Rule for Divisibility by 8 A number with at least 3 digits is divisible by 8 if its last three digits form a number divisible by 8. Examples A.) 7,120 is divisible by 8 because its last three digits, 120, form a number divisible by eight. B.) 9,389 is not divisible by 8 because its last three digits, 389, form a number not divisible by eight. Proof For any integer x written as a_{n}a_{n-1}a_{n-2}...a_{2}a_{1}a_{0}, we will show that x is divisible by 8 if a_{2}a_{1}a_{0 }is divisible by 8. If we write x as a_{n}a_{n-1}a_{n-2}...a_{2}a_{1}a_{0}, then we can also write: x=
a_{0} +
a_{1}(10)
+ a_{2}(10^{2} )+
a_{3}(10^{3})...
+ a_{n-2}(10^{n-2})^{
}+_{
}a_{n-1}(10^{n-1}_{
})_{
}+
a_{n}(10^{n}) =
(a_{n}×10^{n}_{
}+_{
}a_{n-1}×10^{n-1}_{
}+_{
}a_{n-2}×10^{n-2
}+
.... + a_{3})×1000_{
}+
a_{2}a_{1}a_{0} = 8×125× (a_{n}×10^{n}_{ }+_{ }a_{n-1}×10^{n-1}_{ }+_{ }a_{n-2}×10^{n-2 }+ .... + a_{3}) + a_{2}a_{1}a_{0} |
Divisibility Rules >