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Find the remainder of 2^100/12

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by GMATGuruNY » Mon Jan 29, 2018 4:21 am
Vincen wrote:Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none
Test easy exponents and look for a pattern.

2²/12 = 4/12 = 0 R4.
2³/12 = 8/12 = 0 R8.
2�/12 = 16/12 = 1 R4.
2�/12 = 32/12 = 2 R8.
2�/12 = 64/12 = 5 R4.
2�/12 = 128/12 = 10 R8.

The cases above illustrate the following:
When the exponent is EVEN, the remainder is 4.
When the exponent is ODD, the remainder is 8.
Since 2¹��/12 has an even exponent, the remainder will be 4.

The correct answer is A.
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by Scott@TargetTestPrep » Tue Jan 30, 2018 10:32 am
Vincen wrote:Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none
Let's find a remainder pattern:

2^1/12 = 0 remainder 2

2^2/12 = 0 remainder 4

2^3/12 = 0 remainder 8

2^4/12 = 1 remainder 4

2^5/12 = 2 remainder 8

2^6/12 = 5 remainder 4

We see that, besides 2^1/12, we have this pattern: when 2 is raised to an even power, the remainder is 4, and when 2 is raised to an odd power the remainder is 8. Thus, 2^100/12 has a reminder of 4.

Answer: A

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by EconomistGMATTutor » Wed Jan 31, 2018 1:11 pm
Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none

The OA is the option A.

Is there a simple way to solve this PS question? I can I solve it? I'd appreciate your help.
Hi Vincen,
Let's take a look at your question.

We need to find the remainder of
$$\frac{2^{100}}{12}$$
$$=\frac{2^{2\times50}}{12}$$
$$=\frac{4^{50}}{12}$$

Let's check what happens when powers of 4 is divided by 12, starting from 4^2.
$$4^2=16$$
$$\frac{16}{12}=1\ remainder\ 4$$

$$4^3=64$$
$$\frac{64}{12}=5\ remainder\ 4$$

$$4^4=256$$
$$\frac{256}{12}=21\ remainder\ 4$$

$$4^5=1024$$
$$\frac{1024}{12}=85\ remainder\ 4$$

Which shows that remainder is 4 for all the positive powers of 4.
Therefore, option A is correct.

Hope it helps.
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