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What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?

A. 897 B. 1,64,850 C. 1,64,749 D. 1,49,700 Answer: Option B
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Solution(By Apex Team)

The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101. The next number that will leave a remainder of 2 when divided by 3 is 104, 107, …. The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998. So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3. Sum of an AP = $\left(\Large\frac{\text { First Term }+\text { Last Term }}{2}\right) \times \text { Number of Terms }$ We know that in an A.P., the nth term $a_{n}=a_{1}+(n-1) \times d$ In this case, therefore, 998 = 101 + (n – 1) × 3 i.e., 897 = (n – 1) × 3 Therefore, n – 1 = 299 Or n = 300 Sum of the AP will therefore, be $\begin{array}{l}=\left(\Large\frac{101+998}{2}\right)\times300\\ =1,64,850\end{array}$

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