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# Math: Number Theory

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Re: Math: Number Theory [#permalink]  01 Jun 2012, 12:27
Thanks a lot for the detailed post ! This is sure to help with my recurrent mistakes on number properties
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Re: Math: Number Theory [#permalink]  04 Sep 2012, 18:34
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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Re: Math: Number Theory [#permalink]  04 Sep 2012, 23:13
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

You are right. $$k$$ is the highest power of 5 not exceeding $$n.$$
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Re: Math: Number Theory [#permalink]  22 Sep 2012, 19:15
Hey..Awesome post !!
Like number theory, can u pls share links for other topics as well ?
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Re: Math: Number Theory [#permalink]  24 Sep 2012, 03:01
Expert's post
154238 wrote:
Hey..Awesome post !!
Like number theory, can u pls share links for other topics as well ?

Check all topics in our Math Book: gmat-math-book-87417.html

Hope it helps.
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Re: Math: Number Theory [#permalink]  12 Dec 2012, 02:26
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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Re: Math: Number Theory [#permalink]  12 Dec 2012, 02:33
Expert's post
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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Re: Math: Number Theory [#permalink]  25 Dec 2012, 07:33
Thanks for the post!
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Re: Math: Number Theory [#permalink]  27 Dec 2012, 10:02
Expert's post
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 !
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Re: Math: Number Theory [#permalink]  10 Jul 2013, 23:05
Expert's post
Bumping for review*.

*New project from GMAT Club!!! Check HERE

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Re: Math: Number Theory [#permalink]  12 Jul 2013, 02:12
Great Post. thanks a lot
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Re: Math: Number Theory [#permalink]  15 Sep 2013, 09:04
Hi!!! can you explain the number line?? and how we can figure out the radicals on it???
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Re: Math: Number Theory [#permalink]  02 Oct 2013, 19:39
Awesome post.
Bunuel, you are great. Love your posts.
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Re: Math: Number Theory [#permalink]  14 Oct 2013, 09:47
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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Re: Math: Number Theory [#permalink]  14 Oct 2013, 10:18
Quite informative and descriptive... all at one place
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Re: Math: Number Theory [#permalink]  17 Oct 2013, 02:33
Expert's post
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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Re: Math: Number Theory [#permalink]  17 Oct 2013, 15:52
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!!!
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Re: Math: Number Theory [#permalink]  29 Oct 2013, 20:06
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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Re: Math: Number Theory [#permalink]  29 Oct 2013, 23:31
Expert's post
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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Re: Math: Number Theory [#permalink]  06 Nov 2013, 17:51
this is insanely helpful..thank you so much!!
Re: Math: Number Theory   [#permalink] 06 Nov 2013, 17:51

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# Math: Number Theory

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