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If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit

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If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 25 Feb 2013, 14:13
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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

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?

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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 25 Feb 2013, 15:31
4
2
roygush 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

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?

Dear Roygush

I must say, starting from 35^n = (5^n)(7^n), it's not clear to me how you wound up with only two possibilities. Your reasoning is not clear to me. I would say --- start on the left side ---

The factor of (3^4) doesn't play into the (35^n) at all --- that will have to go into the x, no choice.

We have three factors of 7, so that means we could have as many as three powers of 35 ------ we could have 35^1, 35^2, or 35^3. We can't have 35^0 = 1, because even though 1 is positive integer, that would make n = 0, and zero is not a positive integer. So, those three values of n, 1 and 2 and 3, are the only possibilities. Therefore, there are three possibilities.

n = 1 -------> x = (3^4)(5^5)(7^2)
n = 2 -------> x = (3^4)(5^4)(7^1)
n = 3 -------> x = (3^4)(5^3)

Answer = C

Does all that make sense?

Mike :-)
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 25 Feb 2013, 18:45
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(3^4) (5^3) (5^3) (7^3) = (3^4) (5^3) (35^3) => n = 3, x = (3^4) (5^3)
=> answer C
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 26 Feb 2013, 03:02
3
2
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

Notice that the power of 7 in LHS is limiting the value of n, thus n cannot be more than 3 and since n is a positive integer, then n could be 1, 2, or 3.

Answer: C.
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 26 Feb 2013, 08:40
thank you all.
Its clear now and i should have thought about it myself.
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 26 Feb 2013, 11:33
1
This question turned out to be relatively easy, but it inspired me to create a similar question that's a little more challenging:
if-a-and-b-are-positive-integers-and-147953.html

Mike :-)
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 06 Jul 2013, 12:34
Bunuel 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

Notice that the power of 7 in LHS is limiting the value of n, thus n cannot be more than 3 and since n is a positive integer, then n could be 1, 2, or 3.

Answer: C.


hi, i differ on this,

here is the reason

35^0 is also possible as we are not given any limitation to x

we can have values of 35^0, 1, 2, 3, so n can take 4 values .......... and the answer would be 4 D
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 06 Jul 2013, 12:37
krrish wrote:
Bunuel 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

Notice that the power of 7 in LHS is limiting the value of n, thus n cannot be more than 3 and since n is a positive integer, then n could be 1, 2, or 3.

Answer: C.


hi, i differ on this,

here is the reason

35^0 is also possible as we are not given any limitation to x

we can have values of 35^0, 1, 2, 3, so n can take 4 values .......... and the answer would be 4 D


Please read the stem carefully: "x and n are both positive integers..." 0 is NOT a positive integer.
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 26 Feb 2014, 23:20
1
Re-arrange the equation:

x = 3^4 . 5^6. 7^3 / 35^n

n may be 0,1,2 3........ however n is +ve, so can be 1,2,3

Answer = 3 = C
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 08 Mar 2014, 10:32
Option C.
RHS:7^n * 5^n * x
On the LHS,7 has max. Power of 3
Since n is a +ve integer,it can't be zero.So n=1,2,&3 only.

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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 27 Mar 2014, 15:37
How is this problem solved? I don't understand why the powers of 7 limits the answer choices. Maybe I am thinking about this the wrong way.
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 27 Mar 2014, 16:14
1
frenchwr wrote:
How is this problem solved? I don't understand why the powers of 7 limits the answer choices. Maybe I am thinking about this the wrong way.

Dear frenchwr,
I'm happy to respond. :-)

How well do you understand the idea of the prime factorization of a number? See:
http://magoosh.com/gmat/2012/gmat-math-factors/

When you know the prime factorization of a number, it's as if you know the DNA of the number. You have such profound knowledge about it.

For example, 35 = 5*7. That's the prime factorization. This means, for every factor of 35 we have, we need another factor of 7. There is absolutely no way we can make a factor of 35 without using a factor of 7 as one of the ingredients. If we only get three factors of 7, as is evident from the left side of the equation, that means we could make, at most, only three factors of 35. Once those three factors of 7 are used up, building the three factors of 35, then there is absolutely no possible of creating any more factors of 35, because we are out of one of the essential ingredients.

Does this make sense?
Mike :-)
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 03 Jun 2015, 06:27
1
roygush 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

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?


TIP FOR SUCH QUESTIONS: All such Questions Require PRIME FACTORISATION i.e. Breaking the number into Prime factors and their power form on both sides of the equation

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

i.e. (3^4)(5^6)(7^3) = (5*7)^n *(x)

i.e. (3^4)(5^6)(7^3) = (5^n)*(7^n) *(x)

This clearly depicts one thing which is x must be a multiple of (3^4) for sure and other factors of x will depend on the value of n

Please not that n is a positive integer

i.e. at n=1, (3^4)(5^6)(7^3) = (5^1)*(7^1) *(x) i.e. x=(3^4)(5^5)(7^2)
i.e. at n=2, (3^4)(5^6)(7^3) = (5^2)*(7^2) *(x) i.e. x=(3^4)(5^4)(7^1)
i.e. at n=3, (3^4)(5^6)(7^3) = (5^3)*(7^3) *(x) i.e. x=(3^4)(5^3)(7^0)

Only 3 possibilities

Answer: Option
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 06 Mar 2016, 21:12
roygush 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

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?


here => 3^4 is immaterial
now values of n can be found from the following expressions
on LHS there can be (35)^1 or (35)^2 or (35)^3 but beyond that we dont have the excess power of 5 to utilize
thus => N=1,2,3 => 3 values.
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 07 Mar 2016, 12:22
1
Chiragjordan wrote:
roygush 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

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?


here => 3^4 is immaterial
now values of n can be found from the following expressions
on LHS there can be (35)^1 or (35)^2 or (35)^3 but beyond that we dont have the excess power of 5 to utilize
thus => N=1,2,3 => 3 values.

Dear Chiragjordan,
Yes, you are quite right. The 3^4 is entirely immaterial, and the factors of 7 limit the possibilities. This is a relatively straightforward problem. This problem inspired me to create a slightly more challenging problem:
if-a-and-b-are-positive-integers-and-147953.html
Enjoy!
Mike :-)
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 19 Apr 2017, 19:04
oh man, i miss the word positive, so i thought n can be o, so i chose D
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 03 May 2018, 21:46
Hi !
If in the case the value of n =4 ; we still get a value which +ve decimal..
Why cant we say then it has more values
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 03 May 2018, 23:26
amitpandey25 wrote:
Hi !
If in the case the value of n =4 ; we still get a value which +ve decimal..
Why cant we say then it has more values


If n = 4, then from (3^4)(5^6)(7^3) = (35^4)(x), x turns out to be x = 2025/7 but the stem says that x is a positive integer, thus n cannot be 4.
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit  [#permalink]

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New post 03 May 2018, 23:28
Oh ! Thanks Bunuel :)
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Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit &nbs [#permalink] 03 May 2018, 23:28
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