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Updated on: 04 Feb 2021, 09:38
Originally posted by alltimeacheiver on 11 Feb 2011, 02:45.
Last edited by Bunuel on 04 Feb 2021, 09:38, edited 4 times in total.
Last edited by Bunuel on 04 Feb 2021, 09:38, edited 4 times in total.
Edited the question.
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A
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If x is a positive integer, is the remainder 0 when \(3^x + 1\) is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4
11 Feb 2011, 03:34
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If x is a positive integer, is the remainder 0 when \(3^x + 1\) is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
Last digit of \(3^x\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4n+2}\) will have the same last digit as \(3^2\) (remainder upon division 4n+2 upon cyclicity 4 is 2, which means that 3^{4n+2} will have the same last digit as 3^2). Last digit of \(3^2\) is \(9\). So \(3^{4n+2}+1\) will have the last digit \(9+1=0\). Number ending with 0 is divisible by 10 (remainder 0). Sufficient.
(2) x > 4. Clearly insufficient.
Answer: A.
(1) x = 4n + 2, where n is a positive integer.
Last digit of \(3^x\) repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4n+2}\) will have the same last digit as \(3^2\) (remainder upon division 4n+2 upon cyclicity 4 is 2, which means that 3^{4n+2} will have the same last digit as 3^2). Last digit of \(3^2\) is \(9\). So \(3^{4n+2}+1\) will have the last digit \(9+1=0\). Number ending with 0 is divisible by 10 (remainder 0). Sufficient.
(2) x > 4. Clearly insufficient.
Answer: A.
26 May 2016, 05:32
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This question checks divisibility of 3.
Concentrate only on last digit of product and not the entire value.
Now \(3^4\)= 81, i.e. last digit is 1. \(3^5\) has last digit 3.
So \(3^{4x}\) will have "1" as its last digit. [\(3^{4x}\) can be written as (\(3^4\))^x
Statement 1
\(3^{4n+2}\) is nothing but (\(3^{4n}\))* (\(3^2\)). The last digit of this product will be 9.
Adding 1 to this sum will yield a zero and so the entire product will be divisible by 10.
(1) is sufficient
Statement 2
x>4.
x=5 --> This will have last digit of 3's power as 3 [(\(3^4\))*(3) => Last digit will be 3].
Add 3+1 = 4. Not divisible by 10.
x=6 --> This will have last digit of 3's power as 9 [(\(3^4\))*(9) => Last digit will be 9].
Add 9+1 = 10. Will be divisible by 10.
So Statement (B) is not sufficient.
Answer is A
Concentrate only on last digit of product and not the entire value.
Now \(3^4\)= 81, i.e. last digit is 1. \(3^5\) has last digit 3.
So \(3^{4x}\) will have "1" as its last digit. [\(3^{4x}\) can be written as (\(3^4\))^x
Statement 1
\(3^{4n+2}\) is nothing but (\(3^{4n}\))* (\(3^2\)). The last digit of this product will be 9.
Adding 1 to this sum will yield a zero and so the entire product will be divisible by 10.
(1) is sufficient
Statement 2
x>4.
x=5 --> This will have last digit of 3's power as 3 [(\(3^4\))*(3) => Last digit will be 3].
Add 3+1 = 4. Not divisible by 10.
x=6 --> This will have last digit of 3's power as 9 [(\(3^4\))*(9) => Last digit will be 9].
Add 9+1 = 10. Will be divisible by 10.
So Statement (B) is not sufficient.
Answer is A
