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Thanks for your words! 1.6.0 update is available for download. Just get it, go to Store and you can buy any of 10 famous Manhattan GMAT books.

Let me know if you have any questions.

OrenY wrote:

thank you for the great post. I currently use the GMAT Toolkit app, which I highly recommend, when can I expect this update? In addition, when will the Manhattan GMAT books be updated to the app?

is this always true? The product of n consecutive integers is always divisible by n!. Given consecutive integers: . The product of 3*4*5*6 is 360, which is divisible by 4!=24.

for example, n=10 and the first number starts at 100000, then this rule doesn't hold.

Hi, thanks for the great summary. BTW, do you have a list of questions (just question number) in OG12 + Quant Review 2nd edition to practice, just like the Triangles and Circle section?

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

Exponents and divisibility: \(a^n-b^n\) is ALWAYS divisible by \(a-b\). \(a^n-b^n\) is divisible by \(a+b\) if \(n\) is even. \(a^n + b^n\) is divisible by \(a+b\) if \(n\) is odd, and not divisible by a+b if n is even.

Hello, Bunuel. Great post!

Do you have an example problem in which this applies. I plugged in numbers to understand the concept I was just curious about the application and seeing this in action. Thanks.

Exponents and divisibility: \(a^n-b^n\) is ALWAYS divisible by \(a-b\). \(a^n-b^n\) is divisible by \(a+b\) if \(n\) is even. \(a^n + b^n\) is divisible by \(a+b\) if \(n\) is odd, and not divisible by a+b if n is even.

Hello, Bunuel. Great post!

Do you have an example problem in which this applies. I plugged in numbers to understand the concept I was just curious about the application and seeing this in action. Thanks.

Exponents and divisibility: \(a^n-b^n\) is ALWAYS divisible by \(a-b\). \(a^n-b^n\) is divisible by \(a+b\) if \(n\) is even. \(a^n + b^n\) is divisible by \(a+b\) if \(n\) is odd, and not divisible by a+b if n is even.

Hello, Bunuel. Great post!

Do you have an example problem in which this applies. I plugged in numbers to understand the concept I was just curious about the application and seeing this in action. Thanks.

Check this:

Awesome! Thanks. That's definitely above my level but good practice no doubt.

Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).

It's easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\)

Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).

It's easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\)

no of trailing zeros in 25! =6

\(\frac{25}{5}+\frac{25}{5^2}= 5+1\); Please correct me, clarify if i'm wrong. Thanks

You are right. \(k\) is the highest power of 5 not exceeding \(n.\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).

It's easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\)

no of trailing zeros in 25! =6

\(\frac{25}{5}+\frac{25}{5^2}= 5+1\); Please correct me, clarify if i'm wrong. Thanks

The highest power of a prime number "k" that divides any number "n!" is given by the formula n/K + n/k^2+n/k^3.. (until numerator becomes lesser than the denominator). Remember to truncate the remainders of each expression

E.g : The highest number of 2's in 10! is 10/2 + 10/4 + 10/8 = 5 + 2 + 1 = 8 (Truncate the reminder of each expression)

As a consequence of this, the number of zeros in n! is controlled by the presence of 5s. Why ? 2 reasons

a) 10 = 5 x 2, b) Also in any n!, the number of 5's are far lesser than the number of 2's.

Think about this example. The number of cars that you make depends on the number of engines. You can have 100 engines and 1000 cars, but you can only make 100 cars (each car needs an engine !)

10 ! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 Lets factorize each term ... 10! = (5 x 2) x(3x3)x(2x2x2)x7x(2x3)x(5)X(2x2)x1 the number of 5s = 2 The number of 2s = 7 The number of zeros in 10! = the total number of 5s = 2 (You may use a calc to check this10! = 3628800)

hence in any n! , the number of 5's control the number of zeros.

As a consequence of this, the number of 5's in any n! is n/5 + n/25 + n/125 ..until numerator becomes lesser than denominator.

Again, i want to emphasize that this formuala only works for prime numbers !! So to find the number of 10's in any n!, DO NOT DIVIDE by 10 ! (10 is not prime !) i.e DONT do n/10 + n/100 + n/1000 - THIS IS WRONG !!!
_________________

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gmatclubot

Re: Math: Number Theory
[#permalink]
25 Sep 2012, 09:43

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