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)
2 | 3, 4, 5, 6 |
3 | 3, 2, 5, 3 |
1, 2, 5, 1 |
⇒ 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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