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Bunuel
What is the remainder when \(2^{2006}\) is divided by 7?

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



Solution


    • We know that when (\(2^3 = 8\)) is divided by 7 remainder is 1.
      o So, when \(2^{3n}\) will divided by 7 remainder will be 1, where n is a positive integer.
    • So, lets express the power of \(2^{2006}\) (i.e. 2006 ) as multiple of 3.
    • \([\frac{2^{2006}}{7}]_{remainder} = [\frac{(2^{2004}*2^2)}{7}]_{remainder} = 4*[\frac{(2^{2004})}{7}]_{remainder} = 4*1 = 4\)
Thus, the correct answer is Option B
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I'm really struggling in the method you have used? Can you please explain me this concept in detail?
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Bunuel
What is the remainder when \(2^{2006}\) is divided by 7?

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


We see that 2^2006 = (2^3)^668 * 2^2 = 8^668 * 4. Since 8 has a remainder of 1 when divided by 7, 8^668 also has a remainder of 1 when divided by 7 (notice that 1^668 = 1). Therefore, the remainder when 2^2006 = 8^668 * 4 is divided by 7 is 1 * 4 = 4.

Answer: B
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2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
this pattern goes on.

now 2006/4 = r2
our digit unit with r2 = 2^2 = 4
ans 4/7
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We need to find the the remainder when \(2^{2006}\) is divided by 7 ?

Let's solve the problem using two Methods:

Method 1: Cyclicity of Remainder of power of 2 by 7

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7

Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1

=> Cycle is 3

Remainder of 2006 by 3 = 2
=> Remainder of \(2^{2006}\) by 7 = Remainder of 2^2 by 7 =4


Method 2: Binomial Theorem

\(2^{2006}\) = \(2^{2004+2}\) = 2^2004 * 2^2 = 2^(3*668) * 4 = 8^668 * 4

Remainder of \(2^{2006}\) by 7 = Remainder of 8^668 by 7 * Remainder of 4 by 7 = (Remainder of 8^668) * 4

Remainder of 8^668 = Remainder of (7 + 1)^668 by 7

Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7

=> Problem is reduced to what is the remainder when the last term (i.e. 668C668 * 7^0 * 1^668) is divided by 7
=> Remainder of 1 * 1 * 1 is divided by 7
=> Remainder of 1 divided by 7 = 1
=> Remainder of \(2^{2006}\) by 7 = (Remainder of 8^668) * 4 by 7 = 1 * 4 = 4

So, Answer will be B
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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