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Re: Factorial Problem [#permalink]
01 Dec 2005, 18:23

believe2 wrote:

laxieqv wrote:

rlevochkin wrote:

!1856/70^m is an integer. What is the maximum possible value of m.

since the fraction is an integer ---> 70^m is a factor of 1856! 1856: 70 = 26 26: 70 = 0 ---> the highest power of 70 in the factorial is 26

Agreed there are 26 70's in 1856 but 1876 is a factorial don't we want to consider numbers like 2,5,7 which can form another 70 ???

1.2.3.4.5.6.7.8...70...140.........1876

it seems like a loaded question!

you don't need to do this way, do like what i do is a short-cut which is used by several members here . Or you can use gamjatang's way, it's more understandable
doing my way, for example: what is the highest power n of 3 so that 3^n is a factor of 10!
take 10 : 3 = 3 , take this divisor 3 : 3 = 1 , take 1 : 3= 0 , add all the results to get the final answer : 3 +1+0=4 (you can check it)

Re: Factorial Problem [#permalink]
01 Dec 2005, 20:15

laxieqv wrote:

believe2 wrote:

laxieqv wrote:

rlevochkin wrote:

!1856/70^m is an integer. What is the maximum possible value of m.

since the fraction is an integer ---> 70^m is a factor of 1856! 1856: 70 = 26 26: 70 = 0 ---> the highest power of 70 in the factorial is 26

Agreed there are 26 70's in 1856 but 1876 is a factorial don't we want to consider numbers like 2,5,7 which can form another 70 ???

1.2.3.4.5.6.7.8...70...140.........1876

it seems like a loaded question!

you don't need to do this way, do like what i do is a short-cut which is used by several members here . Or you can use gamjatang's way, it's more understandable doing my way, for example: what is the highest power n of 3 so that 3^n is a factor of 10! take 10 : 3 = 3 , take this divisor 3 : 3 = 1 , take 1 : 3= 0 , add all the results to get the final answer : 3 +1+0=4 (you can check it)

..this method is good but do you think it only works in case of divisions by prime numbers?

consider the following case of :
what is the highest power n of 4 so that 4^n is a factor of 10!
by this method the ans would be 2, i think, but the actual answer should be 4.

=1 * 2 * 3 * 4 *5*6*7*8*9*10/4^4

i.e. 4,8 and then 2s from 2,6, 10 and rem(8/4) also make up 2 more 4s.

Re: Factorial Problem [#permalink]
01 Dec 2005, 20:25

believe2 wrote:

laxieqv wrote:

believe2 wrote:

laxieqv wrote:

rlevochkin wrote:

!1856/70^m is an integer. What is the maximum possible value of m.

since the fraction is an integer ---> 70^m is a factor of 1856! 1856: 70 = 26 26: 70 = 0 ---> the highest power of 70 in the factorial is 26

Agreed there are 26 70's in 1856 but 1876 is a factorial don't we want to consider numbers like 2,5,7 which can form another 70 ???

1.2.3.4.5.6.7.8...70...140.........1876

it seems like a loaded question!

you don't need to do this way, do like what i do is a short-cut which is used by several members here . Or you can use gamjatang's way, it's more understandable doing my way, for example: what is the highest power n of 3 so that 3^n is a factor of 10! take 10 : 3 = 3 , take this divisor 3 : 3 = 1 , take 1 : 3= 0 , add all the results to get the final answer : 3 +1+0=4 (you can check it)

..this method is good but do you think it only works in case of divisions by prime numbers?

consider the following case of :
what is the highest power n of 4 so that 4^n is a factor of 10!
by this method the ans would be 2, i think, but the actual answer should be 4.

=1 * [b]2 * 3 * 4 *5*6*7*8*9*10/4^4

i.e. 4,8 and then 2s from 2,6, 10 and rem(8/4) also make up 2 more 4s.[/[/b]quote]

you're considering 2^n , not 4^n , buddy ..is 2 a power of 4?!

Re: Factorial Problem [#permalink]
02 Dec 2005, 10:26

laxieqv wrote:

believe2 wrote:

laxieqv wrote:

believe2 wrote:

laxieqv wrote:

rlevochkin wrote:

!1856/70^m is an integer. What is the maximum possible value of m.

since the fraction is an integer ---> 70^m is a factor of 1856! 1856: 70 = 26 26: 70 = 0 ---> the highest power of 70 in the factorial is 26

Agreed there are 26 70's in 1856 but 1876 is a factorial don't we want to consider numbers like 2,5,7 which can form another 70 ???

1.2.3.4.5.6.7.8...70...140.........1876

it seems like a loaded question!

you don't need to do this way, do like what i do is a short-cut which is used by several members here . Or you can use gamjatang's way, it's more understandable doing my way, for example: what is the highest power n of 3 so that 3^n is a factor of 10! take 10 : 3 = 3 , take this divisor 3 : 3 = 1 , take 1 : 3= 0 , add all the results to get the final answer : 3 +1+0=4 (you can check it)

..this method is good but do you think it only works in case of divisions by prime numbers?

consider the following case of : what is the highest power n of 4 so that 4^n is a factor of 10! by this method the ans would be 2, i think, but the actual answer should be 4.

=1 * [b]2 * 3 * 4 *5*6*7*8*9*10/4^4

i.e. 4,8 and then 2s from 2,6, 10 and rem(8/4) also make up 2 more 4s.[/[/b]quote]

you're considering 2^n , not 4^n , buddy ..is 2 a power of 4?!

i learned from a teacher that when the number is not a prime number, you have to express it in its prime factors. in this case its 7*2*5. then you have to divide 1856 by each prime (the method mentioned above). the lowest of the values is the largest power of 70 than can divide 1856 !. _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

I read this question in number property book. Sorry, I think they gave a wrong explanation. This one can be solved using 3-step method:

1. Factorize the number in the product of primes
2. Identify the prime with the lowest power.
3. Find that power, and that is a greatest power for the number.

1. 70=2*5*7
2. !1856/2^m, !1856/5^m, and !1856/7^m. Power of 7 has the lowest value.
3. !1856/7^m m=265+37+5+0=307
!1856/70^m, the greatest power of m=307 _________________

I read this question in number property book. Sorry, I think they gave a wrong explanation. This one can be solved using 3-step method:

1. Factorize the number in the product of primes 2. Identify the prime with the lowest power. 3. Find that power, and that is a greatest power for the number.

1. 70=2*5*7 2. !1856/2^m, !1856/5^m, and !1856/7^m. Power of 7 has the lowest value. 3. !1856/7^m m=265+37+5+0=307 !1856/70^m, the greatest power of m=307

uhm, let's take a simple example:
4^n as a factor of 10!
In your method, you'll write 4^n= 2^2n
10! has 2^8 as a maximum power ------> 2n=8--->n=4
now let's see: 10!= 1*2*3*4*5*6*7*8*9*10
only 4 and 8 contain power of 4 , that is to say n= 2

Clearly factorizing 4 into prime factors is wrong.

I read this question in number property book. Sorry, I think they gave a wrong explanation. This one can be solved using 3-step method:

1. Factorize the number in the product of primes 2. Identify the prime with the lowest power. 3. Find that power, and that is a greatest power for the number.

1. 70=2*5*7 2. !1856/2^m, !1856/5^m, and !1856/7^m. Power of 7 has the lowest value. 3. !1856/7^m m=265+37+5+0=307 !1856/70^m, the greatest power of m=307

uhm, let's take a simple example: 4^n as a factor of 10! In your method, you'll write 4^n= 2^2n 10! has 2^8 as a maximum power ------> 2n=8--->n=4 now let's see: 10!= 1*2*3*4*5*6*7*8*9*10 only 4 and 8 contain power of 4 , that is to say n= 2

Clearly factorizing 4 into prime factors is wrong.

All right, in this case, I would do it this way

!10/4^n
1. 4=2^2 I'll find greatest value for 2 Since 2 is already in the power of 2, the number has to be divided by 2 to avoid double count.

So, !10/2^m; m=5+2+1+0=8 Therefore, !10/4^n n=8/2=4
Which is not any different from what you did.

Well, basically there is no need to apply method that I was describing above in this case, cause there is only one prime factor, which is 2 _________________

Hard work is the main determinant of success

Last edited by rlevochkin on 02 Dec 2005, 20:23, edited 1 time in total.

12= 2*2*3 as your method, consider 3 15! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15

--->biggest n = 6 but clearly 15! has only 12 as a power of 12

Are you trying to tell me that !15/12^n and n=1. I can't see how, we are looking for the greatest power of 12. 1 is not the greatest power of 12 here. I don't think you should really look at it this way. _________________

I read this question in number property book. Sorry, I think they gave a wrong explanation. This one can be solved using 3-step method:

1. Factorize the number in the product of primes 2. Identify the prime with the lowest power. 3. Find that power, and that is a greatest power for the number. 1. 70=2*5*7 2. !1856/2^m, !1856/5^m, and !1856/7^m. Power of 7 has the lowest value. 3. !1856/7^m m=265+37+5+0=307 !1856/70^m, the greatest power of m=307

Well, i reconsider and you're right! ....for non-prime number, we should factorize them. Anyway, thank you for pointing that out ..which source is this one from?! ...what if ETS also provides 26 as OE ...what should we do then?!

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