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If x is a positive integer, is the remainder 0 when 3^(x) + [#permalink]
 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
Last edited by Bunuel on 12 Feb 2011, 13:57, edited 1 time in total.
Edited the question.
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Re: divisiblity with 10 [#permalink]
11 Feb 2011, 03:34
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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: math-number-theory-88376.html
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Re: divisiblity with 10 [#permalink]
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: divisiblity with 10 [#permalink]
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: divisiblity with 10 [#permalink]
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: divisiblity with 10 [#permalink]
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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
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Re: divisiblity with 10 [#permalink]
02 Jan 2013, 09:02
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Re: divisiblity with 10
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