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What is the remainder when 2^28 is divided by 3?

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by [email protected] » Wed Nov 01, 2017 12:37 pm
Hi swerve,

The GMAT will NEVER expect you to actually calculate the value of 2^28, so you should instead 'play around' with the math a bit and see if a pattern emerges (because there's almost always a pattern).

We're asked for the REMAINDER when 2^28 is divided by 3. By definition, the remainder can only be 0, 1 or 2, so we can eliminate Answers C, D and E. Let's look for a pattern when we divide smaller 'powers of 2' by 3....

2^1 = 2 and 2/3 = 0r2
2^2 = 4 and 4/3 = 1r1
2^3 = 8 and 8/3 = 2r2
2^4 = 16 and 16/3 = 5r1
2^5 = 32 and 32/3 = 10r2
2^6 = 64 and 64/3 = 21r1

Notice how each ODD exponent gives us a remainder of 2 and each EVEN exponent gives us a remainder of 1. That's the pattern we need. Since 28 is an EVEN number, the remainder will be 1.

Final Answer: A

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by Brent@GMATPrepNow » Thu Nov 02, 2017 8:20 am
swerve wrote:What is the remainder when 2^28 is divided by 3?

A. 1
B. 2
C. 3
D. 4
E. 5
Rich's approach is exactly the same as the approach I'd use.
That said, here's another way to look solve it.

RULE: When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

In this question, we're dividing by 3, so the only possible remainders are 0, 1 and 2
So, we can ELIMINATE C, D and E

Now let's examine answer choice A
If the remainder (when dividing 2^28 by 3) is 1, then 2^28 - 1 would leave remainder 0 when divided by 3
In other words, 2^28 - 1 would be divisible by 3
If it's true that 2^28 - 1 is divisible by 3, then we can be certain that 2^28 divided by 3 leaves remainder 1
Let's find out if 2^28 - 1 is divisible by 3
Since 2^28 - 1 is a DIFFERENCE of squares, we can factor it as follows:
2^28 - 1 = (2^14 + 1)(2^14 - 1
2^28 - 1 = (2^14 + 1)(2^7 + 1)(2^7 - 1)) [since 2^14 - 1 is ALSO a difference of squares]
2^28 - 1 = (2^14 + 1)(128 + 1)(128 - 1)
2^28 - 1 = (2^14 + 1)(129)(127)
2^28 - 1 = (2^14 + 1)(3)(43)(127)

This means 2^28 - 1 IS divisible by 3, which means 2^28 divided by 3 leaves a remainder of 1

Answer: A

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Brent
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by Scott@TargetTestPrep » Fri Nov 01, 2019 6:55 pm
swerve wrote:What is the remainder when 2^28 is divided by 3?

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is A.

Please, can any expert assist me with this PS question? I don't have it clear.
Let's determine the pattern of remainders when a base of 2 is divided by 3.

2^1 / 3 = 0 remainder 2

2^2 / 3 = 1 remainder 1

2^3 / 3 = 2 remainder 2

2^4 / 3 = 5 remainder 1

We see that when 2 is raised to an even power and divided by 3, the remainder is 1. Thus, the remainder when 2^28 is divided by 3 is 1.

Answer: A

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