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What is the rightmost nonzero digit of 20!?
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14 Aug 2009, 11:56
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Q1) What is the rightmost nonzero 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
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14 Aug 2009, 12:17
sergbov123 wrote: Q1) What is the rightmost nonzero 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: factorofp82381.html?viewpost=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 nonzero 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 => 4You sure the options are correct?
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Re: PS  Rightmost nonzero digit & product of factors
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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
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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
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18 Aug 2009, 00:46
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
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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
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18 Aug 2009, 08:25
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
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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
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04 Oct 2009, 15:36
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
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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
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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
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30 Apr 2010, 23:57
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
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01 May 2010, 00:46
sergbov123 wrote: Q1) What is the rightmost nonzero 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
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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 nonzero 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 2The 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
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23 Sep 2010, 03:34
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 nonperfect 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 = \(sqrt(n) * n^{\frac{f1}{2}} = n^{\frac{f}{2}}\)
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Re: PS  Rightmost nonzero digit & product of factors
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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 nonperfect 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 = \(sqrt(n) * n^{\frac{f1}{2}} = n^{\frac{f+1}{2}}\) What is the answer for 20!'s last digit?
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Re: PS  Rightmost nonzero digit & product of factors
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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 nonperfect 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 = \(sqrt(n) * n^{\frac{f1}{2}} = n^{\frac{f+1}{2}}\) Even in the perfect square case, it seems that the answer is the same? What am I missing? \(sqrt(n) * n^{\frac{f1}{2}} = n^{\frac{f}{2}}\)
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Re: PS  Rightmost nonzero digit & product of factors
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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? \(sqrt(n) * n^{\frac{f1}{2}} = n^{\frac{f}{2}}\)
doh ! my bad ... edited post
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Re: PS  Rightmost nonzero digit & product of factors
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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 nonzero digit of 20! = 4
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Re: PS  Rightmost nonzero digit & product of factors
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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 nonzero digit of 20! = 4Can you please explain how you calculating the last digit for Set 4 ?




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