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Math Expert V
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What is the sum of the cubes of the first ten positive integers?  [#permalink]

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Question Stats: 68% (01:48) correct 32% (02:19) wrong based on 264 sessions

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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

Kudos for a correct solution.

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Re: What is the sum of the cubes of the first ten positive integers?  [#permalink]

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Bunuel 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

Kudos for a correct solution.

CONCEPT: Sum of Cubes of First n Natural Numbers = (Sum of First n Natural Numbers)^2

i.e. $$(1^3 + 2^3 + 3^3 ----- + 10^3) = (1+2+3+---+10)^2$$

i.e. $$(1^3 + 2^3 + 3^3 ----- + 10^3) = (1/2 * 10 * 11)^2$$

i.e. $$(1^3 + 2^3 + 3^3 ----- + 10^3) = (55)^2$$

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Re: What is the sum of the cubes of the first ten positive integers?  [#permalink]

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Sum of cubes of first n natural numbers = $$(n(n+1)/2)^2$$

Sum of cubes of first 10 positive integers =$$(10(10+1)/2)^2$$ = $$(55)^2$$

Ans : C
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: What is the sum of the cubes of the first ten positive integers?  [#permalink]

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Hi All,

The answer choices to this question are "spread out" enough that we can avoid most of the "math" involved and get to the correct answer. We also don't need any special math knowledge or formulas beyond basic multiplication and some estimation.

We're asked for the sum of the CUBES of the first 10 positive integers.

First off, the GMAT would NEVER ask us to physically add up those 10 numbers, so there has to be another way to get to the solution. Let's do a little work and use the patterns in the math...

1^3 = 1
10^3 = 1,000

Since we're adding up 10 cubes, we know that the sum will be GREATER than 1,001 and since 10^3 is the biggest (and the other 9 cubes are SMALLER than 10^3), the sum MUST be LESS than 10,000. With this deduction, we can eliminate Answer A (1000), Answer D (10,000), and Answer E (1,000,000).

Between Answers B and C, we just have to do a little more work....

50^2 = 2500, so....
45^2 is LESS than 2500
55^2 is GREATER than 2500

9^3 = 81(9) = about 700
8^3 = 64(8) = about 500

Considering what we know about 10^3, we're already at 2200+, with 6 more cubes to go. This will certainly end up summing to a total that is GREATER than 2500. Eliminate Answer B.

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Math Expert V
Joined: 02 Sep 2009
Posts: 55715
Re: What is the sum of the cubes of the first ten positive integers?  [#permalink]

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Bunuel 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

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Set up a table to list the first few cubes and track the cumulative sum:
Attachment: 2015-06-15_1416.png [ 18.72 KiB | Viewed 2833 times ]

By this point, you should start to recognize that the cumulative sums are all perfect squares. Extend the table to the right to list the square roots.
Attachment: 2015-06-15_1416_001.png [ 57.05 KiB | Viewed 2839 times ]

Not only do we see a pattern emerging for the cumulative sum (always a perfect square), but also for the square root of the cumulative sum, which is always the sum of consecutive integers.

Thus, the sum of the first 10 cubes will be the square of the sum of the first 10 positive integers:
1^3 + 2^3 + 3^3 + 4^3 + … + 10^3 = (1 + 2 + 3 + 4 + … + 10)^2

Now, the sum of the first 10 positive integers can be computed by matching pairs:
1 + 2 + … + 9 + 10 = 11 + 11 + 11 + 11 + 11 = 5(11) = 55.

So, the sum of the first 10 cubes is 55^2.

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_________________ Re: What is the sum of the cubes of the first ten positive integers?   [#permalink] 20 Jan 2019, 12:19
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