Difficulty: Easy
Correct Answer: 945
Explanation:
Introduction:
This problem asks you to find the sum of all two digit numbers that are divisible by 5. Recognizing that these numbers form an arithmetic progression allows you to apply the standard formula for the sum of an arithmetic series instead of adding each number individually.
Given Data / Assumptions:
Concept / Approach:
Two digit numbers divisible by 5 must end in 0 or 5. The smallest such number is 10, and the largest such two digit number is 95. These numbers form an arithmetic progression with a common difference of 5. For an arithmetic progression with first term a, last term l, and n terms, the sum is S = n * (a + l) / 2.
Step-by-Step Solution:
First term a = 10. Common difference d = 5. Last term l = 95. Find the number of terms n using l = a + (n − 1)d. 95 = 10 + (n − 1) * 5. 95 − 10 = (n − 1) * 5 ⇒ 85 = 5(n − 1). Divide by 5: 17 = n − 1 ⇒ n = 18. Now compute the sum: S = n * (a + l) / 2. S = 18 * (10 + 95) / 2 = 18 * 105 / 2. 105 / 2 = 52.5, so S = 18 * 52.5. 18 × 50 = 900 and 18 × 2.5 = 45, so total S = 900 + 45 = 945. Therefore, the required sum is 945.
Verification / Alternative check:
You can quickly verify by grouping terms from both ends: (10 + 95) + (15 + 90) + (20 + 85) + ... . Each pair sums to 105. With 18 terms, there are 9 such pairs. So S = 9 * 105 = 945, which confirms our earlier result.
Why Other Options Are Wrong:
The values 678, 439, and 568 do not match the sum produced by the arithmetic progression formula and typically arise from miscounting the terms or omitting numbers ending in 0 or 5. Only 945 is consistent with all two digit multiples of 5 from 10 through 95 inclusive.
Common Pitfalls:
Common mistakes include forgetting that 95 is the largest two digit multiple of 5, using 100 incorrectly, or miscalculating n by not using the formula for the n-th term correctly. Careful use of the A.P. formulas avoids these errors.
Final Answer:
The sum of all two digit numbers divisible by 5 is 945.
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