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# If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w

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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, 07:37
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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.

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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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21 Jan 2017, 00:05
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2
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.

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, 23:41
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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.

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)

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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, 13:49
"Need to verify if x = 5k i.e if unit digit is 0 or 5 then it is divisible.

S-1) 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

S-2) consider n = 2, 12, 22 from S-1 . Sufficient

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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, 01:09
2
1
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, 06:50
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  [#permalink]

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14 Jul 2018, 22:41
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.
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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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15 Jul 2018, 02:51
1
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.
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Re: If x and n are integers such that x = 1^2 + 2^2 + 3^2 + . . . + n^2, w &nbs [#permalink] 15 Jul 2018, 02:51
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