\(B\) is a two-digit number, which can be expressed as the product of \(a\) and \(e\), where both \(a\) and \(e\) are positive integers. What is the remainder when \(B\) is divided by 12?
1) \(a\) is an even number greater than 5.
2) \(a, b, c, d\), and \(e\) are 5 consecutive integers, in increasing order.
The OA is C
Source: e-GMAT
\(B\) is a two-digit number, which can be expressed as the
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So, we have B = ae.swerve wrote:\(B\) is a two-digit number, which can be expressed as the product of \(a\) and \(e\), where both \(a\) and \(e\) are positive integers. What is the remainder when \(B\) is divided by 12?
1) \(a\) is an even number greater than 5.
2) \(a, b, c, d\), and \(e\) are 5 consecutive integers, in increasing order.
The OA is C
Source: e-GMAT
We have to find out the remainder when B = ae is divided by 12.
Let's take each statement one by one.
1) \(a\) is an even number greater than 5.
a is one among 6, 8, 10, 12, ... 98. These values would ensure that B is a 2-digit number. a = 98 is possible if e = 1.
Since we do not know the unique value of a and e, we can't get the value of B. Insufficient.
2) \(a, b, c, d\), and \(e\) are 5 consecutive integers, in increasing order.
=> e = a + 4
Thus, B = ae = a(a + 4)
Case 1: Say a = 2
We have B = ae = a(a + 4) = 2*(2 + 4) = 12. We see that 12 divided by 12 leave 0 as reminder.
Case 2: Say a = 3
We have B = ae = a(a + 4) = 3*(3 + 4) = 21. We see that 21 divided by 12 leave 9 as reminder.
No unique answer. Insufficient.
(1) and (2) together
Case 1: Say a = 6
We have B = ae = a(a + 4) = 6*(6 + 4) = 60. We see that 60 divided by 12 leave 0 as reminder.
Case 2: Say a = 8
We have B = ae = a(a + 4) = 8*(8 + 4) = 96. We see that 96 divided by 12 leave 0 as reminder.
Case 3: Say a = 10
We have B = ae = a(a + 4) = 10*(10 + 4) = 140, not a 2-digit number. Invalid value of a.
Since at higher values of a, B would not be a 2-digit number, we must take a = 6 or 8. Thus, with each case, the remainder when B is divided by 12 is 0. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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