The least multiple of 13 which when divided by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is:

3362

Option verification

But first, subtract 2 from each option and check their divisibility.

(a) 2518   x not divisible by 5, so cannot be answer

(b) 838    x  not divisible by 5, so cannot be answer

(c) 840    x  Divisible by all but not multiple of 13.

(d) 2520  √ divisible by all and multiple as well.

Alternate:

LCM of 4, 5. 6, 7, 8 is = 840

So, ATQ,

Required number = 840K + 2, which is divisible by 13.

Now, let find the value of K:

⇒ \(\frac{840K\;+\;2}{13}\)

⇒ \(\frac{832K\;+8K\;+\;2}{13}\)

⇒ \(\frac{832K}{13}\) + \(\frac{8K\;+\;2}{13}\)

⇒ \(\frac{8K\;+\;2}{13}\), K = 3

Therefore,

Required number = (840× 3) + 2 = 2522