Free
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.
India’s #1 Learning Platform
Start Complete Exam Preparation
Daily Live MasterClasses
Practice Question Bank
Mock Tests & Quizzes
Get Started for Free Download App
Trusted by 3.3 Crore+ Students
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.