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Math: Number Theory

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Re: Math: Number Theory  [#permalink]

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New post 18 Aug 2015, 13:58
1
how do you get the 13 and the 11 in the question?
If the sum of two positive integers is 24 and the difference of their squares is 48, what is the product of the two integers?

x+y=24
and
x2−y2=48 --> (x+y)(x−y)=48, as x+y=24 --> 24(x−y)=48 --> x−y=2 --> solving for x and y --> x=13 and y=11 --> xy=143.

Answer: E.
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Re: Math: Number Theory  [#permalink]

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New post 14 Sep 2016, 06:53
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Quote:
Similarly, all prime numbers above 3 are of the form 6n−16n−1 or 6n+16n+1, because all other numbers are divisible by 2 or 3.
In the prime number section it says prime numbers greater than 3 are of the form 6n-1 or 6n+1. However, this is not necessarily true. eg: n=36, then 6n-1 is 215 and 6n+1 is 217, divisible by 5 and 7 respectively.
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New post 28 Feb 2018, 18:18
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Adding Few Formulas

Sum of First "n" natural nos = n(n+1)/2
Sum of First "n" ODD natural nos = n^2
Sum of First "n" EVEN natural nos = n (n+1)
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New post 04 May 2018, 06:03
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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
Attachment:
gmatbusters3.jpg
gmatbusters3.jpg [ 27.32 KiB | Viewed 2595 times ]


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.


adkikani wrote:

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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Re: Math: Number Theory  [#permalink]

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New post 04 May 2018, 07:03
1
yes you got it right
Attachment:
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adkikani wrote:
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?

Sorry, coming from engineering bakground, am still
not able to erase why aspects of logic :lol:

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New post 08 Jan 2019, 07:50
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nitesh50 wrote:
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.

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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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New post 08 Jan 2019, 07:56
1

The remainder /factor theorem


If you divide a polynomial f(x) by (x - h), then the remainder is f(h).
Hence if f(h) is 0, remainder = 0. hence (x-h) is a factor of f(x).


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nitesh50 wrote:
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
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chetan2u
Gladiator59
gmatbusters
MathRevolution
AjiteshArun

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Re: Math: Number Theory  [#permalink]

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New post 31 Jan 2010, 10:00
Bunuel wrote:
All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. 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.


Awesome post, thank you so much! +1

What is the quickest way to figure out whether a number is prime? I usually check if it's odd or even, then sum its digits to figure out if it's divisible by 3, then look if it ends in 5 and if all else fails divide it by 7. Is this the recommended approach?

What might be a bit confusing is that while all prime numbers are of the form 6n-1 or 6n+1, not all numbers of that form are in fact prime. I think this is crucial. For instance, the number 49 is 6n+1, but is not prime.

Any insight on a quicker check (if one exists) would be much appreciated and thank you again for your efforts. They make a real difference!
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New post 31 Jan 2010, 10:19
ariel wrote:
What is the quickest way to figure out whether a number is prime?


Unfortunately, there is no such quick way to say that this number is prime. You can remember all numbers till 50 and then use rule:

Rule: To check whether a number is prime or not, we try to divide it by 2, 3, 5 and so on. You can stop at \(\sqrt{number}\) - it is enough. Why? Because if there is prime divisor greater than \(\sqrt{number}\), there must be another prime divisor lesser than \(\sqrt{number}\).

Example,

n = 21 -- > \(\sqrt{21}\)~ 4-5
So, we need to check out only 2,3 because for 7, for instance, we have already checked out 3.

n = 101 --> 2,3,5 is out (the last digit is not even or 5 and sum of digits is not divisible by 3). we need to check out only 7
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New post 31 Jan 2010, 11:06
Appreciate the very prompt response, walker. To your point re divisibility by 7: I'm having a hard time proving this algebraically, is it a fair statement to say that the only non-prime numbers of the form 6n-1 and 6n+1 are the ones that are divisible by 7?

If so, a quick way to check whether a big number is prime would be to: 1) check whether it's of the form 6n-1 or 6n+1 2) check whether it's divisible by 7

Is this correct?
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New post 06 Apr 2010, 10:12
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!
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New post 10 Jun 2010, 13:14
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 !!!!!
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New post 10 Jun 2010, 15:00
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, \(gcd(a,b)=1\)), \(c=6\) \(bc=30\) --> \(a=3\) is a factor of \(c=6\).
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New post 14 Sep 2010, 05:33
Bunuel wrote:
NUMBER THEORY
Consecutive Integers

Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers.

• Sum of \(n\) consecutive integers equals the mean multiplied by the number of terms, \(n\). Given consecutive integers \(\{-3, -2, -1, 0, 1,2\}\), \(mean=\frac{-3+2}{2}=-\frac{1}{2}\), (mean equals to the average of the first and last terms), so the sum equals to \(-\frac{1}{2}*6=-3\).

• If n is odd, the sum of consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If n is even, the sum of consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• 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 Set

Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers \(\{9,13,17,21\}\) is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set.

• 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:
\(a_ n=a_1+d(n-1)\)

• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula \(mean=median=\frac{a_1+a_n}{2}\), where \(a_1\) is the first term and \(a_n\) is the last term. Given the set \(\{7,11,15,19\}\), \(mean=median=\frac{7+19}{2}=13\).

• The sum of the elements in any evenly spaced set is given by:
\(Sum=\frac{a_1+a_n}{2}*n\), the mean multiplied by the number of terms. OR, \(Sum=\frac{2a_1+d(n-1)}{2}*n\)

• Special cases:
Sum of n first integers: \(1+2+...+n=\frac{1+n}{2}*n\)

Sum of n first odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n-1\). Given \(n=5\) first odd integers, then their sum equals to \(1+3+5+7+9=5^2=25\).

Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\).

• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65.


There seems to be a discrepancy in what some study guides consider consecutive integers. In Kaplan Premier 2011 consecutive integers are defined as numbers that occur at a fixed interval or exhibit a fixed pattern. However, on the Kaplan Free Practice Test I got a DS question wrong because it didn't consider evenly spaced numbers to necessarily be consecutive. Your definition also separates the two. Could anyone clarify which is correct so I know for the actual GMAT. Thanks!
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New post 20 Sep 2010, 00:35
Kronax wrote:
7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.


Note that you can perform this operation number of times. Also you won't need to check divisibility by 7 for huge numbers on GMAT.
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New post 28 Oct 2010, 18:42
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Quote:
Perfect Square

A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square.

There are some tips about the perfect square:.
Perfect square always has even number of powers of prime factors.


Bunuel : can you please give an example of bold statement

Thanks
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New post 01 Nov 2010, 11:09
If n is a positive integer greater than 1, then there is always a prime number P with n<P<2n

n<p<2n can someone please explain this with example .

Thanks
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New post 18 Nov 2010, 23:47
Hello Bunuel - thank you so much for this fantastic post!

with regards to checking for primality:

Quote:
Verifying the primality (checking whether the number is a prime) of a given number can be done by trial division, that is to say dividing by all integer numbers smaller than , thereby checking whether is a multiple of .
Example: Verifying the primality of : is little less than , from integers from to , is divisible by , hence is not prime.


Would it be accurate to say that a number is prime ONLY if it gives a remainder of 1 or 5 when divided by 6?
i.e, for eg. 10973/6 gives a remainder of 5, so it has to be prime...

i found the reasoning behind this in one of the OG solutions:
prime numbers always take the form: 6n+1 or 6n+5 ....

the only possible remainders when any number is divided by 6 are [0,1,2,3,4,5] ...
A prime number always gives a remainder of 1 or 5, because:
a) if the remainder is 2 or 4, then the number must be even
b) if the remainder is 3, then it is divisible by 3 ...

hence, if a number divided by 6 yields 1 or 5 as its remainder, then it must be prime
...?

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New post 05 Dec 2010, 16:36
\((a^m)^n=a^{mn}\) ----------1

\((2^2)^2 = 2^2*^2 =2^4\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) ------------------2

\(2^2^2 = 2^(2^2) = 2^4\)

If above example is correct then whats the difference 1 & 2. Please clarify
thanks
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New post 05 Dec 2010, 20:36
shrive555 wrote:
\((a^m)^n=a^{mn}\) ----------1

\((2^2)^2 = 2^2*^2 =2^4\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) ------------------2

\(2^2^2 = 2^(2^2) = 2^4\)

If above example is correct then whats the difference 1 & 2. Please clarify
thanks



I think its just that you have taken a bad example here. Consider a=2, m=3, b=2

\((a^m)^n=(2^3)^2=8^2=64\)
\(a^{(m^n)}=2^{(3^2)}=2^9=512\)
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