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Remainder of 1 but divisible by 7


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It is indeed straightforward, so let me start with..

 

 

Take the LCM of 2, 3, 4, 5 and 6 = 4 x 3 x 5 = 60.

 

We want to find a multiple of 60 that will give "-1 mod 7" or equivalently, "6 mod 7". 

 

It is easy to see that,

 

60 = 4 mod 7.

60 x 5 = (4x5) mod 7 = 6 mod 7.

 

So, (60 x 5) + 1 = 301 is the number we are looking for.

 

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