December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL. December 14, 2018 December 14, 2018 09:00 AM PST 10:00 AM PST 10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51123

If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
20 Jan 2017, 06:37
Question Stats:
43% (02:25) correct 57% (02:38) wrong based on 260 sessions
HideShow timer Statistics



Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98

Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
20 Jan 2017, 23:05
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.



Senior Manager
Joined: 03 Apr 2013
Posts: 275
Location: India
Concentration: Marketing, Finance
GPA: 3

If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
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
_________________
Spread some love..Like = +1 Kudos



Intern
Joined: 10 Jun 2016
Posts: 48

Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
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
_________________
Thank You Very Much, CoolKl Success is the Journey from Knowing to Doing
A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.



Senior Manager
Joined: 04 Oct 2015
Posts: 257
Location: Viet Nam
Concentration: Finance, Economics
GPA: 3.56

If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
14 Apr 2017, 00:09
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
_________________
Do not pray for an easy life, pray for the strength to endure a difficult one  Bruce Lee



Director
Joined: 31 Jul 2017
Posts: 504
Location: Malaysia
GPA: 3.95
WE: Consulting (Energy and Utilities)

Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
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.
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!



Intern
Joined: 26 Jan 2016
Posts: 35

Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
14 Jul 2018, 21:41
Bit of a hack. The cyclicity of remainders when x/5 is 4. Remainders 1,0,4,0. Also note that when n is even the remainder is 0. So the question is asking what is n?
A and B give us n as even thus both statements are sufficient.



Manager
Joined: 22 Jan 2014
Posts: 176
WE: Project Management (Computer Hardware)

Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w
[#permalink]
Show Tags
15 Jul 2018, 01:51
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. sum of first n integer squares is n(n+1)(2n+1)/6 1) n is even with units digit 0,2,or 4. if n ends in 0, it is div. by 5 if n ends in 2 then (2n+1) will end in 5 and will be div by 5 if n ends in 4 then (n+1) will end in 5 and will be div by 5 2) n ends in 2 same as 1. so both statements are individually sufficient to answer. D.
_________________
Illegitimi non carborundum.




Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w &nbs
[#permalink]
15 Jul 2018, 01:51






