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04 Dec 2022, 11:44
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:15) correct
39%
(01:34)
wrong
based on 336
sessions
History
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What is the remainder when 2^92 is divided by 7 ?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
04 Dec 2022, 11:58
Kudos
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2^1/7, R=2
2^2/7, R=4
2^3/7, R=1
2^4/7, R=2
Remainder repeats after every 3 steps.
92/3, remainder=2.
Hence, the remainder when 2^92 is divided by 7 will be the same as that when 2^2 is divided by 7 i.e. the remainder is 4.
Answer should be D IMO
Posted from my mobile device
2^2/7, R=4
2^3/7, R=1
2^4/7, R=2
Remainder repeats after every 3 steps.
92/3, remainder=2.
Hence, the remainder when 2^92 is divided by 7 will be the same as that when 2^2 is divided by 7 i.e. the remainder is 4.
Answer should be D IMO
Posted from my mobile device
12 Sep 2024, 08:03
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
We need to find the the remainder when 2^92 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
92 = 90 + 2 = Multiple of 3 + 2
=> Remainder of 2^92 by 7 = Remainder of 2^2 by 7 =4
Method 2: Binomial Theorem
2^92 = 2^(90+2) = 2^90 * 2^2 = 2^(3*30) * 4 = 8^30 * 4
Remainder of 2^92 by 7 = Remainder of 8^30 by 7 * Remainder of 4 by 7 = (Remainder of 8^30) * 4
Remainder of 8^30 = Remainder of (7 + 1)^30 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. 30C30 * 7^0 * 1^30) is divided by 7
=> Remainder of 1 * 1 * 1 is divided by 7
=> Remainder of 1 divided by 7 = 1
=> Remainder of 2^92 = (Remainder of 8^30) * 4 = 1 * 4 = 4
So, Answer will be D
Hope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem
We need to find the the remainder when 2^92 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
92 = 90 + 2 = Multiple of 3 + 2
=> Remainder of 2^92 by 7 = Remainder of 2^2 by 7 =4
Method 2: Binomial Theorem
2^92 = 2^(90+2) = 2^90 * 2^2 = 2^(3*30) * 4 = 8^30 * 4
Remainder of 2^92 by 7 = Remainder of 8^30 by 7 * Remainder of 4 by 7 = (Remainder of 8^30) * 4
Remainder of 8^30 = Remainder of (7 + 1)^30 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. 30C30 * 7^0 * 1^30) is divided by 7
=> Remainder of 1 * 1 * 1 is divided by 7
=> Remainder of 1 divided by 7 = 1
=> Remainder of 2^92 = (Remainder of 8^30) * 4 = 1 * 4 = 4
So, Answer will be D
Hope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem
