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GMAT 2: 770 Q50 V47
Posts: 239
03 Apr 2014, 12:39
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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72% (02:42) correct
28%
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based on 605
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Tough and Tricky questions: Number Properties.
A table of numbers has \(n\) rows and \(n\) columns, where \(n\) is an odd integer greater than 4. Each entry in the table is subject to the following rules:
(I) Any entry is the negative of the entry directly above it, if there is an entry directly above it.
(II) Any entry is double the entry directly to its left, if there is an entry directly to its left.
If the entry in the upper left corner (first row and first column) is 1, what is the sum of all the entries in the table?
A. \(0\)
B. \(n^2 - 1\)
C. \(n^2 + 1\)
D. \(2^n - 1\)
E. \(2^n + 1\)
19 Nov 2014, 08:01
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Official Solution:
A table of numbers has \(n\) rows and \(n\) columns, where \(n\) is an odd integer greater than 4. Each entry in the table is subject to the following rules:
(I) Any entry is the negative of the entry directly above it, if there is an entry directly above it.
(II) Any entry is double the entry directly to its left, if there is an entry directly to its left.
If the entry in the upper left corner (first row and first column) is 1, what is the sum of all the entries in the table?
A. \(0\)
B. \(n^2 - 1\)
C. \(n^2 + 1\)
D. \(2^n - 1\)
E. \(2^n + 1\)
The best way to approach this problem is to generate an actual table according to the rules. Let's start with the most concrete information we know: the entry in the upper left corner is 1. Now let's build the table to the right and then down.
According to the second rule, entries double as we go to the right, so the first row looks like this:
\(\text{1, 2, 4, 8, 16 ...}\)
According to the first rule, entries "flipflop" in sign (+/-) as we go down the table. So the first five rows and columns look like this:
\(\text{1, 2, 4, 8, 16}\)
\(\text{-1, -2, -4, -8, -16}\)
\(\text{1, 2, 4, 8, 16}\)
\(\text{-1, -2, -4, -8, -16}\)
\(\text{1, 2, 4, 8, 16}\)
Since \(n\) is an odd integer greater than 4, we can choose \(n = 5\) and stop here.
The process of adding up all the entries benefits from cancellation. The first four rows cancel each other out completely. All that is left is the last row, which sums up as follows: \(1 + 2 + 4 + 8 + 16 = 31\). Plugging \(n = 5\) into the answer choices, we get the following:
(A) \(0\) INCORRECT
(B) \(n^2 - 1 = 52 - 1 = 24\) INCORRECT
(C) \(n^2 + 1 = 52 + 1 = 26\) INCORRECT
(D) \(2^n - 1 = 25 - 1 = 31\) CORRECT
(E) \(2^n + 1 = 25 + 1 = 33\) INCORRECT
We can prove that \(2^n - 1\) is the general formula for the sum of 1, 2, 4, etc. up to \(2^{n - 1}\) (which will be the last entry in the last row), but there is no need to do so.
Answer: D.
A table of numbers has \(n\) rows and \(n\) columns, where \(n\) is an odd integer greater than 4. Each entry in the table is subject to the following rules:
(I) Any entry is the negative of the entry directly above it, if there is an entry directly above it.
(II) Any entry is double the entry directly to its left, if there is an entry directly to its left.
If the entry in the upper left corner (first row and first column) is 1, what is the sum of all the entries in the table?
A. \(0\)
B. \(n^2 - 1\)
C. \(n^2 + 1\)
D. \(2^n - 1\)
E. \(2^n + 1\)
The best way to approach this problem is to generate an actual table according to the rules. Let's start with the most concrete information we know: the entry in the upper left corner is 1. Now let's build the table to the right and then down.
According to the second rule, entries double as we go to the right, so the first row looks like this:
\(\text{1, 2, 4, 8, 16 ...}\)
According to the first rule, entries "flipflop" in sign (+/-) as we go down the table. So the first five rows and columns look like this:
\(\text{1, 2, 4, 8, 16}\)
\(\text{-1, -2, -4, -8, -16}\)
\(\text{1, 2, 4, 8, 16}\)
\(\text{-1, -2, -4, -8, -16}\)
\(\text{1, 2, 4, 8, 16}\)
Since \(n\) is an odd integer greater than 4, we can choose \(n = 5\) and stop here.
The process of adding up all the entries benefits from cancellation. The first four rows cancel each other out completely. All that is left is the last row, which sums up as follows: \(1 + 2 + 4 + 8 + 16 = 31\). Plugging \(n = 5\) into the answer choices, we get the following:
(A) \(0\) INCORRECT
(B) \(n^2 - 1 = 52 - 1 = 24\) INCORRECT
(C) \(n^2 + 1 = 52 + 1 = 26\) INCORRECT
(D) \(2^n - 1 = 25 - 1 = 31\) CORRECT
(E) \(2^n + 1 = 25 + 1 = 33\) INCORRECT
We can prove that \(2^n - 1\) is the general formula for the sum of 1, 2, 4, etc. up to \(2^{n - 1}\) (which will be the last entry in the last row), but there is no need to do so.
Answer: D.
General Discussion
03 Apr 2014, 13:52
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Note that as any entry is the negative of the entry directly above it, the sum of all rows except the last one will cancel out.
Therefore the problem reduces to finding the sum of the last row.
This row will be 1,2,4,8,......n. This is a GP with sum = (2^n)-1
Option (D).
Therefore the problem reduces to finding the sum of the last row.
This row will be 1,2,4,8,......n. This is a GP with sum = (2^n)-1
Option (D).
