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16 Sep 2014, 00:37
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If \(t\) represents the tens digit of the integer \(n = 194,7t6\), what is the value of \(t\)?
(1) \(n\) is divisible by 4
(2) \(n\) is divisible by 3
(1) \(n\) is divisible by 4
(2) \(n\) is divisible by 3
16 Sep 2014, 00:37
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Official Solution:
If \(t\) represents the tens digit of the integer \(n = 194,7t6\), what is the value of \(t\)?
(1) \(n\) is divisible by 4.
A number is divisible by 4 if its last two digits form a number divisible by 4. For \(n\) to be divisible by 4, \(t6\) must be equal to 16, 36, 56, 76, or 96, which means that \(t\) can be 1, 3, 5, 7, or 9. Not sufficient.
(2) \(n\) is divisible by 3.
A number is divisible by 3 if the sum of its digits is divisible by 3. Since the sum of the digits of \(n\), excluding \(t\), is \(1+9+4+7+6=27\), for \(n\) to be divisible by 3, \(t\) must be equal to 0, 3, 6, or 9. Not sufficient.
(1)+(2) Combining both statements, we find that \(t\) can still be either 3 or 9. Not sufficient.
Answer: E
If \(t\) represents the tens digit of the integer \(n = 194,7t6\), what is the value of \(t\)?
(1) \(n\) is divisible by 4.
A number is divisible by 4 if its last two digits form a number divisible by 4. For \(n\) to be divisible by 4, \(t6\) must be equal to 16, 36, 56, 76, or 96, which means that \(t\) can be 1, 3, 5, 7, or 9. Not sufficient.
(2) \(n\) is divisible by 3.
A number is divisible by 3 if the sum of its digits is divisible by 3. Since the sum of the digits of \(n\), excluding \(t\), is \(1+9+4+7+6=27\), for \(n\) to be divisible by 3, \(t\) must be equal to 0, 3, 6, or 9. Not sufficient.
(1)+(2) Combining both statements, we find that \(t\) can still be either 3 or 9. Not sufficient.
Answer: E
07 Jul 2016, 09:38
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I think this is a high-quality question and I agree with explanation. Good question !!! I was just about to choose Ans C .
Since I identified two sets S2 = { 0,3,6,9} and S1= {1,3,5,7}. At this moment I thought all possible values are exhausted, and would go for Ans C as 3 is common in both, but while cross checking I identified another element 9 for Statement 1, which saved the life to choose Ans E. Good trap !!!!
Since I identified two sets S2 = { 0,3,6,9} and S1= {1,3,5,7}. At this moment I thought all possible values are exhausted, and would go for Ans C as 3 is common in both, but while cross checking I identified another element 9 for Statement 1, which saved the life to choose Ans E. Good trap !!!!
