What is the remainder of 2^51 divided by 7?

2 and 7 are co - primes

=> R[2^6/7] = 1
=> R[2^51/7] = R[2^3/7] = 1

=> Answer: A
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Asked: What is the remainder of 2^51 divided by 7?

\(2^{51}mod7 = 2^{3*17} = 8^{17}mod7 =1^{17}mod7 = 1mod7\)

IMO A
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Using Euler theorem,

When A^x is divided by N, where A and N are co prime.
Since N here has only one factor other than 1
Ø(n) = n(1-1/n)
Ø(7) = 7(1-1/7)
Ø(7) = 6

So A^Ø(n) mod n = 1 as remainder
2^6 mod 7 = 1
(2^6)8 mod 7 * 2^3 mod 7
1*8 mod 7
8 mod 7
1 as remainder

Answer is A

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\(2^3\) Divided by 7 leaves remainder = 1

Taking 17 powers both sides

\((2^3)^{17}\) Divided by 7 leaves remainder = \(1^{17}\)

\((2)^{51}\) Divided by 7 leaves remainder = \(1\)

Answer: Option B

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Tell me if my method is wrong. I doubt it but it landed me on the correct answer.

2^1=2, 2^2=4, 2^3=8, 2^4=16. We've got cyclicity of 2, 4, 8, 6.

Count 7 times through the cyclicity of 2 and you land on 8.

8/7 leaves you with a remainder of 1.

Answer A.

Can anyone check my reasoning and confirm my calculations are accurate?

I used the cycles of 2 in exponents (2,4,8,6), and divided 51 by 4 to give 12r3. This indicates that we are in the third cycle, so we would be on 8 within the cycle. Therefore when divided by 7 the remainder would be 1.

Answer - A

\(2^3*2^(16)*3/(7)\)
1*1^(16)*3
=1
A:)

Coxy4972

That's absolutely correct way.
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No, it is not correct.

If you're looking at cyclic patterns in units digits, then you're answering the question "what remainder will I get if I divide by 10?" It's only when we divide a number by 10 that the units digit tells us the remainder.

We can also use units digits to quickly deduce remainders when we divide by 2 or by 5 (and units digits give us some information when we divide by related numbers like 20), but if you're dividing by a number unrelated to 2 and 5, units digits actually tell you nothing about what remainder you will get. So if you're dividing, say, by 3 or by 7 or by 9 or by 11, you actually don't learn anything useful by finding the units digit of your number, and if the units digit does give you the right answer, it's just pure coincidence.

You can see why that's true. In this question, it is correct that 2^51 has a units digit of 8. But if you look at numbers with a units digit of 8, you can see that you can have any remainder at all when you divide by 7. So 18 gives us a remainder of 4, 28 gives a remainder of 0, 38 gives a remainder of 3, and so on. Notice these numbers do not give a remainder of 1 when you divide by 7; we cannot simply divide the units digit by 7 to get the answer.

So while that method appears to give the right answer to this question, that just works out by luck, and the same method wouldn't work if the exponent in the problem were changed. The concept tested in this particular question is not one I've seen tested in any authentic GMAT question, so I wouldn't worry about it at all. It can be solved using remainder arithmetic (also called "modular arithmetic"), but that's not something the GMAT expects test takers to know. Essentially, since we get a remainder of 1 when we divide 2^3 by 7, we'll get a remainder of 1 when we divide any power of 2^3 by 7, including (2^3)^17 = 2^51.
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2^51
Let's divide 51 by 4 and take the remainder as the final powerinstead of 51.
Because unit digit of 2 is repeated after every 2^4 term
Example
2^1=2
2^2=4
2^3=8
2^4=6
2^5=2
Similarly now onwards unit digit keeps on repeating.
Coming back to 51/4 , the remainder is 3.
Now our question becomes 2^3/7
2^3 =8.
Hence the remainder when 8/7 is 1.
Our answer is A

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What is the remainder of 2^51 divided by 7?

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

Solution:
\(2^{51}=2^{(3)(17)}=8^{17}\)
\((7+1)^{17}\) -> \(1^{17}/7\) = remainder of 1

Answer choice A IMO

Determining the cyclicity of the remainders (not the unit digits) - Please understand there is a difference in the two.

\(2^1 = 2\) So, \(2^1/7\) = remainder 2
\(2^2 = 4\) So, \(2^2/7\) = remainder 4
\(2^3 = 8\) So, \(2^3/7\) = remainder 1
\(2^4 = 16\) So, \(2^4/7\) = remainder 2
\(2^5 = 32\) So, \(2^5/7\) = remainder 4
\(2^6 = 64\) So, \(2^6/7\) = remainder 1

So we see the cyclicity of the remainders is 3 and in the order 2--> 4 --> 1
51/3 = 17 and leaves no remainder, so it is going to be the third remainder (2--> 4 --> 1)

Answer: A

Solution

To Find

• Find the remainder when \(2^{51}\) is divided by 7.

Approach and Working Out

• \(2^{51} \) can be manipulated as \((2^3)^{17}\).

o We do that as \(2^3\) (i.e, 8) leaves a remainder of 1, when divided by 7.

• \((2^3)^{17}\) divided by 7 will leave a remainder of

= 1 × 1 × 1 … 17 times
= 1

The answer is 1.


Correct Answer: Option A
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What are co primes?
N1/N2 — remainder is x and N2/N1 remainder is x ?

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Solution:

Notice that 2^51 = (2^3)^17 = 8^17. Since the remainder when 8 is divided by 7 is 1, the remainder when 8^17 is divided by 7 is 1^17 = 1.

Answer: A
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