Divisibility Rule for #11#
If the last four digits of a number are divisible by #16#, the number is divisible by #16#. For example, in #79645856# as #5856# is divisible by #16#, #79645856# is divisible by #16#
Divisibility Rule for #16#
Although for any power of #2# such as #2^n#, the simple formula is to check last #n# digits and if the number formed by just last #n# digits is divisible by #2^n#, entire number is divisible by #2^n# and hence for divisibility by #16#, one should check last four digits. For example, in #4373408#, as last four digits #3408# are divisible by #16#, entire number is divisible by #16#.
If this is complicated, one can also try the rule - if the thousands digit is even, take the last three digits, but if the thousands digit is odd, add #8# to the last three digits. Now with this #3#-digit number, multiply hundreds digit by #4#, then add to the last two digits. If the result is divisible by #16#, the entire number is divisible by #16#.
Divisibility Rule for #17#
Divisibility rules for somewhat larger primes are not of much help and many times they get complicated. Nevertheless, rules have been designed and for #17# one is, subtract 5 times the last digit from the rest.
For example in the number #431443#, subtract #3xx5=15# from #43144# and we get #43129# and as it is divisible by #17#, number #431443# is also divisible by #17#.
One can also perform series of such action. In above example to check whether #43129# is divisible by #17# or not, subtract #9xx5=45# from #4312# and we get #4267# and to check for this, subtract #7xx5=35# from #426# and we get #391# and finally #1xx5=5# from #39# to get #34#, which is divisible #17# and
hence #431443#, #43129#, #4267# and #391# all are divisible by #17#
