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The product of two consecutive positive integers is 156. Wha
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20 Mar 2020, 11:42
20 Mar 2020, 11:42
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The product of two consecutive positive integers is 156. What is the larger of the two integers?
A. 11
B. 12
C. 13
D. 14
E. 15
Kudos for the right answer and explanation
A. 11
B. 12
C. 13
D. 14
E. 15
Kudos for the right answer and explanation
Re: The product of two consecutive positive integers is 156. Wha
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20 Mar 2020, 21:13
20 Mar 2020, 21:13
1
There are two methods to solve this question.
Let's try the easy one first.
\(12 \times 12 = 144\)
Now, 156 is a little greater than 144. And we need two consecutive positive integers.
So, we try \(12 \times 13\) which gives us \(156\).
The larger integer is 13.
OA, C
Let two consecutive positive integers be N and N+1
Now, \(N \times (N+1) = 156\)
\(N^2 + N = 156\)
\(N^2 + N - 156 = 0\)
\(N^2 +13N - 12N -156 = 0\)
\(N(N+13) -12(N+13) = 0\)
\((N-12)(N+13) = 0\)
\(N = 12\) or \(N = -13\)
Now, since N is a positive integer so the only possible value for N is 12.
\(N+1 = 13\).
The larger integer becomes 13.
OA, C
Let's try the easy one first.
Method 1:
\(12 \times 12 = 144\)
Now, 156 is a little greater than 144. And we need two consecutive positive integers.
So, we try \(12 \times 13\) which gives us \(156\).
The larger integer is 13.
OA, C
Method 2:
Let two consecutive positive integers be N and N+1
Now, \(N \times (N+1) = 156\)
\(N^2 + N = 156\)
\(N^2 + N - 156 = 0\)
\(N^2 +13N - 12N -156 = 0\)
\(N(N+13) -12(N+13) = 0\)
\((N-12)(N+13) = 0\)
\(N = 12\) or \(N = -13\)
Now, since N is a positive integer so the only possible value for N is 12.
\(N+1 = 13\).
The larger integer becomes 13.
OA, C
Carcass wrote:
The product of two consecutive positive integers is 156. What is the larger of the two integers?
A. 11
B. 12
C. 13
D. 14
E. 15
Kudos for the right answer and explanation
A. 11
B. 12
C. 13
D. 14
E. 15
Kudos for the right answer and explanation
gmatclubot
Re: The product of two consecutive positive integers is 156. Wha [#permalink]
20 Mar 2020, 21:13
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