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

1 Answer
MathFact-orials.blogspot.com
Dec 26, 2017

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.

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