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# Math: Number Theory

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Re: Math: Number Theory [#permalink]  25 Sep 2012, 09:43
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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

The highest power of a prime number "k" that divides any number "n!" is given by the formula
n/K + n/k^2+n/k^3.. (until numerator becomes lesser than the denominator). Remember to truncate the remainders of each expression

E.g : The highest number of 2's in 10! is
10/2 + 10/4 + 10/8 = 5 + 2 + 1 = 8 (Truncate the reminder of each expression)

As a consequence of this, the number of zeros in n! is controlled by the presence of 5s.
Why ? 2 reasons

a) 10 = 5 x 2,
b) Also in any n!, the number of 5's are far lesser than the number of 2's.

The number of cars that you make depends on the number of engines. You can have 100 engines and 1000 cars, but you can only make 100 cars (each car needs an engine !)

10 ! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Lets factorize each term ...
10! = (5 x 2) x(3x3)x(2x2x2)x7x(2x3)x(5)X(2x2)x1
the number of 5s = 2
The number of 2s = 7
The number of zeros in 10! = the total number of 5s = 2 (You may use a calc to check this10! = 3628800)

hence in any n! , the number of 5's control the number of zeros.

As a consequence of this, the number of 5's in any n! is
n/5 + n/25 + n/125 ..until numerator becomes lesser than denominator.

Again, i want to emphasize that this formuala only works for prime numbers !!
So to find the number of 10's in any n!, DO NOT DIVIDE by 10 ! (10 is not prime !)
i.e DONT do
n/10 + n/100 + n/1000 - THIS IS WRONG !!!
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Re: Math: Number Theory [#permalink]  27 Oct 2012, 01:00
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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
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Re: Math: Number Theory [#permalink]  27 Oct 2012, 01:34
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$$(a + b)^2 = a^2+ 2ab + b^2$$ 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
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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!!
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Re: Math: Number Theory [#permalink]  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?
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Re: Math: Number Theory [#permalink]  14 Jan 2014, 00:56
Expert's post
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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Re: Math: Number Theory   [#permalink] 14 Jan 2014, 00:56

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