What is the least multiple of 13 which divided by 3 4 5 8 and 10 leaves remainder 2 in each?

The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is : [A]840 [B]842 [C]2520 [D]2522

2522 LCM of 4, 5, 6, 7 and 8 = 840. Let require number be 840 K + 2 which is multiple of 13. Least value of K for which (840 K + 2) is divisible by 13 is K = 3 ∴ Require Number = 840 $latex \times$ 3 + 2 = 2520 + 2 = 2522.

Hence option [D] is correct answer.

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Given:

Multiple of 13 divided by 3, 4, 5, and 6 gives remainder 1, 2, 3, and 4.

Concept used:

LCM of number.

Calculation:

The difference between the divisor and the corresponding remainder is the same in each case

i.e 3 – 1 = 2, 4 – 2 = 2, 5 – 3 = 2, 6 – 4 = 2

⇒ LCM(3, 4, 5, 6)

⇒ LCM = 2 × 3 × 2 × 5

⇒ LCM = 60

Let the required number be 60k – 2 which is a multiple of 13.

The least value of ‘k’ for which (60k – 2) is divisible by 13 is k = 10.

⇒ required number = 60 × 10 – 2

⇒ 598

∴ 598 is the least multiple of 13 divisible by 3, 4, 5, and 6 gives remainder 1, 2, 3, and 4.

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