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