1856/70^m is an integer. What is the maximum possible value : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 21:21

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

1856/70^m is an integer. What is the maximum possible value [#permalink]

### Show Tags

30 Nov 2005, 07:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Hard work is the main determinant of success

SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 292 [0], given: 0

### Show Tags

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
Director
Joined: 14 Sep 2005
Posts: 993
Location: South Korea
Followers: 2

Kudos [?]: 155 [0], given: 0

### Show Tags

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]
_________________

Auge um Auge, Zahn um Zahn !

Director
Joined: 24 Oct 2005
Posts: 659
Location: London
Followers: 1

Kudos [?]: 15 [0], given: 0

### Show Tags

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 ??
Director
Joined: 24 Oct 2005
Posts: 659
Location: London
Followers: 1

Kudos [?]: 15 [0], given: 0

### Show Tags

30 Nov 2005, 07:56
Sorry,
I got it now !!
I forgot it was factorial !!
Manager
Joined: 15 Aug 2005
Posts: 136
Followers: 2

Kudos [?]: 23 [0], given: 0

### Show Tags

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!
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 292 [0], given: 0

### Show Tags

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)
Manager
Joined: 15 Aug 2005
Posts: 136
Followers: 2

Kudos [?]: 23 [0], given: 0

### Show Tags

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.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 292 [0], given: 0

### Show Tags

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?!
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

01 Dec 2005, 23:49
OE is 26.

!1856/70^m

1856/70=26

26/70=0

So, m=26
_________________

Hard work is the main determinant of success

SVP
Joined: 28 May 2005
Posts: 1723
Location: Dhaka
Followers: 7

Kudos [?]: 327 [0], given: 0

### Show Tags

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
_________________

hey ya......

VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 327 [0], given: 0

### Show Tags

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.

Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

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
_________________

Hard work is the main determinant of success

Manager
Joined: 15 Aug 2005
Posts: 136
Followers: 2

Kudos [?]: 23 [0], given: 0

### Show Tags

02 Dec 2005, 13:47
I agree with 307 too.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 292 [0], given: 0

### Show Tags

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.
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 292 [0], given: 0

### Show Tags

02 Dec 2005, 19:35
or take another example: find 12^n of 15!

12= 2*2*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
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

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
_________________

Hard work is the main determinant of success

Last edited by rlevochkin on 02 Dec 2005, 20:23, edited 1 time in total.
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

02 Dec 2005, 20:16
laxieqv wrote:
or take another example: find 12^n of 15!

12= 2*2*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.
_________________

Hard work is the main determinant of success

SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 292 [0], given: 0

### Show Tags

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! ....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?!
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

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.
_________________

Hard work is the main determinant of success

04 Dec 2005, 20:00

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by