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different possible values of n

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by adi_800 » Fri Jul 23, 2010 7:14 am
If (3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x ), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

OA is C
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by kvcpk » Fri Jul 23, 2010 7:28 am
adi_800 wrote:If (3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x ), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

OA is C
(3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x )

3^4 * 5^6 * 7^3 = 7^n * 5^n * x
n can be 1 or 2 or 3
if n=4, then the product will have 7^4 which is not valid

pick C
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by adi_800 » Fri Jul 23, 2010 8:33 am
But cant n be 6? or u have to select the least value of n?
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by selango » Fri Jul 23, 2010 8:40 am
(3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x )

3^4 * 5^6 * 7^3 = 7^n * 5^n * x

n can be 1,2 or 3 and remaining factors are represented by x

n>3 not possible bcoz 7^4 appear in right hand side.The exponent must be equal on both sides to make the eqn valid
--Anand--
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by kvcpk » Fri Jul 23, 2010 2:34 pm
adi_800 wrote:But cant n be 6? or u have to select the least value of n?
No Adi, n cannot be 6 because, if n is 6, then 7 will have a power of 6 which will fail our criteria.

The trick here is to choose maximum n that satisfies both the powers.

Hope this helps!!
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by [email protected] » Sat Jul 24, 2010 9:28 pm
i cant understand the explanation...can someone please elaborate
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by kvcpk » Sat Jul 24, 2010 11:46 pm
[email protected] wrote:i cant understand the explanation...can someone please elaborate
Hi Roopali, it is simple.. Let me give another try..

Given that
(3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x )

This can be rewritten as
3^4 * 5^6 * 7^3 = 7^n * 5^n * x

Any number can be written as a product of primes.
When two numbers are same, the primes involved in their products and their exponents should be same.
Example: 20 = 4 * 5 = 2^2 * 5
20 can also be written as 10*2 = 2*5*2 = 2^2 * 5

Hence, X in the above problem should have a multiple of 3^4. Which means x = (3^4*k) for some k
Now, Let us rewrite..
3^4 * 5^6 * 7^3 = 7^n * 5^n * (3^4*k)
There are 6 fives and 3 sevens on LHS

Now, what is the maximum value n can take??

Suppose n takes 6, then on the right hand side we will get 6 sevens which is wrong.
We can have a maximum of 3 sevens on RHS.

Hence n can take only 3 at maximum.
Eqtn will then look like
3^4 * 5^6 * 7^3 = 7^3 * 5^3 * (3^4*5^3)

Let me know if you have any troubles understanding this.
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by gmatmachoman » Sun Jul 25, 2010 1:49 am
Question asks us to find "How many possible values does n can take".. It doesn't ask thye maximum value of n!!

Soln :

3^4 * 5^6 * 7^3 = (35 ^ n )( x )

case n = 3

7^3 .( 5^2)^ 3 . 3^4

case n = 2

(7V7)^2 .(5^3)2 .3^4

case n=6

(V7)^6 . 5^6 .3^4

So n can take 3 values ( 2,3,6)
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by kvcpk » Sun Jul 25, 2010 2:03 am
gmatmachoman wrote:Question asks us to find "How many possible values does n can take".. It doesn't ask thye maximum value of n!!
Yeah thats right.. I Misread the question.. Thanks for posting back!!

But, N can be 1 or 2 or 3.
How can n be 6?

Am I missing soemthing??
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by gmatmachoman » Sun Jul 25, 2010 2:18 am
kvcpk wrote:
gmatmachoman wrote:Question asks us to find "How many possible values does n can take".. It doesn't ask thye maximum value of n!!
Yeah thats right.. I Misread the question.. Thanks for posting back!!

But, N can be 1 or 2 or 3.
How can n be 6?

Am I missing soemthing??

case n=6

(V7)^6 . 5^6 .3^4

sqrt(7)^6 . 5^6 .3^4

Now Praveen ping me one thanks button..I never got one frm u LOL!![/b]
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by gmatmachoman » Sun Jul 25, 2010 2:19 am
I am not getting how n can be 1??
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by selango » Sun Jul 25, 2010 2:33 am
gmatmachoman,

I also solved it using same approach as praveen.

3^4.5^6.7^3=5^n.7^n.x

n can be 1,2 and 3.Remaining factor ll be expressed by X.

But ur assuming x as 3^4 and solving for n.

I think both ways are correct.
--Anand--
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by gmatmachoman » Sun Jul 25, 2010 2:34 am
selango wrote:gmatmachoman,

I also solved it using same approach as praveen.

3^4.5^6.7^3=5^n.7^n.x

n can be 1,2 and 3.Remaining factor ll be expressed by X.

But ur assuming x as 3^4 and solving for n.

I think both ways are correct.
Anand, if n =1 plz show me the expression....IMO n cant be 1
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by kvcpk » Sun Jul 25, 2010 2:35 am
gmatmachoman wrote:I am not getting how n can be 1??
When n=1
3^4 * 5^6 * 7^3 = 7^1 * 5^1 * (7^2 * 5^5 * 3^4)
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by selango » Sun Jul 25, 2010 2:44 am
gmatmachoman wrote:
selango wrote:gmatmachoman,

I also solved it using same approach as praveen.

3^4.5^6.7^3=5^n.7^n.x

n can be 1,2 and 3.Remaining factor ll be expressed by X.

But ur assuming x as 3^4 and solving for n.

I think both ways are correct.
Anand, if n =1 plz show me the expression....IMO n cant be 1
If n=1

3^4 *5^6*7^3 = 7^1*5^1*(7^2 * 5^5 * 3^4)

Here x=7^2 * 5^5 * 3^4
--Anand--
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