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There are 25 necklaces such that the first necklace contains [#permalink]
 07 Nov 2009, 10:37
Question Stats:
53% (04:28) correct
46% (02:44) wrong based on 4 sessions
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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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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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 necklacesA. 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=3025Answer: B (3025)
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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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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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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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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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I suppose you guys learned by heart most of the sum formulas...
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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]
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]
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: There are 25 necklaces such that the first necklace contains [#permalink]
06 Apr 2013, 04:24
fdecazaux wrote: 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 which formula is applied in the above higlighted step....please elaborate it more....
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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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