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# What is the rightmost non-zero digit of 20!?

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What is the rightmost non-zero digit of 20!?  [#permalink]

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14 Aug 2009, 11:56
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Q1) What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8

Q2) What is the product of the factors of 432?
a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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14 Aug 2009, 12:17
3
sergbov123 wrote:
Q1) What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8

SOL:
Well first we will need to calculate how many zeroes are there in 20!. Since a zero in a factorial is always obtained when a 2 is multiplied with a 5, we need to find out how many powers of 5 are there in 20!. We know that in any factorial, powers of 2 will always be more than the powers of 5.

Note: Find the method for finding out the no of powers of a prime number in a factorial at the following link: factor-of-p-82381.html?view-post=617879#p617879)

Powers of 5 in 20!
20/5 + 20/5^2
= 4 + 0
= 4

Using the above method express 20! as follows:
20! = 2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19

Now from the above remove 4 2's and 4 5's as they are responsible for zeroes at the end. The remaining factors will decide the last non-zero digit.
=> 2^14 * 3^8 * 7^2 * 11 * 13 * 17 * 19

Last digit of this product is decided by the individual last digits of each number in the product.
=> 4 * 1 * 9 * 1 * 3 * 7 * 9
=> 36 * 21 * 9
=> 6 * 1 * 9
=> 54
=> 4

You sure the options are correct?
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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17 Aug 2009, 22:24
sergbov123 wrote:
Q2) What is the product of the factors of 432?

a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

Thats a good & tough question:

a) 2^90 ......... cannot be because it doesnot have any multiple/power of 3.
b) 3^78......... cannot be because it doesnot have any multiple/power of 2.
c) 2^40 * 3^30......... can be because it has multiples/power of 2 & 3.
d) 2^25 * 3^75 ......... can be because it has multiples/power of 2 & 3.
e) 2^5 * 3*10 * (2^25 * 3^30 - 1) probably not.

Lets work on c and d.
432 = (2^4) (3^3)
Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

Product of 432 = 2x3x4x6x8x9x12x16x18x24x27x36x48x54x72x108x144x216x432
= 2x3x 2^2 x 2x3 x 2^3 x 3^3 x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3
= 2^41 x 3^31

Hmmm ........ I got 2^41 x 3^31 instead of 2^40 x 3^30.

Answer should be C but how it is 2^40 x 3^30? I must have made a mistake somewhere....

Can you find it for me?
Thanx....
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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17 Aug 2009, 22:57
GMAT TIGER wrote:
sergbov123 wrote:
Q2) What is the product of the factors of 432?

a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

Thats a good & tough question:

a) 2^90 ......... cannot be because it doesnot have any multiple/power of 3.
b) 3^78......... cannot be because it doesnot have any multiple/power of 2.
c) 2^40 * 3^30......... can be because it has multiples/power of 2 & 3.
d) 2^25 * 3^75 ......... can be because it has multiples/power of 2 & 3.
e) 2^5 * 3*10 * (2^25 * 3^30 - 1) probably not.

Lets work on c and d.
432 = (2^4) (3^3)
Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

Product of 432 = 2x3x4x6x8x9x12x16x18x24x27x36x48x54x72x108x144x216x432
= 2x3x 2^2 x 2x3 x 2^3 x 3^3 x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3
= 2^41 x 3^31

Hmmm ........ I got 2^41 x 3^31 instead of 2^40 x 3^30.

Answer should be C but how it is 2^40 x 3^30? I must have made a mistake somewhere....

Can you find it for me?
Thanx....

2x
3x
2^2 x
2x3 x
2^3 x
3^3(p) x
2^2x3 x
2^4 x
2x3^2 x
2^3x3 x
3^3 x
2^2x3^2 x
2^4x3 x
2x3^3 x
2^3x3^2 x
2^2x3^3 x
2^4x3^2 x
2^3x3^3 x
2^4x3^3

3^3(p) x For 9, you have written 3^3 instead of 3^2. But thts all I could find out.
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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18 Aug 2009, 00:46
1
1
sergbov123 wrote:
Q2) What is the product of the factors of 432?
a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

SOL:
432 = 2^4 * 3^3

Its clear that the final product will only have powers of 2 & 3.

Lets start by finding the total number of powers of 2. First lets consider the factors resulting only from 2^4, i.e. with zero powers of 3:
Product = 2^1 * 2^2 * 2^3 * 2^4 = 2^10

|||ly, the number of powers of 2 with:
One power of 3 i.e. 3^1 = 2^10
Two powers of 3 i.e. 3^2 = 2^10
Three powers of 3 i.e. 3^3 = 2^10

Thus total numbers of powers of 2 = 2^(10 + 10 + 10 + 10) = 2^40

Now lets find the total number of powers of 3. First lets consider the factors resulting only from 3^3, i.e. with zero powers of 2:
Product = 3^1 * 3^2 * 3^3 = 3^6

|||ly, the number of powers of 3 with:
One power of 2 i.e. 2^1 = 3^6
Two powers of 2 i.e. 2^2 = 3^6
Three powers of 2 i.e. 2^3 = 3^6
Four powers of 2 i.e. 2^4 = 3^6

Thus total numbers of powers of 3 = 3^(6 + 6 + 6 + 6 + 6) = 3^30

Thus the product of factors of 432 = 2^40 * 3^30
ANS: C
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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18 Aug 2009, 04:20
Can you pls elaborate:
|||ly, the number of powers of 2 with:
One power of 3 i.e. 3^1 = 2^10
Two powers of 3 i.e. 3^2 = 2^10
Three powers of 3 i.e. 3^3 = 2^10

Couldn't get the idea, because we are concerned with the product of factors.
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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18 Aug 2009, 08:25
1
Economist wrote:
Can you pls elaborate:
|||ly, the number of powers of 2 with:
One power of 3 i.e. 3^1 = 2^10
Two powers of 3 i.e. 3^2 = 2^10
Three powers of 3 i.e. 3^3 = 2^10

Couldn't get the idea, because we are concerned with the product of factors.

Let me explain with the help of 36.
36 = 2^2 * 3^2

Let's count the total powers of 2 in the product of factors of 36:
2^1 * 3^0
2^2 * 3^0
= 3 powers with 3^0

2^1 * 3^1
2^2 * 3^1
= 3 powers with 3^1

2^1 * 3^2
2^2 * 3^2
= 3 powers with 3^2
Total powers of 2 in the product => 2^(3 + 3 + 3)

|||ly we will get 3 powers of 3 with 2^0, 2^1 and 2^2 each => Total of 9 powers of 3 in the product.

Final Product = 2^9 * 3^9

Hope this helps!
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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18 Aug 2009, 09:22
Thats the best approach. I was also considering using combination approach.

Although you did not mention combination in your solution, your approach is similar to that.

Good job...

samrus98 wrote:
sergbov123 wrote:
Q2) What is the product of the factors of 432?
a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

SOL:
432 = 2^4 * 3^3

Its clear that the final product will only have powers of 2 & 3.

Lets start by finding the total number of powers of 2. First lets consider the factors resulting only from 2^4, i.e. with zero powers of 3:
Product = 2^1 * 2^2 * 2^3 * 2^4 = 2^10

|||ly, the number of powers of 2 with:
One power of 3 i.e. 3^1 = 2^10
Two powers of 3 i.e. 3^2 = 2^10
Three powers of 3 i.e. 3^3 = 2^10

Thus total numbers of powers of 2 = 2^(10 + 10 + 10 + 10) = 2^40

Now lets find the total number of powers of 3. First lets consider the factors resulting only from 3^3, i.e. with zero powers of 2:
Product = 3^1 * 3^2 * 3^3 = 3^6

|||ly, the number of powers of 3 with:
One power of 2 i.e. 2^1 = 3^6
Two powers of 2 i.e. 2^2 = 3^6
Three powers of 2 i.e. 2^3 = 3^6
Four powers of 2 i.e. 2^4 = 3^6

Thus total numbers of powers of 3 = 3^(6 + 6 + 6 + 6 + 6) = 3^30

Thus the product of factors of 432 = 2^40 * 3^30
ANS: C

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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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04 Oct 2009, 15:36
4
432 = (2^4) (3^3)
Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

take first and last factor and multiply it and then second and second last like wise go on
i mean 1*432=432=(2^4) (3^3)
2*216=432=(2^4) (3^3)
3*144=432=(2^4) (3^3)

and you go on like wise
you would find there are 10 pairs and all are (2^4) *(3^3)
so if you multiply all these 10 pair you will get (2^40) (3^30).
hope i am clear
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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12 Apr 2010, 04:48
mrohit wrote:
432 = (2^4) (3^3)
Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

take first and last factor and multiply it and then second and second last like wise go on
i mean 1*432=432=(2^4) (3^3)
2*216=432=(2^4) (3^3)
3*144=432=(2^4) (3^3)

and you go on like wise
you would find there are 10 pairs and all are (2^4) *(3^3)
so if you multiply all these 10 pair you will get (2^40) (3^30).
hope i am clear

this looks more simpler. i tried with 36 too. more simpler. this must be 700+ level question(?)
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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12 Apr 2010, 05:10
hello,
I was struggling to get all the factors of 432... may be since not in touch wid maths since few years, but i found easy way, may help others.

just divided 432 by all no. starting from 2, till factors starts to repeat.

2 216
3 144
4 108
6 72
8 54
9 48
12 36
16 27
18 24

as one of our friend said, take a product of 1st and last no.
then second and second last ... so on.

in this tabulation i found its easy to understand.

suggestions/mistakes if any/ any other simpler method, pls share.
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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30 Apr 2010, 23:57
2
432 = 2^4 + 3^3
so no of factors = 5*4 =20

to find the product .... just divide total no. of factors by 2 so = 10
ans is

432^10 = 2^40 * 3 ^30
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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01 May 2010, 00:46
sergbov123 wrote:
Q1) What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8

Q2) What is the product of the factors of 432?
a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

How to solve if in the 1st question 90! is there instead of 20!. It will be really tough.
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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23 Sep 2010, 03:27
gurpreetsingh wrote:
Bump !!

There is a brute force way to solve this. The principle is the same, you have to remove all powers of 5 and a corresponding number of powers of 2 from the product. The product of the rest will give you the non-zero digit.

First of all what I'd do is split the numbers out into four sets. Set 1 is {2,12,22, ... ,82}. Set 2 is {5,15,25,...,85}. Set 3 is {10,20,30,...90} and Set 4 is everything else {1,3,4,6,7,8,9, 11,13,14,16,17,18,19, .... ,81,83,84,86,87,88,89}

Set 1,2,3 have 9 elements each and have all the powers of 5 within them. Set 4 has the remaining 7x9=63 elements

Lets process Set 3 first, I can take out all powers of 10, to get {1,2,3,4,5,6,7,8,9}
Then I take out the 5 and a corresponding 2 to get {1,1,3,4,1,6,7,8,9}

Now lets process Set 1 & 2, For every 5 I take out of set 2, I will take out a 2 from set 1 :
So set 1 becomes : {1,6,11,16,....,36,41} and set 2 : {1,3,5,7,9,11,13,15,17}

We are almost done now, the only powers of 5 left are in set 2 which I take out and against that I take out 4 in set 3 to leave me with with :

Set 1 : {1,6,.....,36,41} : Last digit is 6
Set 2 : {1,3,1,7,9,11,13,3,17} : Last digit is 7
Set 3 : {1,1,3,1,1,6,7,8,9} : Last digit 2
Set 4 : Which has a pattern to it and easy to show last digit is 8 (It has subsets ending in 1,3,4,6,7,8,9 each of which has last digit 8 and 9 such subsets. The "cyclicity" of 8 in a product is 8,4,2,6,repeat so with 9 subsets the answer will be 8)

Multiplying those out, the last digit of 90! is 2

The approach is brute force, but once you get a hang of it, it doesnt take very long to calculate it out. I would be very surprised if this is a GMAT level question, unless of course there is a trick to doing this which has completely evaded me
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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23 Sep 2010, 03:34
4
1
On the first question here, there is also an easy way to think of it and solve it.

Consider any number n, such that it is not a perfect square.
Let x be any factor, then (n/x) is an integer and also another factor.
Therefore, for any non-perfect square, factors are always found in pairs such that their product is n.

Hence if the total number of factors is f, the number of pairs is (f/2)

So the product of factors = $$n^{\frac{f}{2}}$$

For perfect squares, this will change slightly, as all factors except sqrt(n) will form pairs,

So the product of factots = $$sqrt(n) * n^{\frac{f-1}{2}} = n^{\frac{f}{2}}$$
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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23 Sep 2010, 22:31
shrouded1 wrote:
On the first question here, there is also an easy way to think of it and solve it.

Consider any number n, such that it is not a perfect square.
Let x be any factor, then (n/x) is an integer and also another factor.
Therefore, for any non-perfect square, factors are always found in pairs such that their product is n.

Hence if the total number of factors is f, the number of pairs is (f/2)

So the product of factors = $$n^{\frac{f}{2}}$$

For perfect squares, this will change slightly, as all factors except sqrt(n) will form pairs,

So the product of factots = $$sqrt(n) * n^{\frac{f-1}{2}} = n^{\frac{f+1}{2}}$$

What is the answer for 20!'s last digit?
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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23 Sep 2010, 22:36
shrouded1 wrote:
On the first question here, there is also an easy way to think of it and solve it.

Consider any number n, such that it is not a perfect square.
Let x be any factor, then (n/x) is an integer and also another factor.
Therefore, for any non-perfect square, factors are always found in pairs such that their product is n.

Hence if the total number of factors is f, the number of pairs is (f/2)

So the product of factors = $$n^{\frac{f}{2}}$$

For perfect squares, this will change slightly, as all factors except sqrt(n) will form pairs,

So the product of factots = $$sqrt(n) * n^{\frac{f-1}{2}} = n^{\frac{f+1}{2}}$$

Even in the perfect square case, it seems that the answer is the same? What am I missing?
$$sqrt(n) * n^{\frac{f-1}{2}} = n^{\frac{f}{2}}$$
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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23 Sep 2010, 23:48
mainhoon wrote:

Even in the perfect square case, it seems that the answer is the same? What am I missing?
$$sqrt(n) * n^{\frac{f-1}{2}} = n^{\frac{f}{2}}$$

doh ! my bad ... edited post
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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23 Sep 2010, 23:52
mainhoon wrote:
What is the answer for 20!'s last digit?

Using the same approach I showed for 90!

Set 1 : {2,12}
Set 2 : {5,15}
Set 3 : {10,20}
Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19}

This reduces to (canceling 2s&5s in set 1 and set 2 + canceling 10s in set 3) :

Set 1 : {1,6} --> Last digit 6
Set 2 : {1,3} --> Last digit 3
Set 3 : {1,2} --> Last digit 2
Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19} --> Last digit 8*8=4

So overall last digit = 6*3*2*4 = 4

Hence last non-zero digit of 20! = 4
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Re: PS - Rightmost nonzero digit & product of factors  [#permalink]

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24 Sep 2010, 05:12
shrouded1 wrote:
mainhoon wrote:
What is the answer for 20!'s last digit?

Using the same approach I showed for 90!

Set 1 : {2,12}
Set 2 : {5,15}
Set 3 : {10,20}
Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19}

This reduces to (canceling 2s&5s in set 1 and set 2 + canceling 10s in set 3) :

Set 1 : {1,6} --> Last digit 6
Set 2 : {1,3} --> Last digit 3
Set 3 : {1,2} --> Last digit 2
Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19} --> Last digit 8*8=4

So overall last digit = 6*3*2*4 = 4

Hence last non-zero digit of 20! = 4

Can you please explain how you calculating the last digit for Set 4 ?
Re: PS - Rightmost nonzero digit & product of factors &nbs [#permalink] 24 Sep 2010, 05:12

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