Exponential Equations question from ManhattanGMAT CAT

[Guest]

If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?

(1) z is prime

(2) x is prime

The best way to answer this question is to use the exponential rules to simplify the question stem, then analyze each statement based on the simplified equation.

(3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14
(3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27
(5^2)(z) = 3x^y

(1) SUFFICIENT: Analyzing the simplified equation above, we can conclude that z must have a factor of 3 to balance the 3 on the right side of the equation. Statement (1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3.

I am struggling to understand how (1) can be sufficient:

(5^2)(z) = 3x^y
How can we conclude that z MUST have a factor of 3 to balance the right side of the equation ? Why cant z have another factor besides 3 ?

Thanks in advance for your kind help.

[StaceyKoprince]

Think of each side as representing a specific number (if we knew what the variables represented):
(5^2)(z) = 3x^y

So the number on the left side is 25*some unknown number. The number on the right side is 3*some unknown number.

The two numbers equal each other (it's an equation, right?). Think of each side as representing a specific number (if we knew what the variables represented). If I were to break the two numbers down into their prime factors, I would get the exact same set of prime factors on each side (because its the same number).

Right now, the right-hand number has two 5s and some other stuff (whatever primes z contains). The left-hand side has one 3 and some other stuff (whatever primes x^y contain). If the prime factors on each side need to equal, and if I have a 3 on the right-hand side, where am I going to find the 3 on the left-hand side? Is it part of the 5? No. 5 isn't 3 and it can't be simplified to 3. So my only option is for the variable z to provide that factor of 3.

So that's why z must have 3 as a factor. Then, statement 1 also says z is prime. By definition, a prime number has exactly two factors: itself and 1. So if the prime variable z has 3 as a factor, then only 3 and 1 are factors of that number... and that number is 3 itself.

[Guest]

excellent explanation. Thanks Stacy !

[Guest]

what "score" question is this one ? is it a 700+ question ? And how would you try to make an educated guess on it (in case it rattled you on the actual test ?)
Also is there another way of solving this ?

Thanks.

[RonPurewal]

for statement (1) don't forget the final part of the reasoning, which is NOT trivial:
after finding out that z = 3, plug back into the simplified equation to get (5^2)(3) = (3)(x^y), or x^y = 25.
note that, were it not for the restriction that x, y, z > 1, there would be two possible pairs of integers: (x, y) = (5, 2) or (25, 1). the second is invalid, however, because y must be greater than 1.

--

as for the question in the most recent post:
if you see any problem like this one - full of powers of different numbers, some of which are prime and some of which aren't, and all of which are WAY too big to calculate directly - your first instinct should be to factor all the numbers into primes and then cancel what you can. in fact, in any problem involving lots of exponents, your best bet is to break down the numbers (into primes) and then see how many things you can cancel.

if you don't take that sort of approach here, this problem will be tough going (and you'll have a hard time making a dent in it).

this problem is definitely difficult - either 600-700 or 700-800 level - although the systematic approach makes it substantially easier.

[lsyang1212]

I just came across this problem on my CAT 2, and am wondering, if we get the simplified equation (5^2)z = 3(x^y), and we're looking for [prime] factorization, doesn't this equation tell you that z MUST be 3? Since we have a 3 on the right side, and none on the left, before we even get to statement 1, don't we know z MUST BE 3? If so, how does statement 1 validate anything? It seems like it's just solidifying that 3 is prime, not that z must be 3.

Statement 2 is foggy to me as well.

I got to the simplified equation (5^2)z = 3 (x^y), but couldn't apply statements 1 and 2 successfully to get the right answer. Any advice on how to adapt my thinking to understand this type of question more fully?

Thank you!

[RonPurewal]

I just came across this problem on my CAT 2, and am wondering, if we get the simplified equation (5^2)z = 3(x^y), and we're looking for [prime] factorization, doesn't this equation tell you that z MUST be 3? Since we have a 3 on the right side, and none on the left, before we even get to statement 1, don't we know z MUST BE 3? If so, how does statement 1 validate anything? It seems like it's just solidifying that 3 is prime, not that z must be 3.

No. If you don't know that z is prime, then z can be 3 times lots of other stuff.

E.g.,
(5^2)(15) = (3)(5^3) ... in which z = 15
(5^2)(75) = (3)(5^4) ... in which z = 75
(5^2)(27) = (3)(15^2) ... in which z = 27
And so on.

[RonPurewal]

I got to the simplified equation (5^2)z = 3 (x^y), but couldn't apply statements 1 and 2 successfully to get the right answer. Any advice on how to adapt my thinking to understand this type of question more fully?

Thank you!

Hmm. Well, it's easier for me to say what's not the right way of thinking"”namely, traditional, "start-to-finish"-type algebraic thinking. The kind of thinking in which you just plug stuff into a formula, or execute some other known process, until an answer just pops out.

This is more an issue of considering different possibilities. So, the things that come into play include ....
1/ understanding the issue in the first place (what kinds of numbers do I need to try? what kinds of numbers are off the table?)
2/ understanding particular goals / particular values that will accomplish something (e.g., "if I can get one of these and one of those, then 'not sufficient' ")
3/ organizing those possibilities, mentally and/or on paper

There is a fair amount of overlap with number properties (odd/even, pos/neg/zero, etc.) problems, as well as with inequality DS problems. Both of these generally require organized testing of cases; seldom can either of them be solved algebraically.

[mora494]

Hi,

I understand why statement 2 in sufficient.
But I am struggling to understand why stat 1 is.

why can't x= 25 and y=1 z=3 (prime) ??

The explanation is-


(1) SUFFICIENT: Analyzing the simplified equation above, we can conclude that z must have a factor of 3 to balance the 3 on the right side of the equation. Statement (1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3.

Since z = 3, the left side of the equation is 75, so xy = 25. The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so statement (1) is sufficient. Put differently, the expression xy must provide the two fives that we have on the left side of the equation. The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.

[RonPurewal]

y = 1 is not allowed. Go back and read the beginning of the problem again.

Also, slow down. If you aren't noticing conditions imposed on the variables, you're reading too fast.

And write down these conditions on your own paper, just to remind yourself.