GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 05 Dec 2019, 08:13 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Math: Number Theory

Author Message
TAGS:

### Hide Tags

Manager  Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 243

### Show Tags

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 ( )
_________________
I'm the Dumbest of All !!
Intern  Joined: 04 Dec 2010
Posts: 3

### Show Tags

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?

Intern  Joined: 23 Jan 2011
Posts: 7

### Show Tags

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?
Intern  Joined: 23 Jan 2011
Posts: 7

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

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).
_________________
Manager  Joined: 10 Nov 2010
Posts: 168
Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19 GMAT 2: 540 Q44 V21 WE: Information Technology (Computer Software)

### Show Tags

Any nonzero natural number n can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.

Pls give me the example of bold face text because i am not sure what does it exactly means.

Thanks
_________________
The proof of understanding is the ability to explain it.
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

GMATD11 wrote:
Any nonzero natural number n can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.

Pls give me the example of bold face text because i am not sure what does it exactly means.

Thanks

It's called the fundamental theorem of arithmetic (or the unique-prime-factorization theorem) which states that any integer greater than 1 can be written as a unique product of prime numbers.

For example: 60=2^2*3*5 --> 60 can be written as a product of primes (powers of primes) only in this unique way (you can just reorder the multiples and write 3*2^2*5 or 2^2*5*3 ...).
_________________
Manager  Joined: 23 Aug 2011
Posts: 63

### Show Tags

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 _________________
Whatever one does in life is a repetition of what one has done several times in one's life!
If my post was worth it, then i deserve kudos Intern  Joined: 14 May 2011
Posts: 5

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

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: 04 Sep 2013
Posts: 10

### Show Tags

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.

???
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

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: 17 Jan 2012
Posts: 22
Location: India
GMAT 1: 650 Q48 V31 WE: Information Technology (Telecommunications)

### Show Tags

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{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 V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

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{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: 15 Dec 2013
Posts: 3

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

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: 01 Feb 2015
Posts: 1

### Show Tags

First of all, great post! Thanks a ton for creating this resource.

I had a very quick question.

In some DS questions, I have come accross the term - "Range of n integers"

I first assumed, range would mean the number of terms.

I was able to get some of the questions correct, using this but I think I got very lucky. Primarily because when I use different methods to check and practice the question, it leads me to a different answer.

Any chance you can let me know if my assumption was correct? If so, any suggestions how best to tackle these questions in the least amount of time.

Many thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

ud19 wrote:
First of all, great post! Thanks a ton for creating this resource.

I had a very quick question.

In some DS questions, I have come accross the term - "Range of n integers"

I first assumed, range would mean the number of terms.

I was able to get some of the questions correct, using this but I think I got very lucky. Primarily because when I use different methods to check and practice the question, it leads me to a different answer.

Any chance you can let me know if my assumption was correct? If so, any suggestions how best to tackle these questions in the least amount of time.

Many thanks!

The range of a set is the difference between the largest and the smallest numbers of a set. For example, the range of {1, 10, 12} is 12 - 1 = 11 and the range of {-7, 0, 2, 9} is 9 - (-7) = 16.

DS Statistics and Sets problems to practice: search.php?search_id=tag&tag_id=34
PS Statistics and Sets problems to practice: search.php?search_id=tag&tag_id=55

Hope it helps.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59561

### Show Tags

IanStewart wrote:
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.

Than you Ian. Edited the typo.
_________________
Intern  S
Status: Current Student
Joined: 27 Mar 2014
Posts: 26
Concentration: Entrepreneurship, Finance

### Show Tags

In Roots section you wrote:

$$\sqrt{x^2}$$=|x|, when x0, then, $$\sqrt{x^2}$$=-x and when x≥0, then $$\sqrt{x^2}$$=x.

Here, won't it be "when x<0, then $$\sqrt{x^2}$$=−x"? Re: Math: Number Theory   [#permalink] 08 Aug 2015, 23:27

Go to page   Previous    1   2   3   4   5    Next  [ 81 posts ]

Display posts from previous: Sort by

# Math: Number Theory  