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If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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11 Feb 2011, 02:45
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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
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Last edited by Bunuel on 08 Nov 2014, 05:17, edited 2 times in total.
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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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alltimeacheiver wrote: If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10? (1) x = 4n + 2, where n is a positive integer. (2) x > 4 Question should be as follows: 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. Check Number Theory chapter of Math Book for more: mathnumbertheory88376.html
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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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12 Feb 2011, 13:51
If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10? (1) x = 4n + 2, where n is a positive integer. (2) x > 4
looking at the question, we are dealing with even numbers. we know that x has to pos. int. we know that 3x + 1 = odd and we know that 10 is an even number. therefore remainder cannot be zero when divided by 10.
s1, x = 4n + 2. given this eqn in s1, we know that x has to equal an even number. When you substitute the even value of x in 3x + 1, we have an odd number such that it cannot be divisible by 10.
There s1 sufficient.
s2, x > 4. x can have a range of numbers therefore insufficient.
Ans A.



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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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12 Feb 2011, 13:57
maryann wrote: If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10? (1) x = 4n + 2, where n is a positive integer. (2) x > 4
looking at the question, we are dealing with even numbers. we know that x has to pos. int. we know that 3x + 1 = odd and we know that 10 is an even number. therefore remainder cannot be zero when divided by 10.
s1, x = 4n + 2. given this eqn in s1, we know that x has to equal an even number. When you substitute the even value of x in 3x + 1, we have an odd number such that it cannot be divisible by 10.
There s1 sufficient.
s2, x > 4. x can have a range of numbers therefore insufficient.
Ans A. Original question is: If x is a positive integer, is the remainder 0 when [b]3^(x) + 1 is divided by 10?[/b] Solution in my previous post.
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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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02 Jan 2013, 02:05
Bunuel wrote: alltimeacheiver wrote: If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10? (1) x = 4n + 2, where n is a positive integer. (2) x > 4 Question should be as follows: 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. Dear Bunuel, I have another similar question. All info gave me are almost same. The only difference is in Statement (1): \(x= 3^n+1\). Answer goes to E. I knew that approach should be the same reference "\(4^n+2\)". I look at someone started from 1, 3,9,27,81, xxx3,xxx9,xxxx7,xxxx1, xxxx3, xxxx9, xxxx7, xxxx1, and so on.... And then reasoned that reference, which is \(4^n+2\). I'm wondering how to quick approach that reference. Can you answer in this thread or should I submit a new post?



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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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02 Jan 2013, 04:12
curtis0063 wrote: Bunuel wrote: alltimeacheiver wrote: If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10? (1) x = 4n + 2, where n is a positive integer. (2) x > 4 Question should be as follows: 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. Dear Bunuel, I have another similar question. All info gave me are almost same. The only difference is in Statement (1): \(x= 3^n+1\). Answer goes to E. I knew that approach should be the same reference "\(4^n+2\)". I look at someone started from 1, 3,9,27,81, xxx3,xxx9,xxxx7,xxxx1, xxxx3, xxxx9, xxxx7, xxxx1, and so on.... And then reasoned that reference, which is \(4^n+2\). I'm wondering how to quick approach that reference. Can you answer in this thread or should I submit a new post? Please post full question in a separate topic.
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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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02 Jan 2013, 09:02



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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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09 Mar 2014, 13:21



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Re: If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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12 May 2015, 22:02
alltimeacheiver wrote: 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 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 straightaway 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
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If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
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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



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