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05 Aug 2021, 06:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:34) correct
15%
(01:44)
wrong
based on 194
sessions
History
Date
Time
Result
Not Attempted Yet
What is the least positive integer which when divided by 2, 3, 4, 5 and 6 leaves 1, 2, 3, 4 and 5 as remainders, respectively, but when divided by 7 leaves no remainder ?
(A) 109
(B) 119
(C) 126
(D) 154
(E) 210
(A) 109
(B) 119
(C) 126
(D) 154
(E) 210
05 Aug 2021, 06:35
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Bunuel
Let's deal with each piece of information individually.
We are told that, when the number is divided by 2, the remainder is 1.
This tells us that the number must be ODD, which means we can eliminate answer choices C, D and E they are all EVEN.
We're also told that, when the number is divided by 7, the reminder is 0.
In other words, the number must be divisible by 7.
Let's check the two remaining answer choices...
109 is NOT divisible by 7, which means we can eliminate answer choice A.
119 IS divisible by 7, so the correct answer must be B.
General Discussion
05 Aug 2021, 06:43
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My attempt
Looking through the answers, we can get rid of answers C, D, & E because they are even and are divisible by either 2 or 5.
Now for the remaining answers, we can test using 3 and 4. Only choice B works.
Answer: B
Looking through the answers, we can get rid of answers C, D, & E because they are even and are divisible by either 2 or 5.
Now for the remaining answers, we can test using 3 and 4. Only choice B works.
Answer: B
