It is currently 24 Mar 2018, 16:39

### 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

Intern
Status: Active
Joined: 30 Jun 2012
Posts: 37
Location: India

### Show Tags

27 Oct 2012, 02:00
3
KUDOS
I shall like to add one trick to find square of number :

To find square of a number $$(ab)^2$$ where a is tens digit and b is units digit =>

1. Multiply tens digit a by (a+1) i.e. a * (a+1) = A
2. Suffix 25 to the above derived result A and the new number will be A25

Example: what is $$35^2$$?
Solution: 3 * (3+1) = 3*4 = 12 and so answer is 1225
_________________

Thanks and Regards!

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

Intern
Status: Active
Joined: 30 Jun 2012
Posts: 37
Location: India

### Show Tags

27 Oct 2012, 02:34
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.

Intern
Joined: 14 May 2011
Posts: 10

### Show Tags

12 Dec 2012, 03:26
Hi,

I'm not sure whether I undertood the below rule correctly:

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

55^2 = 3025 - the last digit is same as the base (5) so the above rule works.
55^10 = 253295162119141000 - the last digit is not same as the base (5) so the above rule doesn't work.

Math Expert
Joined: 02 Sep 2009
Posts: 44423

### Show Tags

12 Dec 2012, 03:33
klueless7825 wrote:
Hi,

I'm not sure whether I undertood the below rule correctly:

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

55^2 = 3025 - the last digit is same as the base (5) so the above rule works.
55^10 = 25329516211914100[b]0[/b] - the last digit is not same as the base (5) so the above rule doesn't work.

5 in any positive integer power has 5 as the units digit.

5^1=5;
5^2=25;
5^3=125
...
5^10=253,295,162,119,140,625 (your result was just rounded).

Hope it's clear.
_________________
Intern
Joined: 25 Dec 2012
Posts: 2

### Show Tags

25 Dec 2012, 08:33
Thanks for the post!
Retired Moderator
Joined: 27 Aug 2012
Posts: 1168

### Show Tags

27 Dec 2012, 11:02
Now this is a huge man..it's really BIG and I think one of the most critical areas tested in GMAT Quant.

You've really done a mammoth job Bunuel..Hats Off.....

Kudos !
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 44423

### Show Tags

11 Jul 2013, 00:05
Bumping for review*.

*New project from GMAT Club!!! Check HERE

_________________
Intern
Joined: 07 Jul 2013
Posts: 30
Location: India
Concentration: Finance, Economics
GMAT Date: 08-19-2013
GPA: 3.2
WE: Information Technology (Computer Software)

### Show Tags

12 Jul 2013, 03:12
Great Post. thanks a lot
Intern
Joined: 04 Sep 2013
Posts: 19

### Show Tags

15 Sep 2013, 10:04
Hi!!! can you explain the number line?? and how we can figure out the radicals on it???
Intern
Joined: 30 Apr 2013
Posts: 16
Location: India

### Show Tags

02 Oct 2013, 20:39
Awesome post.
Bunuel, you are great. Love your posts.
_________________

GMAT RC Vocab - No nonsense(Only for GMAT)
http://gmatclub.com/forum/gmat-rc-vocab-no-nonsense-only-for-gmat-162129.html#p1283165

Quant Document to revise a week before exam - Mixedbag
http://gmatclub.com/forum/document-to-revise-a-week-before-exam-mixedbag-162145.html

Best questions to revise few days before exam- Mixed bag(25)
http://gmatclub.com/forum/best-questions-to-revise-few-days-before-exam-mixed-bag-162124.html#p1283141

Intern
Joined: 04 Sep 2013
Posts: 19

### Show Tags

14 Oct 2013, 10:47
can someone explain me this property:

If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

???
Intern
Joined: 10 Jun 2013
Posts: 10
GMAT 1: 700 Q49 V35
GPA: 3.54

### Show Tags

14 Oct 2013, 11:18
Quite informative and descriptive... all at one place
Math Expert
Joined: 02 Sep 2009
Posts: 44423

### Show Tags

17 Oct 2013, 03:33
skamran wrote:
can someone explain me this property:

If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

???

Say a=2, b=3 (gcd(a,b)=gcd(2,3)=1), and c=4.

a=2 IS a factor of bc=12, and a=2 IS a factor of c.

OR: if a is a factor of bc and NOT a factor of b, then it must be a factor of c.

Hope it's clear.
_________________
Intern
Joined: 04 Sep 2013
Posts: 19

### Show Tags

17 Oct 2013, 16:52
Bunuel wrote:
skamran wrote:
can someone explain me this property:

If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

???

Say a=2, b=3 (gcd(a,b)=gcd(2,3)=1), and c=4.

a=2 IS a factor of bc=12, and a=2 IS a factor of c.

OR: if a is a factor of bc and NOT a factor of b, then it must be a factor of c.

Hope it's clear.

Yeh Thanks alot!!!
Intern
Joined: 17 Jan 2012
Posts: 26
Location: India
GMAT 1: 650 Q48 V31
WE: Information Technology (Telecommunications)

### Show Tags

29 Oct 2013, 21:06
Hello Bunuel,

• \sqrt{x^2}=|x|, when x\leq{0}, then \sqrt{x^2}=-x and when x\geq{0}, then \sqrt{x^2}=x

• When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.

Isn't both of these points contradict each other?

If I consider second point as valid then how can \sqrt{x^2}=|x|, when x\leq{0}, then \sqrt{x^2}=-x be said ?
Math Expert
Joined: 02 Sep 2009
Posts: 44423

### Show Tags

30 Oct 2013, 00:31
Phoenix22 wrote:
Hello Bunuel,

• \sqrt{x^2}=|x|, when x\leq{0}, then \sqrt{x^2}=-x and when x\geq{0}, then \sqrt{x^2}=x

• When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.

Isn't both of these points contradict each other?

If I consider second point as valid then how can \sqrt{x^2}=|x|, when x\leq{0}, then \sqrt{x^2}=-x be said ?

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$
_________________
Intern
Joined: 03 Nov 2013
Posts: 2

### Show Tags

06 Nov 2013, 18:51
this is insanely helpful..thank you so much!!
Intern
Joined: 15 Dec 2013
Posts: 3

### Show Tags

13 Jan 2014, 20:57
Bunuel wrote:
LAST DIGIT OF A POWER

Determining the last digit of $$(xyz)^n$$:

1. Last digit of $$(xyz)^n$$ is the same as that of $$z^n$$;
2. Determine the cyclicity number $$c$$ of $$z$$;
3. Find the remainder $$r$$ when $$n$$ divided by the cyclisity;
4. When $$r>0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^r$$ and when $$r=0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^c$$, where $$c$$ is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. $$(xy4)^n$$) have a cyclisity of 2. When n is odd $$(xy4)^n$$ will end with 4 and when n is even $$(xy4)^n$$ will end with 6.
• Integers ending with 9 (eg. $$(xy9)^n$$) have a cyclisity of 2. When n is odd $$(xy9)^n$$ will end with 9 and when n is even $$(xy9)^n$$ will end with 1.

Example: What is the last digit of $$127^{39}$$?
Solution: Last digit of $$127^{39}$$ is the same as that of $$7^{39}$$. Now we should determine the cyclisity of $$7$$:

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...

So, the cyclisity of 7 is 4.

Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of $$127^{39}$$ is the same as that of the last digit of $$7^{39}$$, is the same as that of the last digit of $$7^3$$, which is $$3$$.

Congratulation and thank you very much for the post, but in the LAST DIGIT OF A POWER i have an issue, when i try to solve the last digit of (456)^35 with the process i just don't get the correct answers, with the process above gives me 6^4 which is 1296=6 and with calculator its 0, can you explain me that case?
Math Expert
Joined: 02 Sep 2009
Posts: 44423

### Show Tags

14 Jan 2014, 01:56
mandrake15 wrote:
Bunuel wrote:
LAST DIGIT OF A POWER

Determining the last digit of $$(xyz)^n$$:

1. Last digit of $$(xyz)^n$$ is the same as that of $$z^n$$;
2. Determine the cyclicity number $$c$$ of $$z$$;
3. Find the remainder $$r$$ when $$n$$ divided by the cyclisity;
4. When $$r>0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^r$$ and when $$r=0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^c$$, where $$c$$ is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. $$(xy4)^n$$) have a cyclisity of 2. When n is odd $$(xy4)^n$$ will end with 4 and when n is even $$(xy4)^n$$ will end with 6.
• Integers ending with 9 (eg. $$(xy9)^n$$) have a cyclisity of 2. When n is odd $$(xy9)^n$$ will end with 9 and when n is even $$(xy9)^n$$ will end with 1.

Example: What is the last digit of $$127^{39}$$?
Solution: Last digit of $$127^{39}$$ is the same as that of $$7^{39}$$. Now we should determine the cyclisity of $$7$$:

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...

So, the cyclisity of 7 is 4.

Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of $$127^{39}$$ is the same as that of the last digit of $$7^{39}$$, is the same as that of the last digit of $$7^3$$, which is $$3$$.

Congratulation and thank you very much for the post, but in the LAST DIGIT OF A POWER i have an issue, when i try to solve the last digit of (456)^35 with the process i just don't get the correct answers, with the process above gives me 6^4 which is 1296=6 and with calculator its 0, can you explain me that case?

Any integer with 6 as its units digit in any positive integer power has the units digit of 6 (integers ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.). For example, (xxx6)^(positive integer) has the units digit of 6.

The reason you get 0 as the units digit of (456)^35 is because it's a huge number and simple calculator rounds the result.

Exact result is: 1,158,162,485,059,181,044,784,824,077,056,791,483,879,723,809,565,243,305,114,019,731,744,476,935,058,125,438,332,149,170,176.

1 trigintillion 158 novemvigintillion 162 octovigintillion 485 septenvigintillion 59 sexvigintillion 181 quinvigintillion 44 quattuorvigintillion 784 trevigintillion 824 duovigintillion 77 unvigintillion 56 vigintillion 791 novemdecillion 483 octodecillion 879 septendecillion 723 sexdecillion 809 quindecillion 565 quattuordecillion 243 tredecillion 305 duodecillion 114 undecillion 19 decillion 731 nonillion 744 octillion 476 septillion 935 sextillion 58 quintillion 125 quadrillion 438 trillion 332 billion 149 million 170 thousand 176

Hope it's clear.
_________________
Intern
Joined: 19 Jan 2014
Posts: 13

### Show Tags

09 Feb 2014, 09:07
any body can explain that why we need to find the sum of number's factors ????
Re: Math: Number Theory   [#permalink] 09 Feb 2014, 09:07

Go to page   Previous    1   2   3   4   5   6   7   8   9   10   11    Next  [ 214 posts ]

Display posts from previous: Sort by