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There are 25 necklaces such that the first necklace contains [#permalink]

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07 Nov 2009, 10:37

4

This post was BOOKMARKED

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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

52% (03:37) correct
48% (03:37) wrong based on 88 sessions

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There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

Total = 25*5 + 24*2 + 23*3 + ....3*23+ 2*24 + 25 = 26*5 + definae Integral of x(26-x)dx from 2 to 24 = 2961 ( it took me 5 min to do by hand, i checked with Mathcad for the integral part)

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

Im trying to think up a faster way to solve this but nothing seems to be striking Im sure there must be one else this problem is way to time consuming to be on the gmat...
_________________

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The answer is B. I would be happy to skip this problem on the GMAT, if it comes. It has taken me ages to solve this

Any way Please comment whether this solution is faster

We have a sequence of 5, 7, 10, 14, 19…. = (4+1), (4+3), (4+6)……. The sum of the sequence is = (25*4)+1+3+6+….25(25+1)/2 (Generic term is x(x+1)/2)

1+3+6+….25(25+1)/2 = Sum of first 25 triangular number (The formula of sum of n triangular number is ½[Sum [n^2] +Sum[n]], when n=25; ½[Sum [n^2] +Sum[n]] =2925)

N = 5 + (5+2) + (5+2+3) + (5+2+3+4) + + (5+2+3+4+ 25) By rearranging the terms, we find N = (25*5) + (24*2) + (23*3) + (22*4) + (1*25) N = 125 + Sum ((25-k)*(k+1)) from k = 1 to 24 N = 125 + Sum (24k-(k^2)+25) from k = 1 to 24 By developing and applying the formulas Sum (k^2) = 1/3 N*(N+0,5)*(N+1) (for k = 1 to N) We find N = 3025

N = 5 + (5+2) + (5+2+3) + (5+2+3+4) + + (5+2+3+4+ 25) By rearranging the terms, we find N = (25*5) + (24*2) + (23*3) + (22*4) + (1*25) N = 125 + Sum ((25-k)*(k+1)) from k = 1 to 24 N = 125 + Sum (24k-(k^2)+25) from k = 1 to 24 By developing and applying the formulas Sum (k^2) = 1/3 N*(N+0,5)*(N+1) (for k = 1 to N) We find N = 3025

which formula is applied in the above higlighted step....please elaborate it more....

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get: \(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

in the above solution i understood each n every step... but i have only one concern .....from where it wud enter in our minds during the exam that v got to subtract '4' from each term to get the sequence.... is there any way to observe this thing....i know it wud come only by practice but still i wanted to know that how it wud strike our minds to subtract 4 from each term.....phewwww!!!!!!!!

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

Bunuel, how did you think about reducing 4 in each number to find perfect squares? What's the link between perfect squares and that sequence?

Re: There are 25 necklaces such that the first necklace contains [#permalink]

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03 Jul 2014, 17:21

Bunuel wrote:

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

How did you know after 7 to go on to 10 and so forth? I know you probably used the i more beads info but can you show the plugging in the #'s to get 10 if this is what you did?

Re: There are 25 necklaces such that the first necklace contains [#permalink]

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03 Jul 2014, 17:28

Actually Bunuel,

I am also confused on how exactly you arrived at 100 + 1 + (3+6) + (10+15) + (21+28)+... I know you said subtract 4 from each term and then 4 x 25 = 100 but I don't see where all these terms are coming from.

Also I don't get the step after which is 100 + 1^2 + 3^2 etc... where is the 1^2 , 3^2 coming from?

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690 B. 3025 C. 3380 D. 2392 E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence: \(5, 7, 10, 14, 19, 25...\) \((25 terms)\) and we want to determine the sum.

Let's subtract \(4\) from each term it will total \(4*25=100\), and we''l get:

\(100+1+(3+6)+(10+15)+(21+28)+...\) After 100 we have the same 25 terms.

\(100+1^2+3^2+5^2+7^2...\) after \(100\) we have the sum of the squares of the first n odd numbers. As there are \(25\) terms after \(100\) we'll have \(n=13\) squares.

The sum of the squares of the first \(n\) odd numbers=\(\frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925\).

\(100+2925=3025\)

Answer: B (3025)

How did you know after 7 to go on to 10 and so forth? I know you probably used the i more beads info but can you show the plugging in the #'s to get 10 if this is what you did?

I am also confused on how exactly you arrived at 100 + 1 + (3+6) + (10+15) + (21+28)+... I know you said subtract 4 from each term and then 4 x 25 = 100 but I don't see where all these terms are coming from.

Also I don't get the step after which is 100 + 1^2 + 3^2 etc... where is the 1^2 , 3^2 coming from?

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