Question

Find the least number which being divided by 2,3,4,
5, 6 leaves in each case a remainder 1, but when
divided by 7 leaves no remainder.​

Answer 1

Answer:

Therefore, The required number is 301.

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Step-by-step explanation:

we have to find a number that is divisible by 2, 3, 4, 5, and 6 (meaning that it is a multiple of those numbers) and that number + 1 being divisible by 7.

First we find the Lowest Common Multiple of these numbers,

LCM =  2 × 2 × 3 × 5 = 60

Now, we check if 60 + 1 is divisible by 7 or not.

clearly 61 is not divisible by 7.

Now we find a multiple of 60 such that when we add 1 to it, the number becomes divisible by 7.  

After some trial and error, we will found that the lowest number that works is 300, which is  5 × 60 . When we take 300  +  1 = 301 , 301 ÷ 7 = 43.

Therefore, The required number is 301.

Answer 2

Answer:

YOUR ANSWER IS 301.

Step-by-step explanation:

The least common multiple of 2, 3, 4, 5, 6 is their LCM=60.

The least common multiple of 2, 3, 4, 5, 6 is their LCM=60.So, the answer is the least number such that it is a multiple of 7 and a multiple of 60 leaving remainder 1.

The least common multiple of 2, 3, 4, 5, 6 is their LCM=60.So, the answer is the least number such that it is a multiple of 7 and a multiple of 60 leaving remainder 1.∴The required answer is 301.

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