If x and y are positive integers, what is the value of x + y

[rrampriya]

If x and y are positive integers, what is the value of x + y

1. (2^x)(3^y) = 72
2. (2^x)(2^y) = 32

statement 1 leads to 1 solution for x and y :x=3,y=2
however statement 2 leads to 2 solutions: x=3,y=2 and x=1,y=4, u can also add x=2,y=3 and x=4,y=1.

Hence I selected A as my choice.

However, answer is D. How? Can someone explain? Thanks!!

[gurm43]

The question asks to find x+Y not indiviual values of x and y, for all the possible values for x and y in statement 2 the sum (x +y) still equals 5.

[ULHASV]

If x and y are positive integers, what is the value of x + y

1. (2^x)(3^y) = 72
2. (2^x)(2^y) = 32

statement 1 leads to 1 solution for x and y :x=3,y=2
however statement 2 leads to 2 solutions: x=3,y=2 and x=1,y=4, u can also add x=2,y=3 and x=4,y=1.

Hence I selected A as my choice.

However, answer is D. How? Can someone explain? Thanks!!

Hi,

The thing is, no need to be so engraved in what is the individual value of x and y, when the question is asking about x+Y, then be it 3+2, or 4+1(as you suggested), still you are getting definite answer "5", and this is asked. Why to trouble unnecessarily when you already got the answer.
By the way , 2^(x+y) = 2^5, how come you got 4 and 1 thing?

Thanks.
Anurag.

[rrampriya]

Thanks all! It was my bad!! I didnt read the qs properly!! :(!!

[RonPurewal]

read the questions carefully, guys. be sure to double-check exactly what they want you to find.

by the way:

(1) is a PRIME FACTORIZATION of 72.
since 2 and 3 are primes, we know that this is THE prime factorization of 72.

once you make this realization, you can deduce at once that (1) will give unique values of x and y, since it's impossible for a number to have more than one prime factorization.

--

(2) can be combined using basic exponent rules (multiply 2 powers with the same base --> add the exponents) to give 2^(x+y) = 32. at this point you can see that it's possible to find (x + y), even if you don't know the exact value from memory.

so, (d)