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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
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (01:27) correct
23%
(02:02)
wrong
based on 1324
sessions
History
Date
Time
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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.
General Discussion
12 May 2015, 22:02
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alltimeacheiver
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
Though this is quite an easy question, the accuracy statistics suggest that roughly 1 out of every 4 students who attempted this question got it wrong.
This may have happened because the students first straight-away started from the given statements and then, got confused in processing the question statement and the Statement 1 (and 2) together.
Here's an alternate solution that eliminates the chances of such confusion!
Let's first analyze the question statement alone:
The given expression is \(3^{x} + 1\)
The units digit of 3 can be 3 (for powers of the form 4m+1), 9 (for powers of the form 4m+2), 7 (for powers of the form 4m+3) or 1(for powers of the form 4m)
Out of these 4 possible unit digits, the sum \(3^{x} + 1\) will be divisible by 10 only when the units digit is 9.
So, the question is actually asking us to find if the power of 3, that is x, is of the form 4m + 2 or not.
Now that we've simplified the question, analyzing the 2 statements is going to be a cakewalk!
Please note how the analysis in my solution is quite similar to the analysis in the solutions posted above. The point of difference comes in when I did that analysis. I did it before I went to Statements 1 and 2. The benefit of doing this analysis with the question statement itself is that I now have a very clear idea of what I need to look for, in order to answer the question. And because of this clear idea, the chances of my getting confused by irrelevant information in Statements 1 and 2 are also greatly reduced.
Hope this helped!
Japinder
