N, a set of natural numbers is partitioned into subsets: S1 = {1} S2 = {2, 3} S3 = {4, 5, 6}
S4 = {7, 8, 9, 10} and so on. The sum of the elements of the subset S30 is:
(A) 12505 (B) 14115 (C) 13515 (D) 16505 (E) 14000
Please provide your approach to solve this prob.
Tough PS-1
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S1 = {1} (1 number)satishchandra wrote:N, a set of natural numbers is partitioned into subsets: S1 = {1} S2 = {2, 3} S3 = {4, 5, 6}
S4 = {7, 8, 9, 10} and so on. The sum of the elements of the subset S30 is:
(A) 12505 (B) 14115 (C) 13515 (D) 16505 (E) 14000
S2 = {2, 3} (2 numbers)
S3 = {4, 5, 6} (3 numbers)
S4 = {7, 8, 9, 10} (4 numbers)
.
.
.
S29 = {?????} (29 numbers)
Big question: At this point, how many natural numbers have we counted so far?
To find out, we need to find the sum of 1+2+3+4...+27+28+29
There's a nice formula that says: The sum 1+2+3+...+n = (n)(n+1)/2
So, 1+2+3+4...+27+28+29 = (29)(30)/2, which equals 435
This means there are 435 numbers in the subsets from S1 to S29.
In other words, S29 = {407, 408, 409, . . ., 434, 435}
So, S30 = {436, 437, 438, . . ., 464, 465}
We're asked to find the sum of 436+437+438...+464+465
There several different options here, but the fastest is to use the answer choices to our advantage.
Notice that if all 30 numbers in S30 were 436 (the smallest number in the set), then the sum would be (30)(436) = 13,080
This means that the sum of 436+437+438...+464+465 must be greater than 13,080 (eliminate answer choice A)
Now notice that if all 30 numbers in S30 were 465 (the biggest number in the set), then the sum would be (30)(465) = 13,950
This means that the sum of 436+437+438...+464+465 must be less than 13,950 (eliminate answer choices B, D, and E)
This leaves us with C, the correct answer.
Cheers,
Brent
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Hi,
I dont know if there is an easier way to do this and quite frankly I am intrigued to see if there is a better explanation. There would be no way that I can solve this in 2 mins if I got it on the actual test. But my method is as below.
Assume the set number is n ie; S1 where 1 is n. I found a pattern where I could determine the last number of the set
S1 - 1 - n
S2 - 2,3 - n+1 - 2+1 = 3
S3 - 4,5,6 - 2n - 2*3 = 6
S4 - 7,8,9,10 - 2n+2 - 2*4+2 = 10
S5 - 11,12,13,14,15 - 3n
and so on till you get S30 the last number can be determined by 15n+15 which is equal to 465. Then its a matter of finding the sum of all 30 integers from 465 to 436 which will give us a total of 13515. Hence C
I dont know if there is an easier way to do this and quite frankly I am intrigued to see if there is a better explanation. There would be no way that I can solve this in 2 mins if I got it on the actual test. But my method is as below.
Assume the set number is n ie; S1 where 1 is n. I found a pattern where I could determine the last number of the set
S1 - 1 - n
S2 - 2,3 - n+1 - 2+1 = 3
S3 - 4,5,6 - 2n - 2*3 = 6
S4 - 7,8,9,10 - 2n+2 - 2*4+2 = 10
S5 - 11,12,13,14,15 - 3n
and so on till you get S30 the last number can be determined by 15n+15 which is equal to 465. Then its a matter of finding the sum of all 30 integers from 465 to 436 which will give us a total of 13515. Hence C
Regards
Anup
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Here's another way to find the sum of the 30 numbers in S30
Our goal is to add: 436 + 437 + 438 + 439 + . . . + 463 + 464 + 465
Start adding pairs of numbers beginning from the outside and moving towards the center.
436 + 465 = 901
437 + 464 = 901
438 + 463 = 901
and so on.
How many pairs of numbers are there? Well, there are 30 numbers in S30, so there are 15 pairs.
So, 436 + 437 + 438 + 439 + . . . + 463 + 464 + 465 = (15)(901) = [spoiler]13,515 = C[/spoiler]
Cheers,
Brent
Our goal is to add: 436 + 437 + 438 + 439 + . . . + 463 + 464 + 465
Start adding pairs of numbers beginning from the outside and moving towards the center.
436 + 465 = 901
437 + 464 = 901
438 + 463 = 901
and so on.
How many pairs of numbers are there? Well, there are 30 numbers in S30, so there are 15 pairs.
So, 436 + 437 + 438 + 439 + . . . + 463 + 464 + 465 = (15)(901) = [spoiler]13,515 = C[/spoiler]
Cheers,
Brent
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Brent@GMATPrepNow wrote:
S1 = {1} (1 number)
S2 = {2, 3} (2 numbers)
S3 = {4, 5, 6} (3 numbers)
S4 = {7, 8, 9, 10} (4 numbers)
.
.
.
S29 = {?????} (29 numbers)
To find out, we need to find the sum of 1+2+3+4...+27+28+29
This, I suppose, is key to answering the question. Great catch Brent. Thanks
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Brent@GMATPrepNow wrote:S1 = {1} (1 number)satishchandra wrote:N, a set of natural numbers is partitioned into subsets: S1 = {1} S2 = {2, 3} S3 = {4, 5, 6}
S4 = {7, 8, 9, 10} and so on. The sum of the elements of the subset S30 is:
(A) 12505 (B) 14115 (C) 13515 (D) 16505 (E) 14000
S2 = {2, 3} (2 numbers)
S3 = {4, 5, 6} (3 numbers)
S4 = {7, 8, 9, 10} (4 numbers)
.
.
.
S29 = {?????} (29 numbers)
Big question: At this point, how many natural numbers have we counted so far?
To find out, we need to find the sum of 1+2+3+4...+27+28+29
There's a nice formula that says: The sum 1+2+3+...+n = (n)(n+1)/2
So, 1+2+3+4...+27+28+29 = (29)(30)/2, which equals 435
This means there are 435 numbers in the subsets from S1 to S29.
In other words, S29 = {407, 408, 409, . . ., 434, 435}
So, S30 = {436, 437, 438, . . ., 464, 465}
We're asked to find the sum of 436+437+438...+464+465
There several different options here, but the fastest is to use the answer choices to our advantage.
Notice that if all 30 numbers in S30 were 436 (the smallest number in the set), then the sum would be (30)(436) = 13,080
This means that the sum of 436+437+438...+464+465 must be greater than 13,080 (eliminate answer choice A)
Now notice that if all 30 numbers in S30 were 465 (the biggest number in the set), then the sum would be (30)(465) = 13,950
This means that the sum of 436+437+438...+464+465 must be less than 13,950 (eliminate answer choices B, D, and E)
This leaves us with C, the correct answer.
Cheers,
Brent
Thanks Brent for the explanation...Just a small doubt ...
This means there are 435 numbers in the subsets from S1 to S29.
We have found the sum of first 29 numbers to be 435. But not got how we say that the numbers itself are 435 ; I thought its sum and not the number itself Please help !!!
If you cant explain it simply you dont understand it well enough!!!
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GaneshMalkar wrote:Brent@GMATPrepNow wrote:S1 = {1} (1 number)satishchandra wrote:N, a set of natural numbers is partitioned into subsets: S1 = {1} S2 = {2, 3} S3 = {4, 5, 6}
S4 = {7, 8, 9, 10} and so on. The sum of the elements of the subset S30 is:
(A) 12505 (B) 14115 (C) 13515 (D) 16505 (E) 14000
S2 = {2, 3} (2 numbers)
S3 = {4, 5, 6} (3 numbers)
S4 = {7, 8, 9, 10} (4 numbers)
.
.
.
S29 = {?????} (29 numbers)
Big question: At this point, how many natural numbers have we counted so far?
To find out, we need to find the sum of 1+2+3+4...+27+28+29
There's a nice formula that says: The sum 1+2+3+...+n = (n)(n+1)/2
So, 1+2+3+4...+27+28+29 = (29)(30)/2, which equals 435
This means there are 435 numbers in the subsets from S1 to S29.
In other words, S29 = {407, 408, 409, . . ., 434, 435}
So, S30 = {436, 437, 438, . . ., 464, 465}
We're asked to find the sum of 436+437+438...+464+465
There several different options here, but the fastest is to use the answer choices to our advantage.
Notice that if all 30 numbers in S30 were 436 (the smallest number in the set), then the sum would be (30)(436) = 13,080
This means that the sum of 436+437+438...+464+465 must be greater than 13,080 (eliminate answer choice A)
Now notice that if all 30 numbers in S30 were 465 (the biggest number in the set), then the sum would be (30)(465) = 13,950
This means that the sum of 436+437+438...+464+465 must be less than 13,950 (eliminate answer choices B, D, and E)
This leaves us with C, the correct answer.
Cheers,
Brent
Thanks Brent for the explanation...Just a small doubt ...
This means there are 435 numbers in the subsets from S1 to S29.
We have found the sum of first 29 numbers to be 435. But not got how we say that the numbers itself are 435 ; I thought its sum and not the number itself Please help !!!
Ohh I understood sorry...
S1 {1}
S2 {2,3}
S3 {4,5,6}
Taking these 3 subsets we can find the last number in set S3 as 3 * 4 / 2 = 6, so applying the same logic S29 will have 29 * 30 / 2 = 29 * 15 = 435
Got thanks Brent
If you cant explain it simply you dont understand it well enough!!!
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There is 1 number in S1GaneshMalkar wrote:Brent@GMATPrepNow wrote:S1 = {1} (1 number)satishchandra wrote:N, a set of natural numbers is partitioned into subsets: S1 = {1} S2 = {2, 3} S3 = {4, 5, 6}
S4 = {7, 8, 9, 10} and so on. The sum of the elements of the subset S30 is:
(A) 12505 (B) 14115 (C) 13515 (D) 16505 (E) 14000
S2 = {2, 3} (2 numbers)
S3 = {4, 5, 6} (3 numbers)
S4 = {7, 8, 9, 10} (4 numbers)
.
.
.
S29 = {?????} (29 numbers)
Big question: At this point, how many natural numbers have we counted so far?
To find out, we need to find the sum of 1+2+3+4...+27+28+29
There's a nice formula that says: The sum 1+2+3+...+n = (n)(n+1)/2
So, 1+2+3+4...+27+28+29 = (29)(30)/2, which equals 435
This means there are 435 numbers in the subsets from S1 to S29.
In other words, S29 = {407, 408, 409, . . ., 434, 435}
So, S30 = {436, 437, 438, . . ., 464, 465}
We're asked to find the sum of 436+437+438...+464+465
There several different options here, but the fastest is to use the answer choices to our advantage.
Notice that if all 30 numbers in S30 were 436 (the smallest number in the set), then the sum would be (30)(436) = 13,080
This means that the sum of 436+437+438...+464+465 must be greater than 13,080 (eliminate answer choice A)
Now notice that if all 30 numbers in S30 were 465 (the biggest number in the set), then the sum would be (30)(465) = 13,950
This means that the sum of 436+437+438...+464+465 must be less than 13,950 (eliminate answer choices B, D, and E)
This leaves us with C, the correct answer.
Cheers,
Brent
Thanks Brent for the explanation...Just a small doubt ...
This means there are 435 numbers in the subsets from S1 to S29.
We have found the sum of first 29 numbers to be 435. But not got how we say that the numbers itself are 435 ; I thought its sum and not the number itself Please help !!!
There are 2 numbers in S2 (numbers used so far = 1+2=3)
There are 3 numbers in S3 (numbers used so far = 1+2+3=6)
Aside: notice that S4 begins with 7 (one more than 6) S4 = {7, 8, 9, 10}
There are 4 numbers in S4 (numbers used so far = 1+2+3+4=10)
Aside: notice that S5 begins with 11 (one more than 10) S5 = {11, 12, 13, 14, 15}
.
.
.
There are 29 numbers in S29 (numbers used so far = 1+2+3+...+28+29=435)
Cheers,
Brent