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Math: Number Theory

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Manager
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14 Nov 2010, 13:32
Thanks once again Bunuel...
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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
...?

-Raj
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19 Nov 2010, 02:18
Quote:
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.

Understood Sir!
.. i'll just use it one way; i.e, if i'm told that n is a prime number>3, then i can express it as 6n+1 or 6n+5

I think I just got a bit too excited about it that I forgot to thoroughly test it thru...

thx again for the prompt reply!

Raj
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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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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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05 Dec 2010, 22:16
shrouded1 wrote:
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

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

In question would that be given explicitly ... i mean the Brackets ( )
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30 Dec 2010, 10:39
Thanks for this. It made my life easier! Kudossssssss!
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30 Dec 2010, 11:02
awesome work bunnel.I really appreciate u r work
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30 Dec 2010, 11:11
we need more in GMAT math book.It contain awesome post but few.you are providing all this stuff for free.I hardly believe it.but it's true.
+10
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03 Jan 2011, 09:42
Bunuel,

For determining last digit of a power for numbers 0, 1, 5, and 6, I am not clear on how to determine the last digit.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.

What is the last digit of 345^27 ---is the last digit 5?
What is the last digit of 216^32----is the last digit 6?
What is the last digit of 111^56---is the last digit 1?

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03 Jan 2011, 11:22
As always, thank you , Bunuel!
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21 Jan 2011, 06:33
This is THE BEST thing anyone has ever posted. THANK YOU SO MUCH.
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01 Feb 2011, 07:29
Hi Folks,

I am having a small confusion between two concepts for which one of my practice Q went wrong.
During my elementary school I have studied BODMAS
B - Brackets
O - Of
D- Division
M-Mulitplication
S- Substraction

I tried with this approach and it went wrong, while i was going through this again i happened to see a difference between PEMDAS & BODMAS (Multiplication order is different) .

Can somebody help me to understand which one i should follow.

Thanks
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01 Feb 2011, 07:34
gmat709 wrote:
Hi Folks,

I am having a small confusion between two concepts for which one of my practice Q went wrong.
During my elementary school I have studied BODMAS
B - Brackets
O - Of
D- Division
M-Mulitplication
S- Substraction

I tried with this approach and it went wrong, while i was going through this again i happened to see a difference between PEMDAS & BODMAS (Multiplication order is different) .

Can somebody help me to understand which one i should follow.

Thanks
Humble GMAT ASPIRANT

The rule mentioned in the initial post is correct.

Anyway: what difference are you talking about? Can you give an example?
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01 Feb 2011, 10:07
WOW !!! Great !!!! +Kudo
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21 Feb 2011, 18:41
Quote:
If is a prime number and is a factor of then is a factor of or is a factor of .

2 is a prime number. 2 is a factor of 12*16. This implies that 2 is a factor of both 12 and 16. Am I missing something here?
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21 Feb 2011, 18:44
Quote:
If p is a prime number and p is a factor of ab then p is a factor of a or p is a factor of b.

Sorry did not quote the post correctly
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21 Feb 2011, 19:03
bugSniper wrote:
Quote:
If p is a prime number and p is a factor of ab then p is a factor of a or p is a factor of b.

Sorry did not quote the post correctly

This is inclusive *or* (as almost always on the GMAT): p is a factor of a or p is a factor of b (or both).
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27 Feb 2011, 07:44
Hi Bunuel,

I need the list of number theory questions from OG-12 and OG-11 for both Ps and DS.Have you listed them some where.

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27 Feb 2011, 08:00
GMATD11 wrote:
Hi Bunuel,

I need the list of number theory questions from OG-12 and OG-11 for both Ps and DS.Have you listed them some where.

Sorry, don't have such a list.
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Re: Math: Number Theory   [#permalink] 27 Feb 2011, 08:00

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