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Senior Manager  Status: Finally Done. Admitted in Kellogg for 2015 intake
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21 00:00

Difficulty:   55% (hard)

Question Stats: 65% (02:33) correct 35% (02:21) wrong based on 238 sessions

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$$36^2$$ + $$37^2$$ + $$38^2$$ + $$39^2$$ + $$40^2$$ + $$41^2$$ + $$42^2$$+ $$43^2$$ + $$44^2$$ =

(A) 14400
(B) 14440
(C) 14460
(D) 14500
(E) 14520

Guys this is how I am trying to solve this. But after doing it for 15 minutes I gave up and think to post. No idea where I am getting this wrong and i don't have an OA either.

Here we go:

The question can be simplified as:

36^2 +{36+1}^2 +{36+2}^2 +{36+3}^2{36+4}^2{36+5}^2{36+6}^2{36+7}^2{36+8}^2

The above can be further simplified into

$$36^2$$ + {1296+1+72} + {1296+4+44} + ..................... + {1296+64+576} -------> [Using $$(a+b)^2$$ formula] -------------------------------------------(1)

Now the (1) can be simplified into
1296 {1+74+149+226+.................+641} ==> 1296*2805 => This doesn't give me any of the above choices. Any idea guys where I am getting it wrong?

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enigma123 wrote:
$$36^2$$ + $$37^2$$ + $$38^2$$ + $$39^2$$ + $$40^2$$ + $$41^2$$ + $$42^2$$+ $$43^2$$ + $$44^2$$ =

(A)14400
(B)14440
(C)14460
(D)14500
(E)14520

Approach #1:

We have 9 terms, middle term is 40^2. Express all other terms as 40-x:
$$36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=$$
$$=(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+2)^2+(40+3)^2+(40+4)^2$$. 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:

$$(40^2+4^2)+(40^2+3^2)+(40^2+2^2)+(40^2+1^2)+40^2+(40^2+1^2)+(40^2+2^2)+(40^2+3^2)+(40^2+4^2)=$$
$$=9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460$$

Approach #2:
The sum of the squares of the first n positive integers is given by: $$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$ (note that it's highly unlikely that you'll need it on the real GMAT test). For example the sum of the first 3 positive integers: $$1^2+2^2+3^3=\frac{3(3+1)(2*3+1)}{6}=14$$.

Now, we can calculate the sum of the squares of the first 44 positive integers and subtract from it the sum of the squares of the first 35 positive integers to get the answer: $$36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=\frac{44(44+1)(2*44+1)}{6}-\frac{35(35+1)(2*35+1)}{6}=14,460$$.

Hope it helps.
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Bunuel wrote:
enigma123 wrote:
$$36^2$$ + $$37^2$$ + $$38^2$$ + $$39^2$$ + $$40^2$$ + $$41^2$$ + $$42^2$$+ $$43^2$$ + $$44^2$$ =

(A)14400
(B)14440
(C)14460
(D)14500
(E)14520

Approach #1:

We have 9 terms, middle term is 40^2. Express all other terms as 40-x:
$$36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=$$
$$=(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+1)^2+(40+3)^2+(40+4)^2$$. 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:

$$(40^2+4^2)+(40^2+3^2)+(40^2+2^2)+(40^2+1^2)+40^2+(40^2+1^2)+(40^2+2^2)+(40^2+3^2)+(40^2+4^2)=$$
$$=9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460$$

hm, before looking to ur solution, I solved the q. in this way-
let a =36^2
then we have a+(a+1)+(a+2) (a+3) (a+4) (a+5) (a+6) (a+7) (a+8)

I feel that it is just an arithmetic progression with mean=median

so the sum of these numbers are 9*(a+4)=9*40^2=14400

later I saw ur solution -=9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460
we seem to be in the same way, but using my method how to come to +2*(4^2+3^2+2^2+1^2)?
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LalaB wrote:
Bunuel wrote:
enigma123 wrote:
$$36^2$$ + $$37^2$$ + $$38^2$$ + $$39^2$$ + $$40^2$$ + $$41^2$$ + $$42^2$$+ $$43^2$$ + $$44^2$$ =

(A)14400
(B)14440
(C)14460
(D)14500
(E)14520

Approach #1:

We have 9 terms, middle term is 40^2. Express all other terms as 40-x:
$$36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=$$
$$=(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+1)^2+(40+3)^2+(40+4)^2$$. 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:

$$(40^2+4^2)+(40^2+3^2)+(40^2+2^2)+(40^2+1^2)+40^2+(40^2+1^2)+(40^2+2^2)+(40^2+3^2)+(40^2+4^2)=$$
$$=9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460$$

hm, before looking to ur solution, I solved the q. in this way-
let a =36^2
then we have a+(a+1)+(a+2) (a+3) (a+4) (a+5) (a+6) (a+7) (a+8)

I feel that it is just an arithmetic progression with mean=median

so the sum of these numbers are 9*(a+4)=9*40^2=14400

later I saw ur solution -=9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460
we seem to be in the same way, but using my method how to come to +2*(4^2+3^2+2^2+1^2)?

The problem is that 36^2, 37^2, 38^2, 39^2, 40^2, 41^2, 42^2, 43^2, and 44^2 DOES NOT form an AP. You assumed that $$a+1=36^2+1=37^2$$, $$a+2=36^2+2=38^2$$, ... but that's not correct: $$37^2\neq{36^2+1}$$, $$38^2\neq{36^2+2}$$, ... .

Hope it's clear.
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Bunuel, yep, u r right. my idea was wrong. thnx
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LalaB wrote:

I feel that it is just an arithmetic progression with mean=median

It is not an arithmetic progression, so if you assume that it is, you will get the wrong answer. An arithmetic progression is 'equally spaced'. If you just look at the smallest perfect squares, you can see that they are not equally spaced: 1, 4, 9, 16, 25... In fact the spacing gets larger the further you get into this list.

You can use the spacing of perfect squares to answer this question. From the difference of squares, we have that

41^2 - 40^2 = (41 + 40)(41 - 40) = 81

So 41^2 = 40^2 + 81. Similarly, 42^2 = 41^2 + 83, and 39^2 = 40^2 - 79, and so on. Listing all of the values we need to sum:

\begin{align*} 36^2 &= 40^2 - 79 - 77 - 75 - 73\\ 37^2 &= 40^2 - 79 - 77 - 75 \\ 38^2 &= 40^2 - 79 - 77 \\ 39^2 &= 40^2 - 79 \\ 40^2 &= 40^2 \\ 41^2 &= 40^2 + 81 \\ 42^2 &= 40^2 + 81 + 83 \\ 43^2 &= 40^2 + 81 + 83 + 85 \\ 44^2 &= 40^2 + 81 + 83 + 85 + 87 \end{align*}

Now adding these in columns, we get (9)(40^2) + 4(81-79) + 3(83 - 77) + 2(85 - 75) + (87 - 73) = 9*1600 + 4*2 + 3*6 + 2*10 + 14 = 14,400 + 8 + 18 + 20 + 14 = 14,460

I'd still probably use the first method outlined in Bunuel's post above, but you can use the spacing of squares to get the answer if you look at the problem in the right way.
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It would be easier if you think about this problem in this way:

(40-4)^2 + (40-3)^2 + (40-2)^2 + (40-1)^2 + (40-0)^2 + (40+1)^2+ (40+2)^2 + (40+3)^2 + (40+4)^2

all the 4*40*2 or 2* a*b terms will get cancelled out.

1600*9 + (4*4 + 3*3 + 2*2 + 1)*2 = answer = 14460

any other way would be too lengthy.

take middle term in such cases and try to formulate it in ways such as (a-1)^2 + (a+1)^2.
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IanStewart wrote:
LalaB wrote:

I feel that it is just an arithmetic progression with mean=median

It is not an arithmetic progression, so if you assume that it is, you will get the wrong answer. An arithmetic progression is 'equally spaced'. If you just look at the smallest perfect squares, you can see that they are not equally spaced: 1, 4, 9, 16, 25... In fact the spacing gets larger the further you get into this list.

You can use the spacing of perfect squares to answer this question. From the difference of squares, we have that

41^2 - 40^2 = (41 + 40)(41 - 40) = 81

So 41^2 = 40^2 + 81. Similarly, 42^2 = 41^2 + 83, and 39^2 = 40^2 - 79, and so on. Listing all of the values we need to sum:

\begin{align*} 36^2 &= 40^2 - 79 - 77 - 75 - 73\\ 37^2 &= 40^2 - 79 - 77 - 75 \\ 38^2 &= 40^2 - 79 - 77 \\ 39^2 &= 40^2 - 79 \\ 40^2 &= 40^2 \\ 41^2 &= 40^2 + 81 \\ 42^2 &= 40^2 + 81 + 83 \\ 43^2 &= 40^2 + 81 + 83 + 85 \\ 44^2 &= 40^2 + 81 + 83 + 85 + 87 \end{align*}

Now adding these in columns, we get (9)(40^2) + 4(81-79) + 3(83 - 77) + 2(85 - 75) + (87 - 73) = 9*1600 + 4*2 + 3*6 + 2*10 + 14 = 14,400 + 8 + 18 + 20 + 14 = 14,460

I'd still probably use the first method outlined in Bunuel's post above, but you can use the spacing of squares to get the answer if you look at the problem in the right way.

I also used this approach, but it's really calculation intensive.
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(A)14400
(B)14440
(C)14460
(D)14500
(E)14520

Sum of squares = (n)(n+1) (2n+1) /6
Sum of squares (44) - Sum of squares (35)

(44 * 45 * 89) /6 – (35 * 36 * 71)/6

(176220 – 89460) / 6 = 86760 / 6 = 14460
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Bumping for review and further discussion.
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Hi Bunuel,
I have different way but do not know it is a mere coincidence or valid approach. Is this logic ok?

36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2

Taking square of unit digits:
6^2= 36
7^2= 49
8^2=64
9^2=81
0^2=0
1^2=1
2^2=4
3^2=9
4^2=16

Adding all the values 36+49+64+81+1+4+9+16= 260

Only option C has 60 on the last. Hence the answer is C.

Atal Pandit
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atalpanditgmat wrote:
Hi Bunuel,
I have different way but do not know it is a mere coincidence or valid approach. Is this logic ok?

36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2

Taking square of unit digits:
6^2= 36
7^2= 49
8^2=64
9^2=81
0^2=0
1^2=1
2^2=4
3^2=9
4^2=16

Adding all the values 36+49+64+81+1+4+9+16= 260

Only option C has 60 on the last. Hence the answer is C.

Atal Pandit

No, that's not correct. You can only get the units digit with this approach.

For example, 17^2+23^2=818, but 7^2+3^2=58.

Hope it's clear.
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What level question is this please? Its taking more than 2 minutes for me _________________
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PareshGmat wrote:
What level question is this please? Its taking more than 2 minutes for me I'd say above 700, so spending extra 1 minute on those is OK.
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Option C.
Using the formula for calculating sum of squares of n natural nos.:[n(n+1)(2n+1)]/6
First calculate for n=44 then for n=35
And subtract second from first to get the answer.

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36^2+37^2+38^2+ 39^2+40^2+ 41^2+42^2+43^2+44^2=  [#permalink]

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