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26 May 2017, 04:48
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If the sum of the n consecutive positive integers is 68, what is the possible value of n?
A. 3
B. 5
C. 6
D. 8
E. 17
A. 3
B. 5
C. 6
D. 8
E. 17
26 May 2017, 04:49
Kudos
Bookmarks
Official Solution:
If the sum of the n consecutive positive integers is 68, what is the possible value of n?
A. 3
B. 5
C. 6
D. 8
E. 17
In general, you get \(1 + 2 + 3 +\ldots .+(n-1) + n = \frac{n(n+1)}{2}\).
Since it is the sum of n consecutive positive integers, if you assume the smallest integer as x, from
\(x + (x+1) + (x+2) + \ldots \ldots .. + (x+n-1) = 68\), you get
\(nx + [1 + 2 + 3 + \ldots \ldots + (n - 1)] = nx + \frac{n(n-1)}{2} = 68\), and if you multiply 2 on both sides, you get \(2nx + n(n-1) = 68 * 2 = 136\), or \(n(2x + n - 1) = 136\).
In the case of \(n(2x + n - 1) = 136 = 8 * 17, n = 8\) and \(2x + 8 - 1 = 17\), then \(2x = 10, x = 5,\) hence correct. In other words, \(5+6+7+8+9+10+11+12 is 68.\)
In the case of \(n(2x+n - 1) = 136 = 4*34\), there is no integer that satisfies this in \(n = 4\) and \(2x + 4 - 1 = 34\), hence not true.
In the case of \(n(2x+n - 1) = 136 = 17*8, n = 17\) and \(2x + 17 - 1 = 8,\) then \(x = -4\), since \(x\) is not a positive integer, hence not true. The answer is D.
Answer: D
If the sum of the n consecutive positive integers is 68, what is the possible value of n?
A. 3
B. 5
C. 6
D. 8
E. 17
In general, you get \(1 + 2 + 3 +\ldots .+(n-1) + n = \frac{n(n+1)}{2}\).
Since it is the sum of n consecutive positive integers, if you assume the smallest integer as x, from
\(x + (x+1) + (x+2) + \ldots \ldots .. + (x+n-1) = 68\), you get
\(nx + [1 + 2 + 3 + \ldots \ldots + (n - 1)] = nx + \frac{n(n-1)}{2} = 68\), and if you multiply 2 on both sides, you get \(2nx + n(n-1) = 68 * 2 = 136\), or \(n(2x + n - 1) = 136\).
In the case of \(n(2x + n - 1) = 136 = 8 * 17, n = 8\) and \(2x + 8 - 1 = 17\), then \(2x = 10, x = 5,\) hence correct. In other words, \(5+6+7+8+9+10+11+12 is 68.\)
In the case of \(n(2x+n - 1) = 136 = 4*34\), there is no integer that satisfies this in \(n = 4\) and \(2x + 4 - 1 = 34\), hence not true.
In the case of \(n(2x+n - 1) = 136 = 17*8, n = 17\) and \(2x + 17 - 1 = 8,\) then \(x = -4\), since \(x\) is not a positive integer, hence not true. The answer is D.
Answer: D
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