Question

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

Answer 1

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take lcm of 2,3,4,5,6
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lcm=60
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let add 1
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=61 it is not. divided by 7
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so we try 2×60,3×60,4×60,5×60............
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we get 301 fully divided by 7
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301 is your answer
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hope it help you

Answer 2

Answer:

The required number is 301.

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.