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# What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2

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Math Expert
Joined: 02 Sep 2009
Posts: 46307
What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 04:31
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Difficulty:

65% (hard)

Question Stats:

71% (01:49) correct 29% (02:38) wrong based on 295 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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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 04:50
2
Bunuel wrote:
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

I am stuck to 6328!
This is what I did, all the nos are close to 30, and squared, so 30^2 plus there are 7 numbers, so (30)^2 x 7 =6300,
now squaring the last digits of each no individually comes up to (9+4+1+0+1+4+9) = 28.

when you add the 2 up it gives you a (6300+28) =6328, Answer is D IMO.
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What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 06:38
4
Bunuel wrote:
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

Last 2 digit of 27^2 is=29
28^2=84
29^2=41
30^2=00
31^2=61
32^2=24
33^2=89

Sum them up to get 328. Answer D.
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Math Expert
Joined: 02 Aug 2009
Posts: 5938
What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 07:20
1
Bunuel wrote:
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

hi,
the choices being very close, any guess work may not be fruitful...

1) a straight formula for this would be $$\frac{n(n+1)(2n+1)}{6}$$.....
we have to find $$1^2+2^2......+33^2-(1^2+2^2......+27^2)$$
so n in two case is 33 and 26...
$$\frac{33*34*67}{6}-(\frac{26*27*53}{6})$$.... thereafter we require calculations which will give us 6328 as the ans...

2)the method would seem to be lengthy as it is being explained but is very short actually
second would be to take advantage of proximity to 30 and a set of two numbers giving us 60 as sum..

$$(27^2 + 33^2) + (29^2 + 31^2) + (32^2 + 28^2)+ 30^2$$...

$$(60^2-2*27*33) +(60^2-2*29*31)+ (60^2-2*28*32)+ 30^2$$..

$$3*60^2+30^2-2(27*33+29*31+28*32)$$..

$$3*60^2+30^2-2((30-3)*(30+3)+(30-1)*(30+1)+(30-2)*(30+2))$$..

$$3*60^2+30^2-2((30^2-3^2)+(30^2-1^2)+(30^2-2^2))$$..

$$3*60^2+30^2-6(30^2)+2(3^2+1^2+2^2)$$

we are interested only in terms where square of multiple of 10 is not involved..

2(9+1+4)=28 so the answer should be something*100+28....this gives us the answer as 6328

ans D
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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 07:29
6
1
My attempt:

arranged the series in the form

(30-3)^2 + (30-2)^2 + (30-1)^2+
(30+3)^2 + (30+2)^2 + (30+1)^2+
30^2

= 2(30^2 + 3^2 + 30^2 + 2^2 + 30^2 + 1^2) + 30^2
= 7*30^2 + 28
= 6328
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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 11:25
1
Lucky2783 wrote:
Bunuel wrote:
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

Last 2 digit of 27^2 is=29
28^2=84
29^2=41
30^2=00
31^2=61
32^2=24
33^2=89

Sum them up to get 328. Answer D.

i found this trick useful in exam so just want to share it..
how to get last 2-digits in multiplication of 2 two digit numbers.
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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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20 May 2015, 11:41
1
1
took a different route :

27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2

arranged the series in the form pairs of 60 .....(27+33) , (28+32) , (29+31) and using formula $$a^2+b^2 = (a+b)^2 - 2ab$$

$$(27+33)^2 - 2 (27 . 33) + (28+32)^2 - 2 (28 . 32) + (29+31)^2 - 2 (29 . 31) + 30^2$$

$$(27+33)^2 + (28+32)^2 + (29+31)^2 - 2 (27 . 33 + 28 . 32 + 29 . 31 ) + 30^2$$

$$3(60)^2 - 2 (27 . 33 + 28 . 32 + 29 . 31 ) + 30^2$$

=3 (3600) - 2 ( 2686) + 900

=10800 - 5372 + + 900 = 6328

Alternate way , I like the way Jackal did it

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

$$7(30^2) + 2(3^2) + 2(2^2) + 2(1^2) + (2.30.3 + 2.30.2 + 2.30.1 - 2.30.3 - 2.30.2 - 2.30.1)$$

=7*30^2 + 28 = 6328

Ans D
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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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23 May 2015, 02:28
1
27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2

We can solve by taking pairs:
29^2 = (30-1)^2 = 30^2 - (2)(30)(1) + 1^2
31^2 = (30+1)^2 = 30^2 + (2)(30)(1) + 1^2
The sum is 30^2+30^2+2

28^2 = (30-2)^2 = 30^2 - (2)(30)(2) + 2^2
32^2 = (30+2)^2 = 30^2 +(2)(30)(2) + 2^2
The sum is 30^2+30^2+8

27^2 = (30-3)^2 = 30^2 - (2)(30)(3) + 3^2
33^2 = (30+3)^2 = 30^2 + (2)(30)(3) + 3^2
The sum is 30^2+30^2+18

So we got 7 of the 30^2 and 2+8+18
900(7)+28=6328

Math Expert
Joined: 02 Sep 2009
Posts: 46307
Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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25 May 2015, 08:00
6
8
Bunuel wrote:
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

OFFICIAL SOLUTION:

We have 7 terms, where the middle term is 30^2. Express all other terms as 30-x:
$$27^2 + 28^2 + 29^2 + 30^2 + 31^2 + 32^2 + 33^2=$$
$$=(30-3)^2+(30-2)^2+(30-1)^2+30^2+(30+1)^2+(30+2)^2+(20+3)^2$$. 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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What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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16 Oct 2016, 00:53
I would use the last digit analysis

7^2=49 - 9 is the last digit
8^2=64 - 4 is the last digit
9^2=81 - 1 is the last digit
0^2=0 0 is the last digit
1^2 =1- 1
2^2=4 -4
3^2=9- 9

Take the units digit and then just add 9+4+1+0+1+4+9 =28 --> Check the answers only 6328 matches !!

Easy and quick !!
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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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31 Oct 2017, 17:58
Lucky2783 wrote:
Bunuel wrote:
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

Last 2 digit of 27^2 is=29
28^2=84
29^2=41
30^2=00
31^2=61
32^2=24
33^2=89

Sum them up to get 328. Answer D.

how you calculated the last 2 digits , please explain the method used.
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Re: What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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11 Nov 2017, 10:01
I would use the last digit analysis

7^2=49 - 9 is the last digit
8^2=64 - 4 is the last digit
9^2=81 - 1 is the last digit
0^2=0 0 is the last digit
1^2 =1- 1
2^2=4 -4
3^2=9- 9

Take the units digit and then just add 9+4+1+0+1+4+9 =28 --> Check the answers only 6328 matches !!

Easy and quick !!

This makes no sense. Just right in this case.
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What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2 [#permalink]

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11 Nov 2017, 23:13
Bunuel wrote:
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

In problems that are computation intensive, one clue could be symmetry. Here we see on the left of 30^2,
(30-1)^2, (30-2)^2 and (30-3)^2 and on the right (30+1)^2. (30+2)^2 and (30 +3)^2

We can easily see the form (a-b)^2 and (a+b)^2 and that -2ab and +2ab terms cancel out.
Sum of a^2 terms is 6300 and b^2 terms is 28. Total is 6328.
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What is the value of 27^2+28^2+29^2+30^2+31^2+32^2+33^2   [#permalink] 11 Nov 2017, 23:13
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