These numbers form an arithmetic progression (AP). The first 3-digit number leaving a remainder of 2 when divided by 3 is 101 (101 = 3 33 + 2). The last such number is 998 (998 = 3 332 + 2). The common difference is 3. The number of terms is N = ((998 - 101)/3) + 1 = (897/3) + 1 = 299 + 1 = 300. The sum of an AP is (N/2) (First term + Last term) = (300/2) (101 + 998) = 150 1099 = 164850.
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?
Asked in GPSC Prelims CSP 1 2024 · Topic: National Income & Gross Domestic Product (Methods of National Income Estimation: Income, Expenditure, and Production/Value Added Methods)
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