Question

Q1. FIND THE LEAST NUMBER WHICH ON BEING DIVIDED BY 3,4,5,7 AND 8 LEAVES IN EACH CASE A REMAINDER OF 2, BUT WHEN DIVIDED BY 11 LEAVES NO REMAINDER ? TCS NINJA MAY,2021​

Answer 1

Answer:

2 in the correct answer

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given FIND THE LEAST NUMBER WHICH ON BEING DIVIDED BY 3,4,5,7 AND 8 LEAVES IN EACH CASE A REMAINDER OF 2, BUT WHEN DIVIDED BY 11 LEAVES NO REMAINDER

• Now we need to find the lcm of 3,4,5,7,8
• We get the factors of these numbers. So the factors are
• 2 x 2 x 2 x 3 x 5 x 7 = 840
• Therefore lcm is 840
• According to the question
• The number leaves a remainder 2 and no remainder when the numbers 3,4,5,7,8 divisible by 11
• So we can write this as 840 m + 2 where 840 m + 2 is a multiple of 11
• Now we have
• 840(1) + 2 = 842 is not divisible by 11
• 840(2) + 2 = 1682 is not divisible by 11
• 840(3) + 2 = 2522 is not divisible by 11
• 840(4) + 2 = 3362 is not divisible by 11
• 840(5) + 2 = 4202 is divisible by 11
• Therefore the least number is 4202 when divided by 3,4,5,7,8 leaves a remainder 2 .

Reference link will be

https://brainly.in/question/520909