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The sum of the squares of the first 15 positive integers
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15 Nov 2009, 22:34
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The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ? (A) 2480 (B) 3490 (C) 6785 (D) 8215 (E) 9255 How do we solve this one?
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Re: Sum of squares
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16 Nov 2009, 02:26
ctrlaltdel wrote: The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480 (B) 3490 (C) 6785 (D) 8215 (E) 9255
How do we solve this one? If you know the sum of squares formula (something I've never seen tested in a real GMAT question), you can use it to sum the first 30 squares, then subtract 1240 to get the answer. The sum of the squares of the first n positive integers is given by (n)(n+1)(2n+1)/6, though I'd be very surprised to find a real GMAT question where this was at all helpful to know. Or, if you're familiar with remainder/modular arithmetic (if not, this solution might seem confusing), you might notice that when you add any three consecutive perfect squares, the remainder will be 2 when you divide by 3 (since one of the squares will be divisible by 3, and the other two will both give a remainder of 1). So here we're adding five sets of three numbers, where each set gives a remainder of 2 when divided by 3. Since we can add remainders, the remainder when we divide 16^2 + 17^2 + ... + 30^2 by 3 will be the remainder when we divide 5*2 = 10 by 3, so the remainder will be 1. Of the answer choices, only B and D give the correct remainder, but B is far too small. You could also estimate here. Notice that 15^2 + 16^2 + 17^2 + 18^2 + 19^2 is quite a bit bigger than 15^2 + 15^2 + ... + 15^2 = 5*15^2, and quite a bit smaller than 20^2 + 20^2 + ... + 20^2. Similarly, the next block of five squares add to something quite a bit bigger than 5*20^2, and quite a bit smaller than 5*25^2, and similarly for the last block. So our sum S must satisfy the following inequality, and shouldn't be anywhere close to the extreme low or high values: 5*15^2 + 5*20^2 + 5^25^2 < S < 5*20^2 + 5*25^2 + 5*30^2 5(15^2 + 20^2 + 25^2) < S < 5(20^2 + 25^2 + 30^2) 5(1250) < S < 5(1925) 6250 < S < 9625 and since answer choice C is too close to the low end of this range, and E too close to the high end, D must be correct. If one is willing to do more elaborate computations, one can easily get an estimate to narrow the above range further, of course. Or you can proceed algebraically, using the (x +y)^2 = x^2 + 2xy + y^2 pattern to relate the sum to the one given in the question: 16^2 + 17^2 + 18^2 + ... + 29^2 + 30^2 = (1 + 15)^2 + (2 + 15)^2 + (3 + 15)^2 + ... + (14 + 15)^2 + (15 + 15)^2 = (1^2 + 2*15 + 15^2) + (2^2 + 2*2*15 + 15^2) + (3^2 + 2*3*15 + 15^2) + ... + (14^2 + 2*14*15 + 15^2) + (15^2 + 2*15*15 + 15^2) = (1^2 + 2^2 + ... + 15^2) + 2*15(1 + 2 + 3 + ... + 14 + 15) + 15*15^2 = 1240 + 3600 + 3375 = 8215
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Re: Sum of squares
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16 Nov 2009, 01:11
ctrlaltdel wrote: The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480 (B) 3490 (C) 6785 (D) 8215 (E) 9255
How do we solve this one? sum of square of n numbers is found using the formula [n (n+1) (2n+1)]/6 we have sum of 1st 15 numbers = 1240 we need to find sum of squares from 16 to 30 which is = sum of squares of 1st 30 +ve integers  sum of squares of 1st 15 +ve integers. we will get D  8215




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Re: Sum of squares
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16 Nov 2009, 03:23
ctrlaltdel wrote: The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480 (B) 3490 (C) 6785 (D) 8215 (E) 9255
How do we solve this one? The simplest approach is to deduct the sum of the squares of first 15 terms from the sum of the squares of the first 30 terms. Formula to find the sum of the squares of n terms=n(n+1)(2n+1)/6 thussum of first 30 terms=30(31)61)/6=9455 so ans is 94551240=8215 i.e. D



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Re: Sum of squares
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16 Nov 2009, 06:56
Kudos Ian n vyomb for sum of squares forumla. Ian, I will keep in mind that if may not be tested in GMAT.



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Re: Squares of 15 numbers
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12 Jan 2012, 23:32
enigma123 wrote: The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ? (A) 2480 (B) 3490 (C) 6785 (D) 8215 (E) 9255
Any idea guys what will be the answer and how to solve? I don't have an OA. sum of squares is n*(n+1)*(2n+1)/6 [ for 1st n natural number's] so for 1st 15 >15*16*31/6 = 5*8*31 = 1240 for 1st 30 > 30*31*61/6 = 5*31*61 = 9455 94551240 = 8215 D



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Re: Squares of 15 numbers
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12 Jan 2012, 23:38
Thanks kp1811. I am not fully awake yet, I think. I have calculated until the second step but forgot to subtract. Thanks very much buddy.
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Re: Squares of 15 numbers
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13 Jan 2012, 02:16
You'll never need a formula for the sums of squares on the actual GMAT. You don't need to use that formula here, though it's not all that straightforward to solve without one. Two different approaches: 16^2 + 17^2 + 18^2 + ... + 30^2 = (15 + 1)^2 + (15 + 2)^2 + (15 + 3)^2 + ... + (15 + 15)^2 Now we can expand each square; they are all in the (x + y)^2 = x^2 + 2xy + y^2 pattern. = (15^2 + 2*15 + 1^2) + (15^2 + 4*15 + 2^2) + (15^2 + 6*15 + 3^2) + ... + (15^2 + 30*15 + 15^2) Now we have fifteen 15^2 terms, so adding these gives 15*15^2 = 15^3 = 3375. We also have the entire sum 1^2 + 2^2 + ... + 15^2, which we know is equal to 1240. Finally adding the middle terms, we have: 2*15 + 4*15 + 6*15 + ... + 30*15 = 15(2 + 4 + 6 + .... + 30) = 15*2*(1 + 2 + 3 + ... + 15) = 15*2*8*15 = 3600 So the sum must be 3375 + 1240 + 3600 = 8215 Alternatively, we can use a different factoring pattern. We want to find the value of 30^2 + 29^2 + ... + 17^2 + 16^2. Well if we subtract 15^2 + 14^2 + .... + 2^2 + 1^2 from this, the answer will be 1240 less than what we want to find. So if we can find the value of 30^2 + 29^2 + ... + 17^2 + 16^2  (15^2 + 14^2 + .... + 2^2 + 1^2) then we can add 1240 to get the answer. Now grouping the terms above to get differences of squares, we have = (30^2  15^2) + (29^2  14^2) + ... + (16^2  1^2) and factoring each of these using x^2  y^2 = (x + y)(x  y), we have = 45*15 + 43*15 + 41*15 + ... + 17*15 = 15(45 + 43 + 41 + ... + 17) In brackets we have an equally spaced sum with fifteen terms, which we can evaluate using the familiar formula. So the above equals 15*15*62/2 = 6975 and adding back the 1240, we get the answer of 8215. I think no matter how you approach the question, the calculation is pretty annoying, so I don't care for the question much.
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Re: The sum of the squares of the first 15 positive integers
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27 Feb 2015, 04:38
I did not know the formula for calculating the sum of the squares of consecutive natural numbers. But I know how to find the sum of consecutive natural numbers. So I used the following method 1^2 + 2^2 + 3^2 + . . . + 15^2= 1240 ( A) 16^2 + 17^2 + 18^2 + . . . + 30^2= S (B) B A = S 1240 (16^2 + 17^2 + 18^2 + . . . + 30^2 )  (1^2 + 2^2 + 3^2 + . . . + 15^2) = S  1240 the above can be written as (16^21^2) + (17^2 2^2).........+(30^2  15^2) = S1240 15* (1+16) + 15 * (2+17)...........+ 15 * (15 + 30) = S 124015 [ 1+ 2+ 3......................................30 ] = S1240 15 * [sum of first 30 natural numbers] = S1240 15 * 30/2 * 31 = S1240 S= 8215 Hope the above helps !!! Kudos if you like the solution....
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Re: The sum of the squares of the first 15 positive integers
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27 Feb 2015, 19:34
Hi All, The answer choices to this question are "spread out" enough that we can use them to our advantage. Since the GMAT does NOT expect us to know this formula: (N)(N+1)(2N+1)/6...there must be another way to get to the correct answer. A bit of Arithmetic, some patternmatching skills and a bit of estimation will get us the answer. First, I'm going to list some "reference" numbers that are easy to calculate: 16^2 = 256 20^2 = 400 25^2 = 625 30^2 = 900 The question asks us for the sum of the squares from 16^2 to 30^2 inclusive. This requires us to think about 15 numbers. The sums of 16^2, 17^2, 18^2 and 19^2 will be MORE than 1,000 but LESS than 1,600. This is because there are 4 values that are all AT LEAST 256 and LESS THAN 400 each. The sums of 20^2 through 24^2 will be MORE than 2,000 but LESS than 3,000. There are 5 values in this group. The sums of of 25^2 through 30^2 will be MORE than 3,600 but LESS than 5,000. there are 6 values in this group. As an estimate, we can use....1300 + 2500 + 4200 = 8,000 give or take. There's only one answer that is close to this... Final Answer: GMAT assassins aren't born, they're made, Rich
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The sum of the squares of the first 15 positive integers
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12 Apr 2015, 14:29
This is mine reasoning, of course there is the precise way but it may be time consuming even if do know the way it will take well over 2 mints. this is what I think, I made 3 groups 16 to 20, 20 to 25 and 25 to 30, the squares will be 256 to 400, 400 to 625, and 625 to 900, now we need to multiiply the numbers with 5 since there are 5 numbers in each group, so we get ( rough estimates in teh interest of the time) 256 *5 = 1250, so we get 1250 to 2000, 2000 to 3200, 3200 to 4500. next step is take the rough middle of the gorups 1250 to 2000 is around 1600, next is around 2600 and last is around 3800, add these 1600 + 2600 + 3800 = 8000 so the matching answer would be 8215 Of course this approach is very similar if not the same as EMPOWERgmat above, but I tried to simplifiy it even more.



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Re: The sum of the squares of the first 15 positive integers
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03 Nov 2015, 15:45
This is how I did this question
Sum of first 15 squares = 1240
Let n = 15, and calculate the sum of the next 15 squares
(n+1)^2 + (n+2)^2 + ....+(n+15)^2
Now expand this
(n^2 + 2n + 1^2) + (n^2 + 4n + 2^2) + ..... + (n^2 + 30n + 15^2)
Now collect like terms
15n^2 + [sum of first 30 even numbers, inclusive]n + [sum of first 15 squares]
plus in what we know
15(15)^2 + [mean * number of terms]n + 1240
3375 + [(30+2)/2 * 15]n + 1240
3375 + 240n + 1240
3375 + 240(15) + 1240
3375 + 3600 + 1240
= 8215



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Re: The sum of the squares of the first 15 positive integers
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08 Nov 2015, 00:59
Just a thought,
\(16^2+.....+30^2\) = 256+.....+900
\((\frac{900+256}{2})\)*15= 578*15 = 8670 (close to D)
Can we consider this approach ?



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Re: The sum of the squares of the first 15 positive integers
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05 Dec 2016, 14:19
Another way to look at this one with a repeatable strategy for problems that look too big to solve... A helpful strategy with big numbers and particularly with exponents is to see if you can find patterns. Here, you don't want to calculate all those squares from 16^2 to 30^2, but since they gave you the relationship to the first 15 squares there's good evidence that there would be some relationship between the ones you're asked to calculate and the ones they gave you. So say you just used a small set of 5 of them...not terrible to calculate, and did: 16^2 = 256 17^2 = 289 (NOTE: you can do this pretty easily by saying that 16^2 = 16*16, so to get to 16*17 you add one more 16 to get 272, and then to get from 16*17 to 17*17 you add one more 17 to get 289) 18^2 = 324 20^2 = 400 (I'll explain why I skipped 19^2 in a second) Here I'm just jotting down a few easyish to calculate numbers to see if I can find a pattern. So then lay these against the first five squares to see: 16^2 vs. 1^2 = 256 vs. 1, so a difference of 255 17^2 vs. 2^2 = 289 vs. 4...which is a difference of 285 (hmm...both end in 5s at least) 18^2 vs. 3^2 = 324 vs. 9...which is a difference of 315 20^2 vs. 5^2 = 400 vs. 25 which is a difference of 375 So...just by testing enough numbers to see if you have a pattern, you can see that the differences between the first square of the old set and the first of the new, the second vs. second, etc. go up 30 every time, from 255 to 285 to 315 to 345 to 375, and with that many data points you can assume (even if it's not guaranteed...you don't want to spend more than 34 minutes on this thing on test day) that that pattern will hold. So you can keep counting up by 30s to the 8th term (the middle of the 15 terms) to see that it will be 465, and then know that your sum will be the original sum of 1240 plus the sum of the differences from old to new (so 15 * 465), and that adds up to the correct 8215. A few clarification points here: I skipped 19^2 because I knew 20^2 easily and I only cared about 19^2 if it was an easy to recognize pattern with the other numbers. Once I saw that the differences all ended in 5 and were evenly spaced, then I was interested b/c I knew I had a pattern I could use. I looked for the difference between terms in the "old" set (1^2  15^2) and the new because they gave us the sum of the "old" set. That's a recognition thing, kind of playing the GMAT's trends  there are a few problems out there where they base a sequence or series on another one, and then give you a piece of information about the one they based it on. Very frequently in those cases, the path is to "fill in the gap" by calculating the difference between the sets and not by trying to calculate the new one alone. So that was a little bit of "I've seen this game before and here's how it's usually played..." to hopefully find a path. Anyway  just thought that might be interesting since I just used that exact strategy to solve this one in ~2:30 or so while playing around in here on my lunch break (ah, the fascinating life of a GMAT teacher). I always like when I can apply the strategies I tell students to use (when in doubt see if you can jot out a few terms and find a pattern, in this case) and see it work just as advertised!
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Re: The sum of the squares of the first 15 positive integers
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23 Oct 2017, 10:16
shyind wrote: Just a thought,
\(16^2+.....+30^2\) = 256+.....+900
\((\frac{900+256}{2})\)*15= 578*15 = 8670 (close to D)
Can we consider this approach ? This is the approach that I used to come up with a logical solution. Not sure if it's good or not.




Re: The sum of the squares of the first 15 positive integers &nbs
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23 Oct 2017, 10:16






