If x and y are positive integers and x + y = 3^x, is y divisible by 6?

Question

(1) x is odd.
(2) x is a multiple of 3.

Kudos for a correct  solution .

Bunuel · Accepted Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

Manipulate:
x + y = 3^x
y = 3^x - x

In order for y to be divisible by 6, it must be divisible by both 2 and 3.

All of the positive powers of 3 (3^x, where x is a positive integer) are divisible by 3. Thus, y will be divisible by 3 only when x is also divisible by 3 (because multiple of 3 – multiple of 3 = multiple of 3).

Moreover, all of the positive powers of 3 (3^x, where x is a positive integer) are odd. Thus, y will be even only when x is also odd (because odd – odd = even).

Combining these number properties, we can rephrase the question as “Is x an odd multiple of 3?”

(1) INSUFFICIENT: x is odd, but we don't know whether it is a multiple of 3 (i.e. x could be 3 or5).

(2) INSUFFICIENT: x is a multiple of 3, but we don't know whether it is odd (i.e. x could be 3 or6).

(1) AND (2) SUFFICIENT: x is an odd multiple of 3: 3, 9, 15, etc.

The correct answer is C.

GMATinsight · Answer

Given : x + y = 3^x

Question : is y divisible by 6?

Statement 1: x is odd

@x=1, y = 3-1 = 2 Not a multiple of 6

@x=3, y = 27-3 = 24 Multiple of 6

Hence,   NOT SUFFICIENT

Statement 2: x is a multiple of 3

y = 3^x - x

for any value of x that is multiple of 3, let, x = 3a

i.e. y =  3*

i.e. y will always be a multiple of 3 but not not necessarily divisible by 6

Hence,   NOT SUFFICIENT

Combining the two Statements

y =  3*   and x is odd
for every odd value of x, "y = 3^x - x" will be even and 
y is always a multiple of 3  
i.e. y will always be divisible by 6 (2 as well as 3). Hence,

SUFFICIENT

Answer: Option  C

shriramvelamuri · Answer

Hi GMATinsight,

I am struggling to understand the answer B, because..

If X is a multiple of 3 then X can be 6,

6+ y= 3^6 => Y= 723. (is not divisible by 6)

So answer should be C. the answer is only correct if x is odd and muliple of 3 such as 3, 9, 15.

Let me know your thoughts.

Harley1980 · Answer

Hello  shriramvelamuri

I think you are right.

In previous post  GMATinsight  gives answer on question "does y multiple of 3"
But question was "does y multiple of 6"

Number 3 in exponent have a repeating pattern in unit digit:  3 ,  9 , 2 7 , 8 1 , 24 3 , 72 9 ... and so on
So every four numbers unit digit will be repeating

As we know from first statement x = odd and we know that 3 in any power will have odd unit digit
so odd unit digit - some odd number x gives us even number that will be divisible by 2
But this is insufficient because we know nothing about divisibilty on 3

And from second statement we know that x is multiple of 3 ( GMATinsight  gives us very nice algebraic proof)
But this is insufficient because we know nothing about divisibilty on 2

So by combining this two statement we can say that y will be divisible by 2 and 3 and so on 6 too

Answer is C

EMPOWERgmatRichC · Answer

Hi shriramvelamuri,

You are correct.

While the error was a relatively simple one, there's a BIG lesson here regarding taking the proper notes and doing all of the work on the PAD - also make sure to write the QUESTION itself on the pad, so you don't accidentally answer a different question. That's ultimately what happened here. Little mistakes tend to cost many Test Takers serious points on Test Day. If you want to maximize your performance, then you have to minimize these types of errors.

GMAT assassins aren't born, they're made,
Rich

bluesquare · Answer

We have:  x>0,y>0,x=Integer,y=Integer,x+y =3^x  which turns into  y=3^x-x .

1: 
If  x=1, 1+y=3; y=2  so no.
If  x=3, 3^3-3  is divisible by 3, and 2 (since it will be even) so yes.
So not sufficient.

2:
If x is an odd multiple of 3:  3^x-x=odd-odd  is even and thus divisible by 6.
If x is an even multiple of 3:  3^x-x=odd-even  is odd and thus not divisible by 6.
So not sufficient.

Together, we have sufficiency, so the answer is C.

LeoN88 · Answer

x,y > 0 and I.

Statement 1. X is odd, let's negate and confirm the stem.
1 + 2 = 3^1, no
3 + 24= 3^3, yes.

Statement 2. x=3k; let's negate and confirm the stem.
6 + 723 = 3^6, no
3 + 24 = 27, yes.

Now combining I and II,
We see when x is odd and multiple of 3 we get y as a multiple of 6, check pattern with another number.
9 + y = 3^9, yes, so  C

CrackverbalGMAT · Answer

There’s already a lot of discussions that has happened about this question. Just trying to add my 2 cents to it.

Analysing the question stem and using the divisibility rule for 6 is a good way of tackling this question. The value method also is equally efficient, so finally, I think it boils down to personal preference as to which method you wish to take up.

If y has to be divisible by 6, it has to be divisible by both 2 and 3. In other words, you should be able to express y as a product of a multiple of 3 and an even number. 
Let’s also not forget that all powers of 3 are odd numbers. Let us analyse the question data now.
x + y =  3^x  implies y =  3^x  – x. 
If y has to be a multiple of 6, the RHS of the above equation must be a multiple of 6. This can happen if x is an odd multiple of 3.
 
If x is an odd multiple of 3, we can express x as x = 3k, where k is an odd integer. Substituting this value of x in the equation for y, we have, 
 y =  3^x  – 3k.

Taking 3 as common, we have, y = 3  3^{x-1}  – k]. Now, both terms inside the bracket are odd. Since, odd – odd = even, y = 3  . 
From the above analysis, we can conclude that x HAS to be an odd multiple of 3, for y to be a multiple of 6.

Clearly, these two pieces of information are given in two separate statements. Therefore, both statements are required together to solve for a unique answer.

The correct answer option is C.

Hope this helps!

akshatvashistha · Answer

The answer is E for sure. For option C if x=6+y=21=27 Y is not divisible by 6. Statement C is gone so E.

akshatvashistha · Answer

Bunuel   CrackVerbalGMAT  @empowergmat The answer is E for sure. For option C if x=6 Y+21 x+y=27 Y is not divisible by 6. Statement C is gone so E.

Bunuel · Answer

x = 6 and y = 21 satisfy neither x + y = 3^x nor the first statement, which says that x is odd.

GMATGuruNY · Answer

y = 3^x - x 
Question stem, rephrased:
Is  3^x - x  a multiple of 6?

Statement 1:
Case 1: x=1, with the result that  3^x - x = 3^1 - 1 = 2 
In this case, the answer to the question stem is NO.
Case 2: x=3, with the result that  3^x - x = 3^3 - 3 = 24 
In this case, the answer to the question stem is YES.
INSUFFICIENT.

Statement 2:
Case 2: x=3, with the result that  3^x - x = 3^3 - 3 = 24 
In this case, the answer to the question stem is YES.
Case 3: x=6, with the result that  3^x - x = 3^6 - 6 = 3(3^5-2) = 3(ODD-EVEN) = 3(ODD) = ODD = NONMULTIPLE OF 6
In this case, the answer to the question stem is NO.
INSUFFICIENT.

Statements combined:
Case 2: x=3, with the result that  3^x - x = 3^3 - 3 = 24   = MULTIPLE OF 6
Case 4: x=9, with the result that  3^x - x = 3^9 - 9 = 3(3^8-3)  = 3(ODD-ODD) = 3(EVEN) = MULTIPLE OF 6
Case 5: x=15, with the result that  3^x - x = 3^{15} - 15 = 3(3^{14}-5)  = 3(ODD-ODD) = 3(EVEN) = MULTIPLE OF 6
In every case, the answer to the question stem is YES.
SUFFICIENT.

C

Basshead · Answer

y = 3^x - x 
Is  3^x - x  a multiple of 6?

(1) x is odd.

y =  3^1 - 1 ; not a multiple of 6.
y =  3^3 - 3 ; multiple of 6.

Insufficient.

(2) x is a multiple of 3.

y = 3^3 - 3 ; multiple of 6.
 y = 3^6 - 6

Cyclicity of 3: 3, 9, 7, 1

9 - 6 = units digit of 3; not a multiple of 6.

Insufficient.

(1&2) x is odd AND a multiple of 3:

y = 3^3 - 3 ; multiple of 6.
 y = 3^9 - 9

3^9  = units digit of 3
 3 - 9  = units digit of 4

y = 3^9 - 9 ; y will be a multiple of 6.

Answer is C.

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If x and y are positive integers and x + y = 3^x, is y divisible by 6?

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If x and y are positive integers and x + y = 3^x, is y divisible by 6?

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