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What is the sum of all three-digit numbers which leave a remainder $2$ , when divided by $6$ ?

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82656

82658

82650

82654

answer is C.

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Detailed Solution

The given three-digit numbers which leave remainder 2 on dividing by 6 are

104, 110, 116, \dots . 998

The numbers from an AP with first term 104 and common difference 6. So, we get
First term,

a = 104

,
Common difference,

d = 6

.
Let there be n terms.
So, we get

a_{n} = 998

.
We know that

n^{th}

term of an AP is given as,

a_{n} = a + (n - 1) d

\Rightarrow a + (n - 1) d = 998

\Rightarrow 104 + (n - 1) 6 = 998

\Rightarrow n = 150

We know that sum of first n terms of an AP is given as,

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

So, the sum of 150 terms will be,

S_{150} = \frac{150}{2} [104 + 998]

\Rightarrow S_{150} = 75 \times 1102

\Rightarrow S_{150} = 82650

Therefore, the sum of required numbers is 82650.
Hence, the correct option is 3.

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