Math: Number Theory : GMAT Quantitative Section - Page 7
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Math: Number Theory

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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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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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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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24 Sep 2012, 03:01
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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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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12 Dec 2012, 02:33
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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25 Dec 2012, 07:33
Thanks for the post!
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27 Dec 2012, 10:02
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 !
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10 Jul 2013, 23:05
Bumping for review*.

*New project from GMAT Club!!! Check HERE

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12 Jul 2013, 02:12
Great Post. thanks a lot
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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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02 Oct 2013, 19:39
Awesome post.
Bunuel, you are great. Love your posts.
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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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14 Oct 2013, 10:18
Quite informative and descriptive... all at one place
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17 Oct 2013, 02:33
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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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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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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29 Oct 2013, 23:31
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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06 Nov 2013, 17:51
this is insanely helpful..thank you so much!!
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13 Jan 2014, 19:57
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?
Re: Math: Number Theory   [#permalink] 13 Jan 2014, 19:57

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