After 5 or 6 dozen oranges were packed in each box, it was found that

Question

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

IanStewart · Answer

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.

luisdicampo · Answer

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:
 	ext{lcm}(5,6,8,9)=360

So:
 N-3=360 
 N=363

Answer: 363

Edskore · Answer

Good problem — this is a disguised Remainders + LCM question, which is exactly the kind that trips students up because the wording buries the actual math structure.

Key concept being tested: Remainders and LCM

The common trap is treating "5 or 6 dozen per box" as two separate scenarios to test individually, rather than recognizing that the remainder condition applies to both simultaneously.

Step 1: Translate the problem into remainder language.
Let N = total dozens of oranges.
- When packed in boxes of 5 dozen: 3 dozen remain → N ≡ 3 (mod 5)
- When packed in boxes of 6 dozen: 3 dozen remain → N ≡ 3 (mod 6)
- When packed in boxes of 8 dozen: 3 dozen remain → N ≡ 3 (mod 8)
- When packed in boxes of 9 dozen: 3 dozen remain → N ≡ 3 (mod 9)

Step 2: Recognize the pattern.
N − 3 is divisible by 5, 6, 8, and 9 simultaneously. So N − 3 = LCM(5, 6, 8, 9).

Step 3: Calculate the LCM.
5 = 5
6 = 2 × 3
8 = 23
9 = 32
LCM = 5 × 23 × 32 = 5 × 8 × 9 = 360

Step 4: Solve for N.
N − 3 = 360 → N = 363

Answer: D (363)

Why students get stuck: The problem mentions "bigger boxes of 8 or 9 dozen" later in the stem, which makes it feel like a two-part scenario. Once you see that every packing size leaves exactly 3 dozen behind, it collapses to a clean LCM problem. The structure is: if N ≡ r (mod a) and N ≡ r (mod b), find the smallest such N. Classic GMAT remainder pattern.

Takeaway: Any time every divisor leaves the same remainder, find LCM of the divisors and add the remainder — that's your answer.

---
Kavya | 725 GMAT Focus | Free gamified GMAT prep: edskore.com

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