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Any nonzero natural number n can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.

Pls give me the example of bold face text because i am not sure what does it exactly means.

Thanks

It's called the fundamental theorem of arithmetic (or the unique-prime-factorization theorem) which states that any integer greater than 1 can be written as a unique product of prime numbers.

For example: 60=2^2*3*5 --> 60 can be written as a product of primes (powers of primes) only in this unique way (you can just reorder the multiples and write 3*2^2*5 or 2^2*5*3 ...).
_________________

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?

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.

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