Find the smallest integer that produces remainder of 2,4,6&1 when it is divided by 3,5,7&11 respectively?
1 Answer
Explanation:
Call the number

In order to give a remainder of
#2# when divided by#3# ,#n# must be one less than a multiple of#3# . 
In order to give a remainder of
#4# when divided by#5# ,#n# must be one less than a multiple of#5# . 
In order to give a remainder of
#6# when divided by#7# ,#n# must be one less than a multiple of#7# .
So
#3*5*7 = 105#
So:
#n = 105k1" "# for some integer#k# .
In order to give a remainder of
So we have:
#11m+1 = 105k1#
That is:
#105k = 11m + 2#
Now:
#105/11 = 9" "# with remainder#6#
So what are the multiples of
#1*6 = 6 = 0*11+6#
#2*6 = 12 = 1*11 + 1#
#3*6 = 18 = 1*11 + 7#
#4*6 = 24 = 2*11 + 2#
So the smallest possible positive value of
So:
#n = 105k1 = 105*41 = 419#