Math: Number Theory

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.

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?

yes you got it right

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

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.

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

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.

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

Bunnel should work in NASA

20,000∗(1.03) ^8

how do we simplify this easily on the GMAT?

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