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There are 25 necklaces such that the first necklace contains

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

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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

A. 2690
B. 3025
C. 3380
D. 2392
E. 3762
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Re: 25 necklaces [#permalink]

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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

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)
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Re: 25 necklaces [#permalink]

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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)

25*4+ 2925 =3025

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Re: 25 necklaces [#permalink]

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New post 07 Nov 2009, 12:09
let me know your comments:
a1 = 5

a1 = a1
a2 = a1 + 2
a3 = a1 + 2 + 3
.
.
.
a24 = a1 + 2 + 3 + ...... + 24
a25 = a1 + 2 + 3 + ...... + 24 + 25

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)

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Re: 25 necklaces [#permalink]

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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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Re: 25 necklaces [#permalink]

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New post 07 Nov 2009, 13:02
I was correct till the definite integral part as I make the curve smooth ...

Anyway, I recalculate the sum above and it it is equal to 3025. Bravo Bunuel. +1 from me.

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Re: 25 necklaces [#permalink]

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New post 07 Nov 2009, 13:08
anyone could come up with a faster way to calculate this other than muscle??:
24*2 + 23*3 + ....3*23+ 2*24 = 2*(24*2 24*2 + 23*3 + ....14*12) + 13*13

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Re: 25 necklaces [#permalink]

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New post 12 May 2011, 08:28
5+ (5+2) + (5+5) + (5+9) .....
25*5 + (2+5+9+14+...)

In (2+5+9+14+ ...) the differences are in AP with a= 3, and common difference = 1

for 2nd term difference = 3
for 3rd term difference = 4
................................
for 25th term difference = 26

adding 3,4,5,,26 = [(3+26)/2]*24(number of terms) = 348

last difference = 348. Last term = 350

I will try to figure out more on this now.
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Re: There are 25 necklaces such that the first necklace contains [#permalink]

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New post 22 Jan 2013, 01:37
We are given T(n) = T(n-1) + n ----------1
and T1 = 5 ------- 2

We apply Summation (S) on and add constant to T(n) that we obtain containing only n

S[T(n)] = S[T(n-1)] S[n]

S[T(n)] - S[T(n-1)] = S[n] ----- difference of Sum of all and and the sum of all but last term is T(n)

T(n) = S[n]

Thus T(n) = n(n+1)/2 + k

from 2, we get T(1) = K = 4

Thus T(n) = (N^2 + N) / 2 + 4

Apply summation on T(n) to get the required sum

S(n) = 1/2[ n*(n+1)*(2n+1)/6 + n(n+1)/2) + 4n

Substitute N = 25 Answer = 3025.

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

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New post 23 Jan 2013, 02:53
Let us name N the total number of beads on the 25 necklaces and ni the number of beads on the ith necklace.

N= n1 + n2 + + n25
n1 = 5
n2 = 5 + 2
n3 = 5 + 2 + 3
n4 = 5 + 2 + 3 + 4

n25 = 5 + 2 + 3 + 4 + 25

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

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Re: 25 necklaces [#permalink]

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New post 06 Apr 2013, 04:33
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)

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!!!!!!!!

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

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New post 30 Jan 2014, 07:04
Here's another way of solving quickly

We have a sequence of 5,7,10,14,19,25 etc

Note that we have pairs O-O-E-E-O-O etc... We are going to have 12 pairs and one left, hence the sum of the terms will be ODD

Only ODD in the answer choices is B

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Re: 25 necklaces [#permalink]

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New post 30 Jan 2014, 07:08
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)


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?

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

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New post 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?

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

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New post 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?


Hope you have a great 4th of July!

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

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New post 04 Jul 2014, 15:53
jlgdr wrote:
Here's another way of solving quickly

We have a sequence of 5,7,10,14,19,25 etc

Note that we have pairs O-O-E-E-O-O etc... We are going to have 12 pairs and one left, hence the sum of the terms will be ODD

Only ODD in the answer choices is B



HOW DID YOU KNOW THE NUMBERS AFTER 7?
CAN YOU PLEASE SHOW HOW YOU PLUGGED IN AND FIGURED OUT? Thanks :)

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

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New post 05 Jul 2014, 06:59
sagnik242 wrote:
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?


1st necklace = 5 beads;
2nd necklace = 5 + 2 = 7 beads;
3rd necklace = 7 + 3 = 10 beads;
4th necklace = 10 + 4 = 14 beads;
...
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Re: There are 25 necklaces such that the first necklace contains [#permalink]

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New post 05 Jul 2014, 07:03
sagnik242 wrote:
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?


Hope you have a great 4th of July!


100 + (5 - 4) + (7 - 4) + (10 - 4) + (14 - 4) + (19 - 4) + ...
100 + 1 + 3 + 6 + 10 + 15 +...
100 + 1 + (3 + 6) + (10 + 15) + ...
100 + 1 + 9 + 25 + ...
100 + 1^2 + 3^2 + 5^2 + ...
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Re: There are 25 necklaces such that the first necklace contains [#permalink]

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Re: There are 25 necklaces such that the first necklace contains   [#permalink] 03 Oct 2017, 14:27
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