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# Math: Number Theory

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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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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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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, 10:04
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
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27 Feb 2011, 10:12
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 ...).
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04 Sep 2012, 19:34
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
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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.

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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.
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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.

???
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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.
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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 ?
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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 $$\sqrt{x^2}=|x|$$
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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?
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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.
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14 Feb 2015, 23:10
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!
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16 Feb 2015, 03:25
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.
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12 Jun 2015, 11:52
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.
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08 Aug 2015, 23:27
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

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