Data Sufficiency

Hello,

Can you please assist with this:

If r is the remainder of (2^(8x + 3) + y )/5 and x is
a positive integer, what is the value of r?

(1) y is even
(2) y = 2

OA: B


When I plug in x = 1 I get 2^(8x1 + 3) = 2^11 = 2048

1) Let y = 2.

So (2048 + 2)/5 => Remainder = 0
Let y = 4
So (2048 + 4)/5 => Remainder = 2

Insufficient

2) y = 2

Sufficient

I was wondering if this approach is correct or if there is a easier way so that we can avoid calculating 2^11 ?

Thanks a lot,
Sri

Hi Sri,
This is kind of question where we need to rephrase the given pattern.

You have to note the pattern here,
2^1=2
2^2=4
2^3=8
2^4=_6
after here pattern repeats like 2,4,8,6,2,4,8,6.......

2^8x always ends with 6(since x is positive integer)

so 2^(8x+3) will always end with 8

now question can be re-written as
what is the reminder when (8+Y)/5 ?

now we need value of Y.

Statement 1: Y is even.
when Y =2
8+2/5 =0
When Y =4
8+4/5 =2
Hence Insufficient.

Statement 2: Y = 2
so 8+2/5 = 0.
Hence Sufficient.

Ans is B

Regards,
Uva.

gmattesttaker2 wrote:Hello,

Can you please assist with this:

I was wondering if this approach is correct or if there is a easier way so that we can avoid calculating 2^11 ?

Thanks a lot,
Sri

Hi Sri!

For remainder questions, we often only need to look at the last 1 or 2 digits of the number. In this case, since we're dividing by 5, only the last digit actually matters.

So, we can recognize the pattern for powers of 2:

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = ..6
2^5 = ..2

and so on. So, the units digit of powers of 2 is a 4 part cycle: 2, 4, 8, 6, ...

Now we just see where 11 falls in the cycle! 11 is one short of 12, which would end the cycle, so 2^11 must end in "8".

As an aside, you didn't actually prove that (2) was sufficient, since you only tested x=1; x could be any positive integer! You should test at least 1 more value of x to be sure.

For x=2:

(2^(19) + y)/5

Hey, 19 is also 2nd last in the cycle and 2^19 has a units digit of 8.

If we're not convinced yet:

For x=3:

(2^(27) + y)/5

and 27 is also 2nd last in the cycle with a units digit of 8.

How could we figure that out without picking numbers? The "8x" in the "8x+3" will always be a multiple of 4. So, 2^8x would always be the last part of the cycle! Since we're adding 3, (2^(8x+3)) will always be 3rd in the cycle and have a units digit of 8.

I hope that helps!

Stuart

r = (2^(8x + 3) + y )/5
= (2^8x * 8 + y) / 5
= (2^4 * 8 + y)/5 (using remainder rule = 8x/4 = 0.. So, 2^4 will give unit digit)
= ( 128 + y)/5 =

Statement 1:
y = EVEN
y = 0, r = 3
y = 2, r = 0
INSUFFICIENT

Statement 2:
y = 2
130/5 = r = 0
SUFFICIENT

Answer [spoiler]{B}[/spoiler]

Uva@90 wrote:Hi Sri,
This is kind of question where we need to rephrase the given pattern.

You have to note the pattern here,
2^1=2
2^2=4
2^3=8
2^4=_6
after here pattern repeats like 2,4,8,6,2,4,8,6.......

2^8x always ends with 6(since x is positive integer)

so 2^(8x+3) will always end with 8

now question can be re-written as
what is the reminder when (8+Y)/5 ?

now we need value of Y.

Statement 1: Y is even.
when Y =2
8+2/5 =0
When Y =4
8+4/5 =2
Hence Insufficient.

Statement 2: Y = 2
so 8+2/5 = 0.
Hence Sufficient.

Ans is B

Regards,
Uva.

Hello Uva,

Thank you very much for your detailed and thorough explanation.

Best Regards,
Sri

Stuart Kovinsky wrote:

gmattesttaker2 wrote:Hello,

Can you please assist with this:

I was wondering if this approach is correct or if there is a easier way so that we can avoid calculating 2^11 ?

Thanks a lot,
Sri

Hi Sri!

For remainder questions, we often only need to look at the last 1 or 2 digits of the number. In this case, since we're dividing by 5, only the last digit actually matters.

So, we can recognize the pattern for powers of 2:

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = ..6
2^5 = ..2

and so on. So, the units digit of powers of 2 is a 4 part cycle: 2, 4, 8, 6, ...

Now we just see where 11 falls in the cycle! 11 is one short of 12, which would end the cycle, so 2^11 must end in "8".

As an aside, you didn't actually prove that (2) was sufficient, since you only tested x=1; x could be any positive integer! You should test at least 1 more value of x to be sure.

For x=2:

(2^(19) + y)/5

Hey, 19 is also 2nd last in the cycle and 2^19 has a units digit of 8.

If we're not convinced yet:

For x=3:

(2^(27) + y)/5

and 27 is also 2nd last in the cycle with a units digit of 8.

How could we figure that out without picking numbers? The "8x" in the "8x+3" will always be a multiple of 4. So, 2^8x would always be the last part of the cycle! Since we're adding 3, (2^(8x+3)) will always be 3rd in the cycle and have a units digit of 8.

I hope that helps!

Stuart

Hello Stuart,

Thank you very much for your detailed and excellent explanation (as always).

Best Regards,
Sri