Question

A chart consists of numbers from 1 to 360. What is the sum of all the numbers remained on the chart after removing all the multiples of 2, multiples of 3 and multiples of 5?

• A. 9876
• B. 17280
• C. 34560
• D. 51840
• E. 69120

Answer 1

The Correct Option is B

Step 1: Understand the problem.
We are given a chart with numbers from 1 to 360. We need to find the sum of all the numbers that remain on the chart after removing all the multiples of 2, multiples of 3, and multiples of 5.

Step 2: Find the total sum of all numbers from 1 to 360.
The sum of all numbers from 1 to 360 is given by the formula for the sum of an arithmetic series: \[ \text{Sum} = \frac{n(n + 1)}{2} \] where \( n \) is the last number in the series (360). Substituting \( n = 360 \): \[ \text{Sum} = \frac{360 \times (360 + 1)}{2} = \frac{360 \times 361}{2} = 64980 \] So, the total sum of numbers from 1 to 360 is 64980.

Step 3: Find the sum of the multiples of 2, 3, and 5.
We will use the inclusion-exclusion principle to find the sum of the numbers that are multiples of 2, 3, or 5.
- Multiples of 2: The multiples of 2 from 1 to 360 are \( 2, 4, 6, \dots, 360 \). This forms an arithmetic sequence with the first term 2, the common difference 2, and the last term 360. The number of terms in this sequence is: \[ n_2 = \frac{360}{2} = 180 \] The sum of the multiples of 2 is: \[ \text{Sum of multiples of 2} = \frac{n_2}{2} \times (\text{first term} + \text{last term}) = \frac{180}{2} \times (2 + 360) = 90 \times 362 = 32580 \] - Multiples of 3: The multiples of 3 from 1 to 360 are \( 3, 6, 9, \dots, 360 \). The number of terms is: \[ n_3 = \frac{360}{3} = 120 \] The sum of the multiples of 3 is: \[ \text{Sum of multiples of 3} = \frac{n_3}{2} \times (\text{first term} + \text{last term}) = \frac{120}{2} \times (3 + 360) = 60 \times 363 = 21780 \] - Multiples of 5: The multiples of 5 from 1 to 360 are \( 5, 10, 15, \dots, 360 \). The number of terms is: \[ n_5 = \frac{360}{5} = 72 \] The sum of the multiples of 5 is: \[ \text{Sum of multiples of 5} = \frac{n_5}{2} \times (\text{first term} + \text{last term}) = \frac{72}{2} \times (5 + 360) = 36 \times 365 = 13140 \] - Multiples of 2 and 3 (LCM = 6): The multiples of 6 from 1 to 360 are \( 6, 12, 18, \dots, 360 \). The number of terms is: \[ n_6 = \frac{360}{6} = 60 \] The sum of the multiples of 6 is: \[ \text{Sum of multiples of 6} = \frac{n_6}{2} \times (\text{first term} + \text{last term}) = \frac{60}{2} \times (6 + 360) = 30 \times 366 = 10980 \] - Multiples of 2 and 5 (LCM = 10): The multiples of 10 from 1 to 360 are \( 10, 20, 30, \dots, 360 \). The number of terms is: \[ n_{10} = \frac{360}{10} = 36 \] The sum of the multiples of 10 is: \[ \text{Sum of multiples of 10} = \frac{n_{10}}{2} \times (\text{first term} + \text{last term}) = \frac{36}{2} \times (10 + 360) = 18 \times 370 = 6660 \] - Multiples of 3 and 5 (LCM = 15): The multiples of 15 from 1 to 360 are \( 15, 30, 45, \dots, 360 \). The number of terms is: \[ n_{15} = \frac{360}{15} = 24 \] The sum of the multiples of 15 is: \[ \text{Sum of multiples of 15} = \frac{n_{15}}{2} \times (\text{first term} + \text{last term}) = \frac{24}{2} \times (15 + 360) = 12 \times 375 = 4500 \] - Multiples of 2, 3, and 5 (LCM = 30): The multiples of 30 from 1 to 360 are \( 30, 60, 90, \dots, 360 \). The number of terms is: \[ n_{30} = \frac{360}{30} = 12 \] The sum of the multiples of 30 is: \[ \text{Sum of multiples of 30} = \frac{n_{30}}{2} \times (\text{first term} + \text{last term}) = \frac{12}{2} \times (30 + 360) = 6 \times 390 = 2340 \] Step 4: Apply the inclusion-exclusion principle.
The sum of the multiples of 2, 3, and 5 is: \[ \text{Sum of multiples of 2, 3, or 5} = 32580 + 21780 + 13140 - 10980 - 6660 - 4500 + 2340 = 49000 \] Step 5: Find the sum of the remaining numbers.
The total sum of numbers from 1 to 360 is 64980, and the sum of the multiples of 2, 3, or 5 is 49000. The sum of the numbers that remain is: \[ 64980 - 49000 = 17280 \] Step 6: Conclusion.
The sum of the numbers that remain after removing the multiples of 2, 3, and 5 is 17280.

Final Answer:
The correct answer is (B): 17280.