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

Question

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?

mikemcgarry · Accepted Answer

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 :-)

Bunuel · Answer

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.

thanhnguyen8825 · Answer

(3^4) (5^3) (5^3) (7^3) = (3^4) (5^3) (35^3) => n = 3, x = (3^4) (5^3)
=> answer C

roygush · Answer

thank you all.
Its clear now and i should have thought about it myself.

mikemcgarry · Answer

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 :-)

krrish · Answer

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

Bunuel · Answer

Please read the stem carefully: "x and n are both  positive  integers..." 0 is NOT a positive integer.

PareshGmat · Answer

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

AKG1593 · Answer

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.

Posted from my mobile device

frenchwr · Answer

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.

mikemcgarry · Answer

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

How well do you understand the idea of the prime factorization of a number? See: https://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 :-)

GMATinsight · Answer

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  C

stonecold · Answer

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.

mikemcgarry · Answer

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 :-)

chesstitans · Answer

oh man, i miss the word positive, so i thought n can be o, so i chose D

amitpandey25 · Answer

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

Bunuel · Answer

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.

amitpandey25 · Answer

Oh ! Thanks Bunuel

bumpbot · Answer

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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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