If the 5 digit number #1364"?"# is divisible by #3# then what are the possible values of the last digit?

1 Answer
Jun 11, 2016

#1#, #4# or #7#.

Explanation:

In the decimal number system that we use, an integer is divisible by #3# if and only if the sum of its digits is also divisible by #3#.

#1+3+6+4+color(blue)(1) = 15# is divisible by #3#.

So #1364color(blue)(1)# is divisible by #3#, and so will be:

#1364color(blue)(1)+3 = 1364color(blue)(4)#

and

#1364color(blue)(4)+3 = 1364color(blue)(7)#

#color(white)()#
Footnote

Why does this method of checking the digits add up to a multiple of #3# work?

Essentially because when you divide #10# by #3# then the remainder is #1#.

So for example:

#color(red)(153) = (100*color(red)(1)) + (10*color(red)(5)) + (1*color(red)(3))#

#= (99+1)*color(red)(1) + (9+1)*color(red)(5) + (0+1)*color(red)(3)#
#= (99*color(red)(1) + 9*color(red)(5) + 0*color(red)(3)) + (1*color(red)(1) + 1*color(red)(5) + 1*color(red)(3))#
#= 3(33*color(red)(1) + 3*color(red)(5) + 0*color(red)(3)) + (color(red)(1)+color(red)(5)+color(red)(3))#

The expression #3(33*color(red)(1)+3*color(red)(5)+0*color(red)(3))# is divisible by #3#.

So we find that #153# is divisible by #3# if and only if #(1+5+3)# is.