For example, say n = 72. Consider it's factor 36 --> \(36 = 2^2*3^2\) --> 36 = product of prime divisors of n raised to some power.
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Bunuel niks18 gmatbusters pushpitkc VeritasPrepKarishma

I have a small query regarding rounding:

How do I interpret nearest ten, nearest hundred etc in

1234.1234

on both LHS and RHS of decimal?
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Hii

You have asked regarding rounding to nearest tens, hundred for 1234.1234 on both LHS and RHS of decimal.

First of all , Lets know what does the place signify:
we denote the digits as units, tens, hundreds, thousands before decimal and tenth, hundredth after decimal. see the sketch

Rounding
Simplifying a number to a certain place value. Drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

It means if we round 47 to nearest ten = 50
whereas, 44 to nearest ten = 40

Now lets come to your question:
1234.1234
Rounding to nearest ten = 1230
Rounding to nearest hundred = 1200
Rounding to nearest tenth =1234.1
Rounding to nearest hundredth= 1234.12

I hope it is clear now. Fell free to tag again.



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gmatbusters

Thanks for the pic , that made me wonder if the difference in position of tens on LHS and RHS
is because of raising base of 10 to exponents say 0,1 and -1. Is this how we arrive at
PLACE VALUES for a number?


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yes you got it right


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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.


Hi

Can some expert please explain this concept more clearly.
What I am looking for is the proof of these statements.

Bunuel
VeritasKarishma
chetan2u
Gladiator59
gmatbusters
MathRevolution
AjiteshArun

Hi nitesh,

It is to do with binomial theorem, which further deals with expansion of a term..
Say you are looking for a^n . I can write a = a-b+b..
\(a^n=(a-b+b)^n=((a-b)+b)^n = (a-b)^n+n(a-b)^{n-1}b^1+....+b^n.....a^n-b^n=(a-b)^n+n(a-b)^{n-1}b^1+...=(a-b)((a-b)^{n-1}+.........)\)
So, Right hand side is multiple of a-b and on left side we have a^n-b^n..
so a^n-b^n is a multiple of a-b

similarly for the other too..

Just take small values to confirm..

Let n = 4.. \(a^4-b^4=(a^2-b^2)(a^2+b^2)=(a-b)(a+b)(a^2+b^2)\).. so multiple of a-b and a+b.
Let n = 3... \(a^3-b^3=(a-b)(a^2+ab+b^2)\)... so multiple of only a-b
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1. Similarly, all prime numbers above 3 are of the form 6n−1 or 6n+1, because all other numbers are divisible by 2 or 3.
2. Verifying the primality (checking whether the number is a prime) of a given number n can be done by trial division, that is to say dividing n by all integer numbers smaller than √n, thereby checking whether n is a multiple of m≤n.

Can we not check the primality of a number by checking if the number is of the form 6n−1 or 6n+1? Do we have to run a trial division?

First of all there is no known formula of prime numbers.

Next:
Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) or \(p=6n-1\), where n is an integer >1.

But:
Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so vise-versa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number.

Hope it's clear.
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I'm curious what the practical application of prime numbers after 3 having 6n−1 or 6n+1 has on the GMAT? Since it can't help us identify whether a number is prime or not, in which context will we find it useful?

Well, it can help us identify that a number is not a prime - but you’re right that it doesn’t help us determine whether a number is a prime. For GMAT purposes, I suggest that you focus on the reasoning behind this rule rather than on the rule itself.

Posted from my mobile device
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Hi,

could someone confirm if my understand is good

for this rule : 'Any positive divisor of n is a product of prime divisors of n raised to some power' the example given is the following:

"For example, say n = 72. Consider it's factor 36 --> \(36=2^2∗3^2 \) --> 36 = product of prime divisors of n raised to some power."

But if I take other factors, like 12 or 18, in those cases the I get different powers

Example #1 : 12 is a divisor of 72 --> \(12=2^2*3\) --> 12 is not a product of prime divisors of n raised to some power.

Example #2 : 18 is a divisor of 72 --> \(18=2*3^2\) --> 18 is not a product of prime divisors of n raised to some power.

is there a concept that I am loosing.

Thanks

\(72 = 2^3*3^2\). So, prime divisors of 72 are 2 and 3.

\(12=2^2*3\) is product of prime divisors of 72 (2 and 3) raised to some power.
\(18=2*3^2\) is product of prime divisors of 72 (2 and 3) raised to some power.
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Thanks! As always, great resource!

Hey Bunuel, in the above it says all primes are either 6n+1 or 6n-1

I just realised that 49 is 6(8)+1 but it is not a prime.

Can you please help me clear this doubt. Thanks.

Posted from my mobile device

I think this is already addressed couple of times in this thread. For example:
https://gmatclub.com/forum/math-number- ... l#p1735203
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Bunnel should work in NASA