GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 14:28 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  What is the sum of the cubes of the first ten positive

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Intern  Joined: 20 Jun 2011
Posts: 43
What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

4
12 00:00

Difficulty:   45% (medium)

Question Stats: 67% (02:04) correct 33% (02:13) wrong based on 271 sessions

HideShow timer Statistics

What is the sum of the cubes of the first ten positive integers?

A. $$10^3$$
B. $$45^2$$
C. $$55^2$$
D. $$100^2$$
E. $$100^3$$
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: What is the sum of the cubes of the first ten positive integ  [#permalink]

Show Tags

12
6
superpus07 wrote:
What is the sum of the cubes of the first ten positive integers?

a) $$10^3$$
b) $$45^2$$
c) $$55^2$$
d) $$100^2$$
e) $$100^3$$

Thanks!

The sum of the cubes of first n positive integers is given by $$[n(n+1)/2]^2$$
If n = 10, the sum is 55^2.

Now, do you need to learn it up?
No. Even if you didn't know it, you should have tried to look at the pattern.
$$1^3 = 1 = 1^2$$
$$1^3 + 2^3 = 9 = 3^2$$
$$1^3 + 2^3 + 3^3 = 36 = 6^2$$
$$1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2$$
You see that when you find the sum of the numbers (not of the cubes) and square the sum, you get the sum of the cubes.
Sum of 1+2+3+4 = 10. Square it to get 100. Sum of $$1^3 + 2^3 + 3^3 + 4^3 = 100$$
This is pattern recognition.

You can also use another method - of averaging.
The numbers look like this: 1, 8, 27, 64, 125 ... etc
The numbers on the right 8^3, 9^3, 10^3 will be much larger than those on the left which are small. The average would lie not in the middle but on the right somewhere between 6^3 and 7^3. I would say around 300 so the sum will be around 300*10 = 3000. This leads us to 55^2.
The only hitch might be the 45^2 given which might make you uncomfortable in using this approximation. 45^2 = 2025

To ensure that avg is around 300 and not around 200, notice that from 10^3, you can make five 200s. From 9^3, you can make about four 200s (using some extra). So overall, you can make many more 200s than the required 10. Therefore, the average must be 300 and not 200.

_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
General Discussion
Intern  Joined: 06 Sep 2012
Posts: 38
Concentration: Social Entrepreneurship
Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

Karishma, how did you get that formula [n(n+1)/2]^2?
Is that something we should be memorizing for the GMAT or is that a derived form of something else?
Can you provide guidance on other formulas that we should know? Thank you!!
_________________
Life begins at the edge of your comfort zone.

Appreciate the +1!
Intern  Joined: 13 Nov 2012
Posts: 1
Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

How does that pattern recognition help solve the problem? How does it translate to 5^3+6^3+7^3... to find 55^2 as the answer?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

1
7
JJ2014 wrote:
Karishma, how did you get that formula [n(n+1)/2]^2?
Is that something we should be memorizing for the GMAT or is that a derived form of something else?
Can you provide guidance on other formulas that we should know? Thank you!!

There are quite a few related formulas:

Sum of first n consecutive positive integers: n(n+1)/2
Sum of first n consecutive positive odd integers: n^2
Sum of first n consecutive positive even integers: n(n+1)
Sum of squares of first n consecutive positive integers: n(n+1)(2n+1)/6
Sum of cubes of first n consecutive positive integers: [n(n+1)/2]^2

Do you need to memorize them for GMAT? No, except for the first one (since it is very useful). GMAT doesn't expect you to 'learn up' these formulas. The question will be such that you can use pattern recognition in such cases. But obviously it wouldn't hurt to know them.

I am no fan of memorizing tons of formulas. I concede that they are very useful in very few cases but I generally don't consider the probability of seeing those cases very high. But knowing these formulas can sometimes speed up your calculations. The question asks you something else - you recognize a pattern, you know a formula, you save time at the intermediate step and quickly arrive at the answer. I know these formulas because I have come across them many times so I ended up retaining them.

All in all, do I suggest you to learn them up? It's up to you. It's certainly not necessary. GMAT will not ask you a question where you must know these formulas. It's all about your commitment to the exam (if you want to leave no stone unturned, you might want to memorize them) and your aptitude (how easy it is for you to recognize patterns and derive a relation in exam conditions).
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

1
nlrosidi wrote:
How does that pattern recognition help solve the problem? How does it translate to 5^3+6^3+7^3... to find 55^2 as the answer?

Revisit the post above. You need to recognize that the first few sums obtained are all squares. 1, 9, 36, 100 ...
The square roots are 1, 3, 6, 10 ...
These are sum of first few numbers:
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10

etc

Find the sum of first ten numbers: 1+ 2 + 3 + ...10 which is 10*11/2 = 55
Sum of cubes will be 55^2
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  Joined: 14 May 2014
Posts: 13
GMAT 1: 760 Q50 V42 Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

3
1
VeritasPrepKarishma wrote:
superpus07 wrote:
What is the sum of the cubes of the first ten positive integers?

a) $$10^3$$
b) $$45^2$$
c) $$55^2$$
d) $$100^2$$
e) $$100^3$$

Thanks!

The sum of the cubes of first n positive integers is given by $$[n(n+1)/2]^2$$
If n = 10, the sum is 55^2.

Now, do you need to learn it up?
No. Even if you didn't know it, you should have tried to look at the pattern.
$$1^3 = 1 = 1^2$$
$$1^3 + 2^3 = 9 = 3^2$$
$$1^3 + 2^3 + 3^3 = 36 = 6^2$$
$$1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2$$
You see that when you find the sum of the numbers (not of the cubes) and square the sum, you get the sum of the cubes.
Sum of 1+2+3+4 = 10. Square it to get 100. Sum of $$1^3 + 2^3 + 3^3 + 4^3 = 100$$
This is pattern recognition.

You can also use another method - of averaging.
The numbers look like this: 1, 8, 27, 64, 125 ... etc
The numbers on the right 8^3, 9^3, 10^3 will be much larger than those on the left which are small. The average would lie not in the middle but on the right somewhere between 6^3 and 7^3. I would say around 300 so the sum will be around 300*10 = 3000. This leads us to 55^2.
The only hitch might be the 45^2 given which might make you uncomfortable in using this approximation. 45^2 = 2025

To ensure that avg is around 300 and not around 200, notice that from 10^3, you can make five 200s. From 9^3, you can make about four 200s (using some extra). So overall, you can make many more 200s than the required 10. Therefore, the average must be 300 and not 200.

I think this problem can be quickly solved by just looking at the answer choices.
(A) This can be eliminated since the answer obviously should be greater than 10^3 or 1000
(D) This can be eliminated since the answer should also definitely be less than 10*10^3 or less than 10000
(E) can be eliminated using similar logic for D above.

Now the contenders are 45^2 and 55^2.
Now 45^2 = 2025. and 55^2 = 3025. We know 10^3 = 1000 and 9^3 = 729. So we can easily see that the total of our required sum would exceed 2025. Hence answer = (C).

Hope this helps.
Intern  B
Joined: 26 Apr 2018
Posts: 22
Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

isn't the question asking for the sum of the cubes of the first 10 intergers? which should be 1^3+2^3...+10^3, and the result should be 5005?
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4015
Re: What is the sum of the cubes of the first ten positive  [#permalink]

Show Tags

Top Contributor
superpus07 wrote:
What is the sum of the cubes of the first ten positive integers?

A. $$10^3$$
B. $$45^2$$
C. $$55^2$$
D. $$100^2$$
E. $$100^3$$

Let's look for a pattern

1³ = 1 = 1²
1³ + 2³ = 9 = 3²
1³ + 2³ + 3³ = 36 = 6²
1³ + 2³ + 3³ + 4³ = 100 = 10²
1³ + 2³ + 3³ + 4³ + 5³ = 225 = 15²

See the pattern yet?
1 + 2 = 3
3 + 3 = 6
6 + 4 = 10
10 + 5 = 15

First we add 2, then we add 3, then 4, then 5, etc

So, continuing the pattern, we get:
1³ + 2³ + 3³ + . . . + 6³ = 21²
1³ + 2³ + 3³ + . . . + 7³ = 28²
1³ + 2³ + 3³ + . . . + 8³ = 36²
1³ + 2³ + 3³ + . . . + 9³ = 45²
1³ + 2³ + 3³ + . . . + 10³ = 55²

Cheers,
Brent
_________________ Re: What is the sum of the cubes of the first ten positive   [#permalink] 08 Apr 2019, 09:37
Display posts from previous: Sort by

What is the sum of the cubes of the first ten positive

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  