Question

If x is an integer less than 1000 that has a remainder of 1 when it is divided by 2, 3, 4, 5, 6, or 7, what is one possible value of x?

Answer is 421 or 841 but I want to know how to do it.
Thanks

Answer 1

See below:

Explanation:

One way we can do this is to see that we can express:

`2a+1=x`
`3b+1=x`
`4c+1=x`
`5d+1=x`
`6e+1=x`
`7f+1=x`

We can approach this one way by finding values of `a -> f` by doing a prime factorization and then having the lowest common multiple of `2 -> 7` be to the left of the `+1`:

`2=2`
`3=3`
`4=2xx2`
`5=5`
`6=2xx3`
`7=7`

Which gives:

`2xx2xx3xx5xx7=420`

`2(2xx3xx5xx7)+1=421`
`3(2xx2xx5xx7)+1=421`
`4(3xx5xx7)+1=421`
`5(2xx2xx3xx7)+1=421`
`6(2xx5xx7)+1=421`
`7(2xx2xx3xx5)+1=421`

We can also multiply the 420 by 2 to get 840. We add 1 to it and end up with 841.