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

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Answer 1 · 2017-05-02T06:19:50

#419#

Call the number #n#.

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 #n# is one less than a multiple of the LCM of #3#, #5# and #7#, namely:

#3*5*7 = 105#

So:

#n = 105k-1" "# for some integer #k#.

In order to give a remainder of #1# when divided by #11#, #n# must be of the form #11m+1# for some integer #m#.

So we have:

#11m+1 = 105k-1#

That is:

#105k = 11m + 2#

Now:

#105/11 = 9" "# with remainder #6#

So what are the multiples of #6# modulo #11#?

#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 #k# which gives us the required remainder is #k=4#

So:

#n = 105k-1 = 105*4-1 = 419#