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14 Dec 2013, 06:14
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What is the remainder when the two-digit, positive integer x is divided by 3?
(1) The sum of the digits of x is 5
(2) The remainder when x is divided by 9 is 5
(1) The sum of the digits of x is 5
(2) The remainder when x is divided by 9 is 5
14 Dec 2013, 06:21
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What is the remainder when the two-digit, positive integer x is divided by 3?
(1) The sum of the digits of x is 5 --> x can be 14, 41, 23, 32, or 50. Each of this numbers gives the remainder of 2 when divided by 3. Sufficient.
(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.
Answer: D.
(1) The sum of the digits of x is 5 --> x can be 14, 41, 23, 32, or 50. Each of this numbers gives the remainder of 2 when divided by 3. Sufficient.
(2) The remainder when x is divided by 9 is 5 --> \(x = 9q + 5 = 9q + 3 + 2 =3(3q+1)+2\) --> the remainder when x is divided by 3 is 2. Sufficient.
Answer: D.
14 Dec 2013, 18:51
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A no is divisile by 3 if the sum of the digits is divisible by 3. The remainder any no with 3 can be found by just dividing the sum of digits(of that no) with 3. If it is zero, the no is divisible by 3, otherwise the remainder obtained is same as what we would have got if we divide the no by 3.
1) sum of digits of x is 5. Hence, the remainder obtained by dividing x by 3 is same as remainder obtained by dividing 5 by 3, i.e. 2.
2) x = 9k + 5. => 3*3k + 3 + 2 = 3(3k + 1)+ 2. Now if we divide this by3, the remainder will be 2.
Hence, each statement is sufficient. Ans- D
1) sum of digits of x is 5. Hence, the remainder obtained by dividing x by 3 is same as remainder obtained by dividing 5 by 3, i.e. 2.
2) x = 9k + 5. => 3*3k + 3 + 2 = 3(3k + 1)+ 2. Now if we divide this by3, the remainder will be 2.
Hence, each statement is sufficient. Ans- D
