Math: Number Theory : GMAT Quantitative Section - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 10:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Math: Number Theory

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 958 [2] , given: 25

### Show Tags

28 Oct 2010, 21:28
2
KUDOS
$$36=6^2=2^2*3^2$$

Powers of 2 & 3 are even (2)

Posted from my mobile device
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 958 [2] , given: 25

### Show Tags

01 Nov 2010, 15:16
2
KUDOS
shrive555 wrote:
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

The result you are referring to is a weak form of what is known as Bertrand's Postulate. The proof of this result is beyond the scope of the GMAT, but it is easy to show some examples.

Choose any n>1, you will always find a prime number between n & 2n.

Eg. n=5, 2n=10 ... p=7 lies in between
n=14, 2n=28 ... p=19 lies in between
n=20, 2n=40 ... p=23 lies in between
_________________
Intern
Status: Active
Joined: 30 Jun 2012
Posts: 37
Location: India
Followers: 5

Kudos [?]: 82 [2] , given: 36

### Show Tags

27 Oct 2012, 01:34
2
KUDOS

$$(a + b)^2 = a^2+ 2ab + b^2$$ Square of a Sum
$$(a - b)^2 = a^2 - 2ab + b^2$$ Square of a Diffe rence

$$a^n - b^n$$ is always divisble by a-b i.e. irrespective of n being odd or even
Proof:
$$a^2 - b^2 = (a-b)(a+b)$$
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$

Thus divisible by a- b in both cases where n = 2 i.e. even and 3 i.e. odd

$$a^n + b^n$$ is divisble by a+b i.e. only if n = odd
Proof:
$$a^3 - b^3 = (a+b)(a^2-ab+b^2)$$
Thus divisible by a + b as n = 3 i.e. odd
_________________

Thanks and Regards!

P.S. +Kudos Please! in case you like my post.

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1183
Followers: 422

Kudos [?]: 1512 [2] , given: 4

### Show Tags

12 Jun 2015, 10:27
2
KUDOS
Expert's post
Bunuel wrote:

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 $$\sqrt{n}$$, thereby checking whether $$n$$ is a multiple of $$m<\sqrt{n}$$.
Example: Verifying the primality of $$161$$: $$\sqrt{161}$$ is little less than $$13$$, from integers from $$2$$ to $$13$$, $$161$$ is divisible by $$7$$, hence $$161$$ is not prime.

A minor point, but the inequalities here should not be strict. If you want to test if some large integer n is prime, then you need to try dividing by numbers up to and including $$\sqrt{n}$$. We must include $$\sqrt{n}$$, in case our number is equal to the square of a prime.

And it might be worth mentioning that it is only necessary to try dividing by prime numbers up to $$\sqrt{n}$$, since if n has any divisors at all (besides 1 and n), then it must have a prime divisor.

It's very rare, though, that one needs to test if a number is prime on the GMAT. It is, computationally, extremely time-consuming to test if a large number is prime, so the GMAT cannot ask you to do that. If a GMAT question asks if a large number is prime, the answer really must be 'no', because while you can often quickly prove a large number is not prime (for example, 1,000,011 is not prime because it is divisible by 3, as we see by summing digits), you cannot quickly prove that a large number is prime.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Intern
Joined: 06 Oct 2009
Posts: 33
Schools: Ryerson University
Followers: 1

Kudos [?]: 10 [1] , given: 7

### Show Tags

24 Jan 2010, 06:33
1
KUDOS
no problem.

Great post by the way, very informative.
Intern
Joined: 06 Oct 2009
Posts: 33
Schools: Ryerson University
Followers: 1

Kudos [?]: 10 [1] , given: 7

### Show Tags

24 Jan 2010, 14:06
1
KUDOS
Quote:
ding is simplifying a number to a certain place value. To round the decimal 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.

Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.249, since the dropped 5 is equal to 5.

I'm assuming it was just a typo for the last part of the example:
Its entered as 5.3485 rounded to the nearest thousandth = 5.249, since the dropped 5 is equal to 5.

I guess you meant 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
Intern
Joined: 06 Oct 2009
Posts: 33
Schools: Ryerson University
Followers: 1

Kudos [?]: 10 [1] , given: 7

### Show Tags

25 Jan 2010, 14:54
1
KUDOS
I have a question regarding number properties, which I found on an old GMAT test paper form. Here it is:

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?

(a) 108
(b) 119
(c) 128
(d) 135
(e) 143

Is there a more efficient way of solving this than choosing two numbers at random?
Intern
Joined: 06 Oct 2009
Posts: 33
Schools: Ryerson University
Followers: 1

Kudos [?]: 10 [1] , given: 7

### Show Tags

26 Jan 2010, 06:15
1
KUDOS
Thank you very much, sorry about that, will do from now on.
Intern
Joined: 12 Oct 2009
Posts: 16
Followers: 1

Kudos [?]: 3 [1] , given: 1

### Show Tags

26 Jan 2010, 20:01
1
KUDOS
Great Post, thanks a lot
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93653 [1] , given: 10583

### Show Tags

31 Jan 2010, 10:34
1
KUDOS
Expert's post
ariel wrote:
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?

Not so. Divisibility by 7 does not check whether the number is prime or not.

Actually this issue is covered in the post. First you should know that all prime numbers except 2 and 5 end in 1, 3, 7 or 9. So if it ends in some other digit it's not prime.

Next, if the above didn't help (meaning that number ends in 1, 3, 7 or 9) there is a way to check whether the number is prime or not. Walker gave an example how to do this, but here it is again:

Verifying the primality of a given number $$n$$ can be done by trial division, that is to say dividing $$n$$ by all integer numbers smaller than $$\sqrt{n}$$, thereby checking whether $$n$$ is a multiple of $$m<\sqrt{n}$$.

Examples: Verifying the primality of $$161$$: $$\sqrt{161}$$ is little less than $$13$$. We should check $$161$$ on divisibility by numbers from 2 to 13. From integers from $$2$$ to $$13$$, $$161$$ is divisible by $$7$$, hence $$161$$ is not prime.

Verifying the primality of $$149$$: $$\sqrt{149}$$ is little more than $$12$$. We should check $$149$$ on divisibility by numbers from 2 to 12, inclusive. $$149$$ is not divisible by any of the integers from $$2$$ to $$12$$, hence $$149$$ is prime.

Verifying the primality of $$73$$: $$\sqrt{73}$$ is little less than $$9$$. We should check $$73$$ on divisibility by numbers from 2 to 9. $$73$$ is not divisible by any of the integers from $$2$$ to $$9$$, hence $$149$$ is prime.

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93653 [1] , given: 10583

### Show Tags

05 Mar 2010, 00:22
1
KUDOS
Expert's post
The topic is done. At last!

I'll break it into several smaller ones in a day or two.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93653 [1] , given: 10583

### Show Tags

30 Apr 2010, 13:30
1
KUDOS
Expert's post
AloneAndInsufficient wrote:
Bunuel wrote:
NUMBER THEORY
• For GMAT it's good to memorize following values:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{7}\approx{2.45}$$
$$\sqrt{8}\approx{2.65}$$
$$\sqrt{10}\approx{2.83}$$

Anyone else notice that these are wrong?
They should be:
• For GMAT it's good to memorize following values:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{6}\approx{2.45}$$
$$\sqrt{7}\approx{2.65}$$
$$\sqrt{8}\approx{2.83}$$
$$\sqrt{10}\approx{3.16}$$

Thanks. Edited. +1 for spotting this.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93653 [1] , given: 10583

### Show Tags

19 Nov 2010, 00:53
1
KUDOS
Expert's post
Araj wrote:
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

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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93653 [1] , given: 10583

### Show Tags

06 Dec 2010, 00:08
1
KUDOS
Expert's post
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

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$, which on the other hand equals to $$a^{mn}$$.

So:
$$(a^m)^n=a^{mn}$$;

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.

Now, there are some specific values of $$a$$, $$m$$ and $$n$$ for which $$a^m^n$$ equals to $$a^{mn}$$. For example:
$$a=1$$: $$1^{m^n}=1=1^{mn}$$;

$$m=0$$: $$a^0^n=a^0=1$$ and $$a^{0*n}=a^0=1$$;

$$m=2$$ and $$n=2$$ --> $$a^{2^2}=a^4$$ and $$a^{2*2}=a^4$$;

$$m=4$$ and $$n=\frac{1}{2}$$ --> $$a^{4^{\frac{1}{2}}}=a^2$$ and $$a^{4*{\frac{1}{2}}}=a^2$$;
...

So, generally $$a^m^n$$ does not equal to $$(a^m)^n$$, but for specific values of given variables it does.

shrive555 wrote:
In question would that be given explicitly ... i mean the Brackets ( )

$$a^m^n$$ ALWAYS means $$a^{(m^n)}$$, so no brackets are needed. For example $$2^{3^4}=2^{(3^4)}=2^{81}$$;

If GMAT wants the order of operation to be different then the necessary brackets will be put. For example: $$(2^3)^4=2^{(3*4)}=2^{12}$$.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93653 [1] , given: 10583

### Show Tags

03 Jan 2011, 08:52
1
KUDOS
Expert's post
resh924 wrote:
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?

First of all: last digit of 345^27 is the same as that of 5^27 (the same for 216^32 and 111^56);

Next:
1 in any integer power is 1;
5^1=5, 5^2=25, 5^3=125, ...
6^1=6, 6^2=36, 5^3=216, ...

So yes, integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base: thus 0, 1, 5, and 6 respectively.

Hope it's clear.
_________________
Manager
Joined: 18 Jan 2012
Posts: 51
Location: United States
Followers: 3

Kudos [?]: 91 [1] , given: 26

### Show Tags

25 Sep 2012, 09:43
1
KUDOS
conty911 wrote:
Bunuel wrote:
NUMBER THEORY

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer $$n$$, can be determined with this formula:

$$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}$$, where k must be chosen such that $$5^k<n$$.

It's easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of $$32!$$?
$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)

Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

I noticed in case the number (n) is multiple of $$5^k$$ and we have to find number of trailing zero zeroes, then it will be $$5^k<=n$$ rather $$5^k<n$$

no of trailing zeros in 25! =6

$$\frac{25}{5}+\frac{25}{5^2}= 5+1$$;
Please correct me, clarify if i'm wrong. Thanks

The highest power of a prime number "k" that divides any number "n!" is given by the formula
n/K + n/k^2+n/k^3.. (until numerator becomes lesser than the denominator). Remember to truncate the remainders of each expression

E.g : The highest number of 2's in 10! is
10/2 + 10/4 + 10/8 = 5 + 2 + 1 = 8 (Truncate the reminder of each expression)

As a consequence of this, the number of zeros in n! is controlled by the presence of 5s.
Why ? 2 reasons

a) 10 = 5 x 2,
b) Also in any n!, the number of 5's are far lesser than the number of 2's.

The number of cars that you make depends on the number of engines. You can have 100 engines and 1000 cars, but you can only make 100 cars (each car needs an engine !)

10 ! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Lets factorize each term ...
10! = (5 x 2) x(3x3)x(2x2x2)x7x(2x3)x(5)X(2x2)x1
the number of 5s = 2
The number of 2s = 7
The number of zeros in 10! = the total number of 5s = 2 (You may use a calc to check this10! = 3628800)

hence in any n! , the number of 5's control the number of zeros.

As a consequence of this, the number of 5's in any n! is
n/5 + n/25 + n/125 ..until numerator becomes lesser than denominator.

Again, i want to emphasize that this formuala only works for prime numbers !!
So to find the number of 10's in any n!, DO NOT DIVIDE by 10 ! (10 is not prime !)
i.e DONT do
n/10 + n/100 + n/1000 - THIS IS WRONG !!!
_________________

-----------------------------------------------------------------------------------------------------
IT TAKES QUITE A BIT OF TIME AND TO POST DETAILED RESPONSES.
YOUR KUDOS IS VERY MUCH APPRECIATED

-----------------------------------------------------------------------------------------------------

Intern
Joined: 01 Apr 2014
Posts: 4
Schools: Ross '17, Tepper '17
GMAT 1: 690 Q49 V34
Followers: 0

Kudos [?]: 1 [1] , given: 27

### Show Tags

17 Sep 2014, 21:09
1
KUDOS
lbnyc13 wrote:
Hi -

Can somebody explain where the '00' part of the 9900 in the denominator is coming from? I understand where the '99' is coming from. I copied this from the number theory section.

Thanks

" Example #2: Convert 0.2512(12) to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300. "

let x =0.2512(12)
100x = 25.(12)
10000x = 2512.(12)

10000x -100x = 2512.(12) - 25.(12) = 2487
9900x = 2487
x = 2487/9900

That is the logic. Hope that helps.
Manager
Joined: 22 Jan 2010
Posts: 121
Followers: 2

Kudos [?]: 5 [0], given: 15

### Show Tags

26 Jan 2010, 21:57
Thanks for sharing.
Current Student
Joined: 12 Nov 2008
Posts: 368
Schools: Ross (R2), Cornell (R3) , UNC (R3) , INSEAD (R1 Jan)
WE 2: FP & Analysis (2 yrs at matriculation)
Followers: 23

Kudos [?]: 105 [0], given: 45

### Show Tags

31 Jan 2010, 09:00
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!
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 550

Kudos [?]: 3566 [0], given: 360

### Show Tags

31 Jan 2010, 09:19
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
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Re: Math: Number Theory   [#permalink] 31 Jan 2010, 09:19

Go to page   Previous    1   2   3   4   5   6   7   8   9   10    Next  [ 187 posts ]

Similar topics Replies Last post
Similar
Topics:
2 A number theory curiosity 5 02 Aug 2011, 13:09
20 If a and b are different positive integers and a+b=a(a+b) 13 14 Sep 2010, 18:46
1 Qs about Number Theory 3 17 May 2010, 20:24
62 Math: Number Theory - Percents 50 22 Mar 2010, 14:24
12 Math: Number Theory (broken into smaller topics) 11 10 Mar 2010, 05:20
Display posts from previous: Sort by