What is the remainder when 2^100 is divided by 15?

Question

A. 1
B. 2
C. 3
D. 5
E. 6

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odiodeserunt · Answer

to find the remainder, find the cyclicity of remainders when dividing powers of 2 by 15

2/15 = 2
2^2/15 = 4
2^3/15= 8
2^4/15 = 1
2^5/15 = 2
2^6/15 = 4

hence the cycle repeats after every 4 numbers (2,4,8,1)
now to find how many cycles are there in 2^100 : 100/25
There are 25 whole cycles.
 The answer must be 1 , Option A

nemoexplicabo · Answer

Great question! This is a classic remainder problem that tests your understanding of cyclicity — one of those concepts that shows up frequently on the GMAT and can save you tons of time once you master it.

The key here is recognizing that when you divide powers of 2 by 15, the remainders follow a repeating pattern. Instead of trying to calculate 2^100 (which would be astronomically large), we just need to find where 100 falls in this cycle.

Step 1: Find the pattern of remainders

Let me calculate the first few powers of 2 divided by 15:

2^1 ÷ 15 = remainder 2
2^2 ÷ 15 = 4 ÷ 15 = remainder 4 
2^3 ÷ 15 = 8 ÷ 15 = remainder 8
2^4 ÷ 15 = 16 ÷ 15 = remainder 1
2^5 ÷ 15 = 32 ÷ 15 = remainder 2
2^6 ÷ 15 = 64 ÷ 15 = remainder 4

Notice what happened? The remainders are: 2, 4, 8, 1, 2, 4...

The pattern repeats every 4 powers! This is the cyclicity of 2 when divided by 15.

Step 2: Determine where 100 falls in the cycle

Since the cycle repeats every 4 numbers, I need to figure out what position 100 occupies in this cycle.

100 ÷ 4 = 25 with remainder 0

When the remainder is 0, that means 100 is divisible by 4, so it falls at the END of a cycle (position 4).

Step 3: Find the remainder at position 4

Looking back at our pattern:
- Position 1 in cycle: remainder 2
- Position 2 in cycle: remainder 4
- Position 3 in cycle: remainder 8
- Position 4 in cycle: remainder 1

Since 100 is at position 4 of the cycle, the remainder is 1.

Answer: A

Common traps to avoid:

One mistake students often make is thinking that since 100 is a large number, the remainder must also be large. But remainders are all about patterns, not magnitude. Another pitfall is miscounting the cycle — make sure to check where your exponent falls carefully. If you got remainder 2 or 4, you likely miscalculated which position in the cycle corresponds to the exponent.

Key takeaway: Cyclicity problems are all about pattern recognition. Once you identify the cycle length, the problem becomes much simpler. This same approach works for any base and divisor — just find the repeating pattern and determine where your exponent lands in that cycle. On test day, this technique can turn a 2-minute calculation into a 30-second problem!

Dereno · Answer

Look for the nearest multiple of 15. We can take either 16 or 32.

16 = 2^4

32 = 2^5

2^4 divided by 15 leaves a remainder 1.

While 2^5 divided by 15 leaves a remainder 2.

To make Calculations more simple, let’s stick with 2^4.

So, 2^100 can be written as  ^(4*25) = (2^4)*25

(2^4)*25 Divided by 15, yields (1)^25 =  1.

Option A

nemoexplicabo · Answer

This is a classic remainder question that relies on the cyclicity of remainders.
Key concept being tested: cyclicity of remainders in Number Properties.
 
 Instead of computing 2^100 directly, examine the pattern of remainders when powers of 2 are divided by 15. 
 Compute the first few powers:
 
 2^1 = 2 gives remainder 2 
 2^2 = 4 gives remainder 4 
 2^3 = 8 gives remainder 8 
 2^4 = 16 gives remainder 1 
 2^5 = 32 gives remainder 2 again  
 
 You can see the remainders repeat every 4 terms: 2, 4, 8, 1. 
 To find where 100 falls in this cycle, divide 100 by 4. The quotient is 25 with a remainder of 0, meaning the exponent corresponds to the last term of the cycle. 
 The 4th term in the cycle has remainder 1. Therefore the remainder when 2^100 is divided by 15 is 1.  
Common trap: Some students try to break 15 into its factors 3 and 5 and handle each separately, which adds unnecessary steps. Others forget that a zero remainder when dividing the exponent by the cycle length means you take the last term in the pattern, not the first.
Takeaway: For large exponent remainder problems, identify the repeating pattern of remainders early. Once you know the cycle length, you can quickly determine the remainder by simple division without heavy computation.

ExpertsGlobal5 · Answer

Video explanation:

https://www.youtube.com/watch?v=fjp1d0ohWe0

luisdicampo · Answer

Deconstructing the Question

We must find the remainder when

2^{100}

is divided by  15 .

A fast GMAT strategy is to look for repeating patterns in powers modulo  15 .

Step-by-step

Compute small powers of  2

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

Since

16 \equiv 1 \pmod{15}

we get

2^4 \equiv 1 \pmod{15}

Now rewrite

2^{100}

= (2^4)^{25}

Using the modular equivalence

(2^4)^{25} \equiv 1^{25}

\equiv 1

So the remainder when dividing  2^{100}  by  15  is

1

Answer A: 1

What is the remainder when 2^100 is divided by 15?

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GMAT Problem Solving (PS) Questions

What is the remainder when 2^100 is divided by 15?

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