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Re: Math: Number Theory [#permalink]
30 Apr 2010, 13:30

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. _________________

Re: Math: Number Theory [#permalink]
10 May 2010, 01:41

2

This post received KUDOS

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.

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. _________________

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%.

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=6bc=30 --> a=3 is a factor of c=6. _________________

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.

Re: Math: Number Theory [#permalink]
11 Aug 2010, 16:12

2

This post received KUDOS

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?

Re: Math: Number Theory [#permalink]
11 Aug 2010, 17:07

2

This post received KUDOS

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. _________________