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# M31-21

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Math Expert
Joined: 02 Sep 2009
Posts: 49251

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14 Jun 2015, 13:09
00:00

Difficulty:

65% (hard)

Question Stats:

72% (01:21) correct 28% (01:45) wrong based on 39 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 49251

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14 Jun 2015, 13:09
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:

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

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

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Joined: 05 Apr 2014
Posts: 3

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05 Sep 2015, 20:53
1
I think this is a high-quality question and I don't agree with the explanation. In the third line, 20 should be 30
Math Expert
Joined: 02 Sep 2009
Posts: 49251

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06 Sep 2015, 05:12
bfwell wrote:
I think this is a high-quality question and I don't agree with the explanation. In the third line, 20 should be 30

Edited the typo. Thank you.
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Intern
Joined: 14 Oct 2015
Posts: 34
GMAT 1: 640 Q45 V33

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05 Nov 2015, 07:51
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:

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

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

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
Intern
Joined: 07 Mar 2011
Posts: 26

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16 Jun 2016, 21:59
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
Intern
Joined: 14 Apr 2013
Posts: 2

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27 Aug 2016, 19:21
1
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
Intern
Joined: 20 Apr 2018
Posts: 36
Location: United States (DC)
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26 Jun 2018, 10:16
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.
Re: M31-21 &nbs [#permalink] 26 Jun 2018, 10:16
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# M31-21

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