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Re: PS - Rightmost nonzero digit & product of factors [#permalink]
14 Aug 2009, 12:17

2

This post received KUDOS

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.

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? _________________

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

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

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

Re: PS - Rightmost nonzero digit & product of factors [#permalink]
18 Aug 2009, 00:46

1

This post received KUDOS

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 _________________

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

Re: PS - Rightmost nonzero digit & product of factors [#permalink]
18 Aug 2009, 08:25

1

This post received KUDOS

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.

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

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

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(?) _________________

"Great people don't do different things, they do things differently"

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 _________________

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}} _________________

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? _________________

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}} _________________

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 _________________

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 ?

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