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30 Dec 2021, 06:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (02:21) correct
39%
(02:45)
wrong
based on 96
sessions
History
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Not Attempted Yet
After 5 or 6 dozen oranges were packed in each box, it was found that three dozen oranges remained unpacked. Therefore, bigger boxes were utilized to pack 8 or 9 dozen oranges. However, three dozen oranges again remained unpacked. What was the least number of dozens of oranges that needed to be packed?
(A) 210
(B) 216
(C) 242
(D) 363
(E) 435
(A) 210
(B) 216
(C) 242
(D) 363
(E) 435
30 Dec 2021, 08:12
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Bunuel
It took me a long time to work what the question (probably) means -- as it's worded, the question isn't asking what it intends to ask. When it says "5 or 6 dozen oranges were packed in each box", I couldn't tell if that meant "either 5 dozen were packed in each box, or 6 dozen were packed in each box, but I'm not telling you which". If that were the interpretation, the answer is 43 (using only 5-dozen and 8-dozen boxes, we'd have 3 left over each time). That wasn't an answer choice, so then I wondered if it meant "5 dozen were packed in some boxes, and 6 dozen in the rest". But then if they were packed arbitrarily, and in the end they discovered 3 left over dozen each time, there could be as few as 19 dozen oranges (using 2 five-dozen and one six-dozen boxes, then two eight-dozen boxes).
Finally I think I figured out what the question is trying to ask (but only by looking at the answer choices -- as the question is worded, it is not asking this). I assume the question is trying to say that when they pack 5 dozen to a box, 3 dozen are left over, when they pack 6 dozen to a box, 3 dozen are left over, when they pack 8 dozen to a box, 3 dozen are left over, and when they pack 9 dozen to a box, 3 dozen are left over. So we're looking for the smallest positive integer greater than 9 that gives us a remainder of 3 when we divide by 5, 6, 8 and 9. That will just be whatever number is 3 greater than the LCM of 5, 6, 8 and 9, and that LCM is 360, so the answer is 363.
20 Feb 2026, 04:54
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Deconstructing the Question
Let total dozens be N.
Given:
\(N\equiv3\pmod{5}\)
\(N\equiv3\pmod{6}\)
\(N\equiv3\pmod{8}\)
\(N\equiv3\pmod{9}\)
Key idea: subtract 3 and find a number divisible by all 5, 6, 8, and 9.
Step-by-step
\(N-3\) must be divisible by \(5,6,8,9\).
Compute LCM:
\(\text{lcm}(5,6,8,9)=360\)
So:
\(N-3=360\)
\(N=363\)
Answer: 363
Let total dozens be N.
Given:
\(N\equiv3\pmod{5}\)
\(N\equiv3\pmod{6}\)
\(N\equiv3\pmod{8}\)
\(N\equiv3\pmod{9}\)
Key idea: subtract 3 and find a number divisible by all 5, 6, 8, and 9.
Step-by-step
\(N-3\) must be divisible by \(5,6,8,9\).
Compute LCM:
\(\text{lcm}(5,6,8,9)=360\)
So:
\(N-3=360\)
\(N=363\)
Answer: 363
