Find the smallest number which when divided by 4 5 and 6 leaves remainder 3 4 5 respectively

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10 Questions 10 Marks 15 Mins

Calculation:

The LCM of the numbers 3, 4, 5, 6 and 7 is 3 × 5 × 2 × 2 × 7 = 420

The first number to satisfy the given conditions is LCM - 1 

⇒ 420 - 1

⇒ 419

⇒ 419 ÷ 3 gives remainders of 2.

Now, the number will follow the arithmetic progression having the common difference = 420

The second number = 420 + 419 = 839

⇒ 839 ÷ 4 gives remainders of 3.

The third number = 420 + 839 = 1259

⇒ 1259 ÷ 5 gives remainders of 4.

The fourth number = 420 + 1259 = 1679

⇒ 1679 ÷ 6 gives remainders of 5.

The fifth number = 420 + 1679 = 2099

⇒ 2099 ÷ 7 gives remainders of 6.

Therefore, the smallest of all such numbers fulfilling the above condition is 419.

Hence, the correct answer is 419.

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The least number which when divided by 4, 5 and 6 leaves remainder 1, 2 and 3 respectively, is : [A]57 [B]59 [C]61 [D]63

57 Here 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3 ∴ The required Number = LCM of (4, 5, 6) – 3 = 60 – 3 = 57.

Hence option [A] is correct answer.