What are are the tests of divisibility of various numbers?

1 Answer
Jun 12, 2018

Answer:

See below

Explanation:

Well, I'll list all the criteria I know of:

  • A number is divisible by #2# if it's even, i.e. if its last digit is #0,2,4,6# or #8#
  • A number is divisible by #3# if the sum of its digits is divisible by #3#. This operation can be recursive. For example, we want to check if #69873# is divisible by #3#. The sum of the digits is #33#. Again, #33# is divisible by #3# if the sum of its digits is, which is #3+3=6#. Thus, #69873# is divisible by #3#.
  • A number is divisible by #4# if its last two digits are (including numbers ending by #00#). For example, #345876345color(green)(44)# is divisible by #4# because #44# is, while #348576color(red)(33)# isn't because #33# isn't
  • A number is divisible by #5# if its last digit is #0# or #5#.
  • #6# doesn't really have its criterion. You can tell if a number is divisible by #6# if it divisible by #2# and #3# at the same time. This technique works for every number without a proper criterion, for example you can tell that a number is divisible by #14# if it's divisible by #2# and #7#.
  • #7# is a bit complicated: assuming your number has more than two digits, you must consider the number without the last digit, and subtract twice the last digit from the truncated number. You proceed until you get a two digit number. For example, start with #color(red)(9567)color(green)(6)#. This number is divisible by #7# if #color(red)(9567)-2*color(green)(6)# is, which is #color(red)(955)color(green)(5)#. We perform another step to get to #color(red)(955)-2*color(green)(5) = color(red)(94)color(green)(5)#. Last step: we have #(94)color(green)-2*(5) = color(red)(84)#, which has two digits and is indeed divisible by #7#. So, #95676# is divisible by #7# as well.
  • A number is divisible by #8# if its last three digits are (including numbers ending by #000#).
  • A number is divisible by #9# if the sum of its digits is divisible by #9#.
  • A number is divisible by #10# if it ends with #0#. Actually, a number is divisible by #10^n# if it ends with #n# zeroes.
  • #11# requires some calculation as well: you must sum all the even-positioned digits, and subtract (absolute values) all the odd-positioned digits. If it is divisible by #11# (or it is #0#), then so is the starting number. For example, start with #color(red)(6)color(green)(2)color(red)(5)color(green)(8)color(red)(3)color(green)(4)#. This leads to #-6+2-5+8-3+4=0#. So, #625834# is divisible by #11#
  • #13# works like #7#, except how have to sum four times the last digit to the rest of the number and check if the number is divisible by #13#. For example, start with #color(red)(730)color(green)(6)#. You have #color(red)(730)+4*color(green)(6) = color(red)(75)color(green)(4)#. Again, you have #color(red)(75)+4*color(green)(4)=91#, which is divisible by #13#.
  • #17# is similar again, but you need to subtract five times the last digit to the rest of the number and check if the number is divisible by #17# (absolute values again). For example, start with #color(red)(258)color(green)(4) \to color(red)(258)-2*color(green)(4)= \color(green)(23)color(green)(8)#. Finally, you have #\color(green)(23)-5*color(green)(8) = -17#. Thus, #2584# is divisible by #17#
  • A number is divisible by #25# if it ends with #00, 25, 50# or #75#.