numbers: $$a_1+a_2+...+a_n=1+3+...+a_n=n^2$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n-1$$. Given $$n=5$$ first odd integers, then their sum equals to $$1+3+5+7+9=5^2=25$$.

Sum of n first positive even numbers: $$a_1+a_2+...+a_n=2+4+...+a_n=n(n+1)$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n$$. Given $$n=4$$ first positive even integers, then their sum equals to $$2+4+6+8=4(4+1)=20$$.

• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65.

There seems to be a discrepancy in what some study guides consider consecutive integers. In Kaplan Premier 2011 consecutive integers are defined as numbers that occur at a fixed interval or exhibit a fixed pattern. However, on the Kaplan Free Practice Test I got a DS question wrong because it didn't consider evenly spaced numbers to necessarily be consecutive. Your definition also separates the two. Could anyone clarify which is correct so I know for the actual GMAT. Thanks!
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14 Sep 2010, 06:38
Expert's post
cheetarah1980 wrote:
There seems to be a discrepancy in what some study guides consider consecutive integers. In Kaplan Premier 2011 consecutive integers are defined as numbers that occur at a fixed interval or exhibit a fixed pattern. However, on the Kaplan Free Practice Test I got a DS question wrong because it didn't consider evenly spaced numbers to necessarily be consecutive. Your definition also separates the two. Could anyone clarify which is correct so I know for the actual GMAT. Thanks!

When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, ...

-7, -6, -5 are consecutive integers.

2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers.

3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers.

2, 5, 8, 11 ARE NOT consecutive integers, they are terms of arithmetic progression with common difference of 3.

All sets of consecutive integers represent arithmetic progression but not vise-versa.

Hope it's clear.
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20 Sep 2010, 00:27
7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.
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20 Sep 2010, 00:35
Expert's post
Kronax wrote:
7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.

Note that you can perform this operation number of times. Also you won't need to check divisibility by 7 for huge numbers on GMAT.
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20 Sep 2010, 00:45
Bunuel wrote:
Kronax wrote:
7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.

Note that you can perform this operation number of times. Also you won't need to check divisibility by 7 for huge numbers on GMAT.

Thank you for the notice! I haven't thought about doing the operation multiple times.
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29 Sep 2010, 23:30
How to print this document
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28 Oct 2010, 18:42
Quote:
Perfect Square

A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square.

There are some tips about the perfect square:.
Perfect square always has even number of powers of prime factors.

Bunuel : can you please give an example of bold statement

Thanks
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01 Nov 2010, 11:09
If n is a positive integer greater than 1, then there is always a prime number P with n<P<2n

n<p<2n can someone please explain this with example .

Thanks
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02 Nov 2010, 08:51
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06 Nov 2010, 13:36
Bunnel!!! you are the rock !
Many thanks for the above post !! It helps me a lot since my exam is very soon !!
Re: Math: Number Theory   [#permalink] 06 Nov 2010, 13:36

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