If the numbers which are divisible by 3 from 1 to 61 are arranged in descending order, then which numbers will be at `5^(th)` place and at `15^(th)` place from above?

Options

A. 45, 18
B. 48, 18
C. 48, 21
D. 45, 21

Answer

To solve the problem of finding the 5th and 15th numbers divisible by 3 from 1 to 61 when arranged in descending order, follow these steps:

### Step 1: Identify the multiples of 3 from 1 to 61
We need to find all the numbers between 1 and 61 that are divisible by 3. The multiples of 3 can be found by multiplying 3 by integers starting from 1 until the product exceeds 61.

- The multiples of 3 are:
- \(3 \times 1 = 3\)
- \(3 \times 2 = 6\)
- \(3 \times 3 = 9\)
- \(3 \times 4 = 12\)
- \(3 \times 5 = 15\)
- \(3 \times 6 = 18\)
- \(3 \times 7 = 21\)
- \(3 \times 8 = 24\)
- \(3 \times 9 = 27\)
- \(3 \times 10 = 30\)
- \(3 \times 11 = 33\)
- \(3 \times 12 = 36\)
- \(3 \times 13 = 39\)
- \(3 \times 14 = 42\)
- \(3 \times 15 = 45\)
- \(3 \times 16 = 48\)
- \(3 \times 17 = 51\)
- \(3 \times 18 = 54\)
- \(3 \times 19 = 57\)
- \(3 \times 20 = 60\)

Thus, the multiples of 3 from 1 to 61 are:
\[ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60 \]

### Step 2: Arrange the multiples in descending order
Now, we will arrange these numbers in descending order:
\[ 60, 57, 54, 51, 48, 45, 42, 39, 36, 33, 30, 27, 24, 21, 18, 15, 12, 9, 6, 3 \]

### Step 3: Find the 5th and 15th numbers
Now we will identify the 5th and 15th numbers in this descending list.

- **5th place**:
- Counting from the top:
1. 60 (1st)
2. 57 (2nd)
3. 54 (3rd)
4. 51 (4th)
5. 48 (5th)

Thus, the number at the 5th place is **48**.

- **15th place**:
- Continuing the count:
1. 60 (1st)
2. 57 (2nd)
3. 54 (3rd)
4. 51 (4th)
5. 48 (5th)
6. 45 (6th)
7. 42 (7th)
8. 39 (8th)
9. 36 (9th)
10. 33 (10th)
11. 30 (11th)
12. 27 (12th)
13. 24 (13th)
14. 21 (14th)
15. 18 (15th)

Thus, the number at the 15th place is **18**.

### Final Answer
The numbers at the 5th and 15th places are **48** and **18**, respectively.

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