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If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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06 Nov 2016, 08:21
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88% (01:44) correct 12% (01:14) wrong based on 112 sessions
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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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06 Nov 2016, 09:18
Bunuel wrote: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 5? (1) n = 10 (2) 5 < n < 11 FROM STATEMENT  I ( SUFFICIENT )Remainder of a number divided by 5 will come only when the Units digit is not 0 or 5 1^1 = 1 2^2 = 4 3^3 = 7 4^4 = 6 5^5 = 5 6^6 = 6 7^7 = 3 8^8 = 6 9^9 = 9 10^10 = 0 Here the units digit will be 7 ( ie, 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 ), thus the remainder will be 2FROM STATEMENT  II ( INSUFFICIENT )Since, we need to know the units value of the series, it is essential to find the value of n for the series 1^1 + 2^2 + 3^3 + . . . + n^nHere , 5 < n < 11 ; n can take multiple values  n = { 6 , 7 , 8 , 9 , 10 } Thus unique value of units digit can not be obtained, hence this statement alone is not sufficient to find the remainde of the series divided by 5 Hence, Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked, answer will be (A)
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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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06 Nov 2016, 23:37
Bunuel wrote: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 5? (1) n = 10 (2) 5 < n < 11 From statement 1 n=10 x= 1+4+7+6+5+6+3+6+9+0= 47/5 = 2 ( taking only the units digit of all powers) Hence Sufficient from statement 2 5<n<11 n=6, x=1+4+7+6+5+6= 29/5= 4 n=7, x= 1+4+7+6+5+6+3= 32/5= 2 n=8, x= 1+4+7+6+5+6+3+6= 38/5= 3 remainder varies hence insufficient Hence A



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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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07 Nov 2016, 02:34
Since the value of x can be determined with n=10 (or whatever be the value of n) we can determine the reminder of the number when divided by 5 Hence A is sufficient. for option B 5<n<11 n=6 \(1^1+2^2+3^3+4^4+5^5+6^6\) =1+4+7+6+5+6 (calculate only the units digits) =29 29/5 reminder 4n=6 29(Calculated previously)+3 =32 32/5 reminder 2different answers for n=6,7 hence not sufficient Regards ARUN Also chekout the review on GMAT Practice testshttp://gmatclub.com/forum/allgmatcatpracticetestslinkspricesreviews77460620.html#p1758429



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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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07 Nov 2016, 06:37
Bunuel wrote: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 5? (1) n = 10 (2) 5 < n < 11 x = 1^1 + 2^2 + 3^3 + . . . + n^n, Question: what is the remainder when x is divided by 5? The remainder depends on the value of n hence the question is basically what is the value of n Statement 1: n = 10 SUFFICIENT Statement 2: 5 < n < 11 n may be 6 or 7 or 8 or 9 or 10 leading to the different remainders. Hence NOT SUFFICIENT Answer: option A
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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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07 Nov 2016, 08:01
from statement 1, we get a constant value and that value divided by 5 gives a fix number. no need of calculation.
second statement does not give a fix value as n is not fixed and n can be anything within that range



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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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07 Nov 2016, 21:26
1)From statement 1 we can get a constant value and you'll get the remainder if you divide the value with number 5. So A is Sufficient
2)From statement 2 you have (6,7,8,9,10) values so we need a specific value to get a sufficient answer. So insufficient



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Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n,
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12 Jul 2018, 19:58
Such kinds of question trick you to perform calculation but in truth you should not be doing it to save some precious time during the exam.
Question here in simple term is can you tell the remainder if you have certain information. Do note that question is NOT that what is the value of remainder.
No matter what is value of N, if you have a confirmed value of N, X can be definitely derived (No calculation needed for this fact). However, if N is not a specific value but a range, you can get multiple values of X and in turn multiple remainders.
Therefore, Option (1) alone is enough and Option 2 in insufficient.
Answer A.




Re: If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, &nbs
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12 Jul 2018, 19:58






