Re: A magician marks-off on a stick of length 1 yard in thirds and fifths
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10 Jun 2025, 04:00
To solve this problem, we need to identify all the unique points where the stick is marked and then calculate the lengths of the segments created by these marks.
Let the total length of the stick be 1 unit (yard).
The stick is marked in:
- Thirds: This creates marks at $\(1 / 3\)$ and $\(2 / 3\)$.
- Fifths: This creates marks at $\(1 / 5,2 / 5,3 / 5\)$, and $\(4 / 5\)$.
We should also consider the start and end points of the stick: 0 and 1 .
Now, let's list all the distinct points on the stick, including 0 and 1 , and convert them to a common denominator to easily compare them. The least common multiple (LCM) of 3 and 5 is 15 .
- $0=0 / 15$
- $1 / 5=3 / 15$
- $1 / 3=5 / 15$
- $2 / 5=6 / 15$
- $3 / 5=9 / 15$
- $2 / 3=10 / 15$
- $4 / 5=12 / 15$
- $1=15 / 15$
Let's order these unique points from smallest to largest: $0 / 15,3 / 15,5 / 15,6 / 15,9 / 15,10 / 15,12 / 15,15 / 15$
Now, we calculate the length of each piece by finding the difference between consecutive marked points:
1. Piece 1: $3 / 15-0 / 15=3 / 15$
2. Piece 2: $5 / 15-3 / 15=2 / 15$
3. Piece 3: $6 / 15-5 / 15=1 / 15$
4. Piece 4: $9 / 15-6 / 15=3 / 15$
5. Piece 5: $10 / 15-9 / 15=1 / 15$
6. Piece 6: $12 / 15-10 / 15=2 / 15$
7. Piece 7 : $15 / 15-12 / 15=3 / 15$
Finally, let's count how many pieces of each length we have:
- Pieces of length $3 / 15$ : There are 3 such pieces.
- Pieces of length $2 / 15$ : There are 2 such pieces.
- Pieces of length $1 / 15$ : There are 2 such pieces.
The question asks for the maximum number of pieces which are equal in length. From our count, the maximum number is 3 (for pieces of length 3/15).
The final answer is 3 .