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

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1856/70^m is an integer. What is the maximum possible value [#permalink] New post 30 Nov 2005, 07:37
!1856/70^m is an integer. What is the maximum possible value of m.
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Re: Factorial Problem [#permalink] New post 30 Nov 2005, 07:41
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
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 [#permalink] New post 30 Nov 2005, 07:42
!1856/70^m is an integer. What is the maximum possible value of m.

1856 / 70 = 26 + 0.xxxxx
1856 / 70^2 = 0 + 0.xxxxx

There are twenty six 70's in 1856.

So I guess the maximum possible value of m is 26.[/b]
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Re: Factorial Problem [#permalink] New post 30 Nov 2005, 07:48
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


Hey I'm not getting this.. can you pls explain ??
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 [#permalink] New post 30 Nov 2005, 07:56
Sorry,
I got it now !!
I forgot it was factorial !!
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Re: Factorial Problem [#permalink] New post 01 Dec 2005, 09:40
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!
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Re: Factorial Problem [#permalink] New post 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 :wink:
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)
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Re: Factorial Problem [#permalink] New post 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 :wink:
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.
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Re: Factorial Problem [#permalink] New post 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 :wink:
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 :wink: ..is 2 a power of 4?!
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 [#permalink] New post 01 Dec 2005, 23:49
OE is 26.

!1856/70^m

1856/70=26

26/70=0

So, m=26
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 [#permalink] New post 02 Dec 2005, 00:07
1856! = 1* 2*........70*......140*.......1820....1856

now 1820 = 70*26

that means there are 26 times we have 70.

so maximum value of m is 26
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Re: Factorial Problem [#permalink] New post 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 :wink:
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 :wink: ..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 !.
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 [#permalink] New post 02 Dec 2005, 13:30
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
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 [#permalink] New post 02 Dec 2005, 13:47
I agree with 307 too.
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 [#permalink] New post 02 Dec 2005, 19:28
rlevochkin wrote:
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.
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 [#permalink] New post 02 Dec 2005, 19:35
or take another example: find 12^n of 15!

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
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 [#permalink] New post 02 Dec 2005, 20:05
laxieqv wrote:
rlevochkin wrote:
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
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Last edited by rlevochkin on 02 Dec 2005, 20:23, edited 1 time in total.
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 [#permalink] New post 02 Dec 2005, 20:16
laxieqv wrote:
or take another example: find 12^n of 15!

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.
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 [#permalink] New post 02 Dec 2005, 20:56
rlevochkin wrote:
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! :wink: ....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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 [#permalink] New post 04 Dec 2005, 20:00
I covered this material in e-book for Number theory and Equalities on the web-site http://www.4gmat.com.
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