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If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w [#permalink]
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20 Jan 2017, 06:37
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Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w [#permalink]
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20 Jan 2017, 23:05
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Bunuel wrote: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, what is the remainder when x is divided by 5? (1) n is an even integer with a units digit less than 6. (2) The units digit of n is 2. (1) n=10 > 1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0 = 1  1  1 + 1 + 0 +1  1  1 + 1 + 0 = 0 n=20 > pattern repeats 2 times 0 + 0 = 0 ... n=2 > 1 + 4 = 5 n=12 > 0 + 0 = 0 ... n=4 > 1 + 4 + 9 + 6 = 1  1  1 + 1 = 0 n=14 > 0 + 0 = 0 ... Sufficient (2) Single case from previous example. Sufficient. Answer D Another approach: \(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + ... =\) \(= 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 0^2 + 1^2 + 2^2 + ... =\) \(= 1^2 + 2^2 + (2)^2 + (1)^2 + 0^2 + 1^2 + 2^2 + (2)^2 + 0^2 + 1^2 + 2^2 + ... =\) \(= 1 + 4 + 4 + 1 + 1 + 4 + 4 + 1 + 0 + 1 + 4 + 4 + 0 + 1 + 4 ... =\) \(= 1  1  1 + 1 + 1  1  1 + 1 + 0 + 1  1  1 + 0 + 1  1 ...\) Units digits 0, 2 and 4 will give us clusters of 2, 4, 10 2 = 1  1 = 0 4 = 1  1  1 + 1 = 0 10 = 1  1  1 + 1 + 1  1  1 + 1 + 0 = 0 All the rest is just a combination of previous three.



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If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w [#permalink]
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01 Mar 2017, 22:41
Bunuel wrote: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, what is the remainder when x is divided by 5? (1) n is an even integer with a units digit less than 6. (2) The units digit of n is 2. Okay..here's what I did.
Think of all the remainders possible with 5..{1,2,3,4,0} Now think of their squares..{1,4,9,16,0} Now think of the remainders these squares give when divided by 5..{1,4,4,1,0} The remainders can also be written as..{1,1,1,1,0}..
Do you see it?.. If n is Even or a Multiple of 5, The remainder of their sum will be 0
In case when n is Even..the remainders can be {1,1} or {1,1,1,1} or {1,1,1,1,1,1}...so on and so forth.. In all these cases..the total is 0.
In case when n is a Multiple of 5..the remainders can be {1,1,1,1,0} or {1,1,1,1,0,1,1,1,1,0}..so on and so forth.. In all the cases..the total is 0.
Let's consider the given statements.. (1) It states us that n is even. Sufficient.
(2) It also states that n is even. Sufficient.
Thus, (D) Bunuel Please review my solution
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Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w [#permalink]
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17 Mar 2017, 12:49
"Need to verify if x = 5k i.e if unit digit is 0 or 5 then it is divisible. S1) If n = 2 then 1+4 = 5 n = 4 then 1 + 4 + 9 + 16 = 30. Div by 5 So generalizing all possible values of n as 10,12,14 or 20,22, 24 gives remainder as 0. Sufficient S2) consider n = 2, 12, 22 from S1 . Sufficient Answer is D
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If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w [#permalink]
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14 Apr 2017, 00:09
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we have formula for total sum of squares x = [n(n+1)(2n+1)]/6 Unit digit = 2 > unit digit of the total sum of squares = 2(2+1)(2*2+1)]/6 = 5 > x is divisible by 5 Unit digit = 4 > unit digit of the total sum of squares = 4(4+1)(2*4+1)]/6 = 0 > x is divisible by 5 Unit digit = 0 > unit digit of the total sum of squares = 0(0+1)(2*0+1)]/6 = 0 > x is divisible by 5 Statement 1: sufficient Statement 2: a special case > sufficient Hence, D
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Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w [#permalink]
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13 Feb 2018, 05:50
Bunuel wrote: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, what is the remainder when x is divided by 5? (1) n is an even integer with a units digit less than 6. (2) The units digit of n is 2. I did the following way  Sum of squares of n integers = \(\frac{n(n+1)(2n+1)}{6}\) So, X = \(\frac{n(n+1)(2n+1)}{6}\) Statement I: Lets take \(n = 10,12,14,20,22,24,30,32,34..... etc.\) For all the above numbers X will be divisible by 5. Statement II: Lets take \(n = 2,12,22,32,42,52\).......etc. X will be divisible by 5. Option, D.
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Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
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13 Feb 2018, 05:50






