47,322 bulbs are to be packed in several boxes. Each box should contain an…
2023
47,322 bulbs are to be packed in several boxes. Each box should contain an equal number of bulbs and no bulb should be unpacked. The number of boxes used can be
- A.
12
- B.
11
- C.
8
- D.
14
Attempted by 1 students.
Show answer & explanation
Correct answer: B
To share a total quantity into boxes that hold an equal whole number of items, with none left over, the number of boxes must be an exact divisor of the total — the total divided by the box count must leave remainder 0. So this question reduces to: which of the given counts exactly divide 47,322?
Prime factorization of 47,322:
47,322 ÷ 2 = 23,661 — one factor of 2; 23,661 is odd, so no more 2's remain.
Digit sum of 23,661 is 18 (divisible by 3): 23,661 ÷ 3 = 7,887.
Digit sum of 7,887 is 30 (divisible by 3): 7,887 ÷ 3 = 2,629.
2,629 = 11 × 239, and both 11 and 239 are prime.
So, 47,322 = 2 × 32 × 11 × 239.
Checking each option:
Option | What it needs | Does 47,322 have it? |
|---|---|---|
12 | 22 × 3 | Only one factor of 2 is available, so 22 is not present — fails. |
11 | 11 | 11 is exactly one of the prime factors — satisfied. |
8 | 23 | Only one factor of 2 is available, so 23 is not present — fails. |
14 | 2 × 7 | 7 does not appear among the prime factors — fails. |
Cross-check:
11 × 4,302 = 47,322, confirming that 11 boxes, each holding 4,302 bulbs, account for every bulb with none left over.
Answer:
The number of boxes used can be 11.