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M31-21

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M31-21  [#permalink]

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New post 14 Jun 2015, 12:09
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

70% (01:21) correct 30% (01:54) wrong based on 40 sessions

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Re M31-21  [#permalink]

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New post 14 Jun 2015, 12:09
1
Official Solution:

What is the value of \(27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2\)?

A. 6298
B. 6308
C. 6318
D. 6328
E. 6338


We have 7 terms, where the middle term is \(30^2\). Express all other terms as \(30-x\):

\(27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2=\)

\(=(30-3)^2+(30-2)^2+(30-1)^2+30^2+(30+1)^2+(30+2)^2+(30+3)^2\).

Now, when you expand these expressions applying \((x-y)^2=x^2-2x+y^2\) and \((x+y)^2=x^2+2x+y^2\) you'll see that \(-2xy\) and \(2xy\) cancel out and we'll get:

\((30^2+3^2)+(30^2+2^2)+(30^2+1^2)+30^2+(30^2+1^2)+(30^2+2^2)+(30^2+3^2)=\)

\(=7*30^2+2*(3^2+2^2+1^2)=6,300+28=6,328\).


Answer: D
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Re: M31-21  [#permalink]

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New post 05 Sep 2015, 19:53
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I think this is a high-quality question and I don't agree with the explanation. In the third line, 20 should be 30
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Re: M31-21  [#permalink]

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New post 06 Sep 2015, 04:12
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Re: M31-21  [#permalink]

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New post 05 Nov 2015, 06:51
2
Bunuel wrote:
Official Solution:

What is the value of \(27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2\)?

A. 6298
B. 6308
C. 6318
D. 6328
E. 6338


We have 7 terms, where the middle term is \(30^2\). Express all other terms as \(30-x\):

\(27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2=\)

\(=(30-3)^2+(30-2)^2+(30-1)^2+30^2+(30+1)^2+(30+2)^2+(30+3)^2\).

Now, when you expand these expressions applying \((x-y)^2=x^2-2x+y^2\) and \((x+y)^2=x^2+2x+y^2\) you'll see that \(-2xy\) and \(2xy\) cancel out and we'll get:

\((30^2+3^2)+(30^2+2^2)+(30^2+1^2)+30^2+(30^2+1^2)+(30^2+2^2)+(30^2+3^2)=\)

\(=7*30^2+2*(3^2+2^2+1^2)=6,300+28=6,328\).


Answer: D


An alternate approach is to find the tens digit, since they are all different. You would get:

29
84
41
00
61
24
89

Which results in a tens digit of 2 and leads you to select D
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Re: M31-21  [#permalink]

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New post 16 Jun 2016, 20:59
Another method, we know sum of squares of first 'n' natural numbers given by n (n+1) ( 2n+1) / 6.

Calculate when n = 33 => X
Calculate when n = 26 => Y

X - Y gives the solution. It is slightly calculation intensive. It took me 4 mins.

X = 12529, Y = 6201, X - Y = 6328
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M31-21  [#permalink]

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New post 27 Aug 2016, 18:21
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I used the following rules to solve this problem in under 1 minute. I'm not sure if this is the right application of the rules but it gets me to the RIGHT answer. I applied the following rules for sum of consecutive integers to this problem which is sums of SQUARES OF consecutive integers.

Rule: The sum of k consecutive integers, with an ODD number of items, k, is always a multiple of the number of items, k.

Example: 4 + 5 + 6 + 7 + 8 = 30 , multiple of 5

Rule: The sum of k consecutive integers, with an EVEN number of items, k, is NEVER a multiple of the number of items, k.

Example: 4 + 5 + 6 + 7 + 8 + 9 = 39 , NOT a multiple of 6

In this question, there are 7 consecutive integers. In the answer choices, the only answer divisible by 7 is 6328. Answer Choice D
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Re: M31-21  [#permalink]

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New post 26 Jun 2018, 09:16
I approached by finding the units digit in each set and adding them together because the numbers all have different last two digits. Thus, 7*7 ends in 9, 8*8 ends 4....3*3 ends in 9.

Thus, 9+4+1+0+1+4+9=28
Only choice D ends with 28, thus it is our answer.
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Re: M31-21 &nbs [#permalink] 26 Jun 2018, 09:16
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