Question

No is divisible by 7 if n is divided by 2 3 4 5 the remainder is 1 what is the smallest value that n could b

Answer 1

Answer: $301$

Step-by-step explanation:

Since, The LCM of numbers given numbers $2$ ,$3$,  $4$ and $5$ is $60$.

Thus, the number that gives 1 as reminder and is the multiple of 7 is  $60 n + 1$

Where n is any positive integer,

Since, For $n = 5$,

The number is, $60\times 5 + 1 = 300 + 1 = 301$

Which is divisible by $7$.

Thus, the required number is 301.

Note : For n = 1, 2 3 and 4, numbers are 61, 121, 181 and 241

But they are not the multiple of 7.

Answer 2

Answer:

The required number is 301

Step-by-step explanation:

Let the number be N.

The number when divided by 2, 3, 4, 5 leaves the remainder 1 so, N-1 is divisible by 2, 3, 4, 5. So, possible cases of N-1 are : multiples of  l.c.m.(2, 3,4, 5) = multiples of 60

1. N-1 = 60 but 61 is not divisible by 7. So rejected
2. N-1 = 120 but 121 is not divisible by 7. So rejected
3. N-1 = 180 but 181 is not divisible by 7. So rejected
4. N-1 = 240 but 241 is not divisible by 7. So, rejected
5. N-1 = 300 and 301 is divisible by 7 So 301 is our required number.