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Math: Number Theory

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Re: Math: Number Theory [#permalink]

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New post 04 May 2010, 05:21
Thank you :)
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Re: Math: Number Theory [#permalink]

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New post 10 May 2010, 02:41
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Example: A company received $2 million in royalties on the first $10 million in sales and then $8 million in royalties on the next $100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next $100 million in sales?

Solution: Percent decrease can be calculated by the formula above: \(Percent=\frac{Change}{Original}*100=\frac{\frac{2}{10}-\frac{10}{100}}{\frac{2}{10}}*100=50%\), so the royalties decreased by 50%.

I could not get this , i think there is some error... Plzz explain..

as the same Q in Percent Part of Math book is giving an answer of 60 %..

Last edited by sag on 10 Jun 2010, 21:15, edited 1 time in total.
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Re: Math: Number Theory [#permalink]

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New post 10 May 2010, 16:51
sag wrote:
Example: A company received $2 million in royalties on the first $10 million in sales and then $8 million in royalties on the next $100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next $100 million in sales?

Solution: Percent decrease can be calculated by the formula above: Percent=\frac{Change}{Original}*100=\frac{\frac{2}{10}-\frac{10}{100}}{\frac{2}{10}}*100=50%, so the royalties decreased by 50%.

I could not get this , i think there is some error... Plzz explain..

as the same Q in Percent Part of Math book is giving an answer of 60 %..


There was a typo. I edited it in Percent section and forgot to edit it here. Now it's OK. Thanks. +1 for spotting this.
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Re: Math: Number Theory [#permalink]

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New post 12 May 2010, 13:53
sag wrote:
Example: A company received $2 million in royalties on the first $10 million in sales and then $8 million in royalties on the next $100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next $100 million in sales?

Solution: Percent decrease can be calculated by the formula above: Percent=\frac{Change}{Original}*100=\frac{\frac{2}{10}-\frac{10}{100}}{\frac{2}{10}}*100=50%, so the royalties decreased by 50%.

I could not get this , i think there is some error... Plzz explain..

as the same Q in Percent Part of Math book is giving an answer of 60 %..


2 million royalties on 10 million in sales is equivalent to 20 million royalties on 100 million sales (multiply both number by 10). Going down from 20 million royalties to 8 million royalties is a decrease of 60%.
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New post 16 May 2010, 00:24
Thanks Bunuel for all the efforts put in creating this. Really appreciate.
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Re: Math: Number Theory [#permalink]

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New post 10 Jun 2010, 13:14
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

Can anyone please explain this rule??? I'm not sure what it means by gcd(a,b)=1.

Thanks a bunch and great summary !!!!!
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New post 10 Jun 2010, 15:00
bely202 wrote:
If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

Can anyone please explain this rule??? I'm not sure what it means by gcd(a,b)=1.

Thanks a bunch and great summary !!!!!


\(gcd(a,b)=1\) means that greatest common divisor of \(a\) and \(b\) is 1, or in other words they are co-prime, the don't share any common factor but 1. So if we are told that \(a\) is a factor of \(bc\) and \(a\) and \(b\) don't share any common factors, then it must be true that \(a\) is a factor of only \(c\).

So if \(a=3\), \(b=5\) (\(a\) and \(b\) don't share any common factors but 1, \(gcd(a,b)=1\)), \(c=6\) \(bc=30\) --> \(a=3\) is a factor of \(c=6\).
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New post 10 Jun 2010, 15:08
Thanks a lot of the detailed explanation !!!! :-D
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New post 12 Jun 2010, 19:11
thanks for sharing!! :lol:
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New post 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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New post 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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New post 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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New post 23 Jul 2010, 06:01
Thanks for this post.
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New post 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.

Thanks in advance.
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Re: Math: Number Theory [#permalink]

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New post 11 Aug 2010, 17:12
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Bunuel wrote:
fruit wrote:
Thanks! It was very very helpful! Kudos!
But I have a question:

How many powers of 900 are in 50!

Make the prime factorization of the number: \(900=2^2*3^2*5^2\), then find the powers of these prime numbers in the n!.

Find the power of 2:
\(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)

= \(2^{47}\)

Find the power of 3:
\(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)

=\(3^{22}\)

Find the power of 5:
\(\frac{50}{5}+\frac{50}{25}=10+2=12\)

=\(5^{12}\)

We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.


Why do we take just 5 from {2,3,5} and why do we need divide 12 by 2 to get the result?


Thanks in advance!


\(50!=900^xa=(2^2*3^2*5^2)^x*a\), where \(x\) is the highest possible value of 900 and \(a\) is the product of other multiples of \(50!\).

\(50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b\), where \(b\) is the product of other multiples of \(50!\). So \(x=6\).

Below is another example:

Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\).

\(12=2^2*3\), so we should calculate how many 2-s and 3-s are in \(18!\).

Calculating 2-s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\).

Calculating 3-s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\).

Now as \(12=2^2*3\) we need twice as many 2-s as 3-s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) --> \(x=8\).



Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere?
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Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere?


You made everything right 21! ends with 21/5=4 zeros. It's excel: it makes rounding with such a huge numbers thus giving incorrect result.
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Great Collection.. kudos..
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New post 07 Sep 2010, 21:20
This was immensely helpful, there's stuff in here not found elsewhere.
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New post 09 Sep 2010, 12:16
great compilation.......
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Re: Math: Number Theory [#permalink]

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New post 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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Re: Math: Number Theory   [#permalink] 14 Sep 2010, 05:33

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