Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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?
_________________

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....
_________________

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.

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 _________________

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.

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.

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 ?

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...