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# Math: Number Theory

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Math Expert
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Re: Math: Number Theory [#permalink]  30 Apr 2010, 13:30
1
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Expert's post
AloneAndInsufficient wrote:
Bunuel wrote:
NUMBER THEORY
• For GMAT it's good to memorize following values:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{7}\approx{2.45}$$
$$\sqrt{8}\approx{2.65}$$
$$\sqrt{10}\approx{2.83}$$

Anyone else notice that these are wrong?
They should be:
• For GMAT it's good to memorize following values:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{6}\approx{2.45}$$
$$\sqrt{7}\approx{2.65}$$
$$\sqrt{8}\approx{2.83}$$
$$\sqrt{10}\approx{3.16}$$

Thanks. Edited. +1 for spotting this.
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Re: Math: Number Theory [#permalink]  04 May 2010, 04:21
Thank you
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Re: Math: Number Theory [#permalink]  10 May 2010, 01:41
2
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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, 20:15, edited 1 time in total.
Math Expert
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Re: Math: Number Theory [#permalink]  10 May 2010, 15:51
Expert's post
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]  12 May 2010, 12:53
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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Re: Math: Number Theory [#permalink]  15 May 2010, 23:24
Thanks Bunuel for all the efforts put in creating this. Really appreciate.
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Re: Math: Number Theory [#permalink]  10 Jun 2010, 12:14
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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Re: Math: Number Theory [#permalink]  10 Jun 2010, 14:00
Expert's post
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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Re: Math: Number Theory [#permalink]  10 Jun 2010, 14:08
Thanks a lot of the detailed explanation !!!!
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Re: Math: Number Theory [#permalink]  12 Jun 2010, 18:11
thanks for sharing!!
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Re: Math: Number Theory [#permalink]  23 Jun 2010, 10:21
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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Re: Math: Number Theory [#permalink]  21 Jul 2010, 17:38
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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Re: Math: Number Theory [#permalink]  21 Jul 2010, 19:33
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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Re: Math: Number Theory [#permalink]  23 Jul 2010, 05:01
Thanks for this post.
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Re: Math: Number Theory [#permalink]  25 Jul 2010, 20:43
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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Re: Math: Number Theory [#permalink]  11 Aug 2010, 16:12
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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?

$$50!=900^xa=(2^2*3^2*5^2)^x*a$$, 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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Re: Math: Number Theory [#permalink]  11 Aug 2010, 17:07
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Expert's post
utfan2424 wrote:
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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Re: Math: Number Theory [#permalink]  21 Aug 2010, 22:03
Great Collection.. kudos..
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Re: Math: Number Theory [#permalink]  07 Sep 2010, 20:20
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Re: Math: Number Theory [#permalink]  09 Sep 2010, 11:16
great compilation.......
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Re: Math: Number Theory   [#permalink] 09 Sep 2010, 11:16

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