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Night reader
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When positive integer a is divided by 5, the remainder is 3. What is the value of a?
(1) When a is divided by 6, the remainder is 1.
(2) When a is divided by 4, the remainder is 1.
OA E
GMAT Hacks' explanation >>>
Given the information in the question, the value of a could be 8, 13,18, etc.-3 greater than any multiple of 5.
Statement (1) is insufficient; a could be 7, 13, 19, 25, etc. There's one overlap so far between that list and the initial lists of possible a's-it could be 13. However, it could be 43 as well. To Â…nd that number quickly, realize that the patterns of multiples of 5 and 6 "restart" every 30 integers (30, because that's the least common multiple of 5 and 6). So a could be 13 greater than any multiple of 30.
Statement (2) is also insufficient. The logic in (1) suggests that, whatever the possible values of a, there will be more than 1. In this case, a again could be 13, and since the LCM of 5 and 4 is 20, the process will restart every 20 integers. So a could be 13 greater than any multiple of 20, meaning that a might be 13, 33, 53, etc.
Taken together, the statements are still insufficient. We've seen that a could be 13, and it could also be 13 greater than the LCM of the three numbers, which is 60, or any multiple thereof. So while a must be of the form 60i + 13, that doesn't give us a single value. Choice (E) is correct.
>>>
My solution:
a=5i+3, where i is a positive integer
Statement (1) a=6i+1, set equation 5i+3=6i+1 <> i=2 <> 2 is a positive integer, a=5*2+3, a=13 <> Satisfies with a=6i+1 (deleted 3, mistyping)
Statement (2) a=4i+1, set equation 5i+3=4i+1 <> i=-2 <> -2 is not a positive integer <> Does not satisfy
Select A as the correct answer.
(1) When a is divided by 6, the remainder is 1.
(2) When a is divided by 4, the remainder is 1.
OA E
GMAT Hacks' explanation >>>
Given the information in the question, the value of a could be 8, 13,18, etc.-3 greater than any multiple of 5.
Statement (1) is insufficient; a could be 7, 13, 19, 25, etc. There's one overlap so far between that list and the initial lists of possible a's-it could be 13. However, it could be 43 as well. To Â…nd that number quickly, realize that the patterns of multiples of 5 and 6 "restart" every 30 integers (30, because that's the least common multiple of 5 and 6). So a could be 13 greater than any multiple of 30.
Statement (2) is also insufficient. The logic in (1) suggests that, whatever the possible values of a, there will be more than 1. In this case, a again could be 13, and since the LCM of 5 and 4 is 20, the process will restart every 20 integers. So a could be 13 greater than any multiple of 20, meaning that a might be 13, 33, 53, etc.
Taken together, the statements are still insufficient. We've seen that a could be 13, and it could also be 13 greater than the LCM of the three numbers, which is 60, or any multiple thereof. So while a must be of the form 60i + 13, that doesn't give us a single value. Choice (E) is correct.
>>>
My solution:
a=5i+3, where i is a positive integer
Statement (1) a=6i+1, set equation 5i+3=6i+1 <> i=2 <> 2 is a positive integer, a=5*2+3, a=13 <> Satisfies with a=6i+1 (deleted 3, mistyping)
Statement (2) a=4i+1, set equation 5i+3=4i+1 <> i=-2 <> -2 is not a positive integer <> Does not satisfy
Select A as the correct answer.
Last edited by Night reader on Wed Dec 01, 2010 12:01 am, edited 1 time in total.
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