Have you ever wanted to know some quick tricks to know if a number is
divisible by another number? I've always loved shorcuts of this sort, and
on a mailing list with some friends, we got to talking about divisibility
and I wrote up most of this information at that time.
Background
When describing the resoning behind these shortcuts, I will ask you to consider
numbers of up to four digits, such as 6745. However, I will not use a
specific number, instead I will write the number as
1000*a + 100*b + 10*c + d.
A specific number like 6745 is obtained when a=6, b=7, c=4, and d=5.
Okay, onto the shortcuts.
Divisibility by 2
A number is evenly divisible by 2 if its ones digit is evenly divisible by 2.
Why?
Consider a two digit number (10*a+b). Factor this to 2*5*a + b. This shows
that all numbers ending a zero are divisible by two, so if the ones digit
is divisible by two, the entire number is as well.
Divisibility by 3
A number is evenly divisible by 3 if the sum of all its digits is evenly
divisible by 3.
Why?
Consider a 2 digit number 10*a + b = 9*a + (a+b).
We know that 9*a is divisible by 3, so 10*a + b will be divisible by 3
if and only if a+b is.
Similarly, 100*a + 10*b + c = 99*a + 9*b + (a + b + c), and 99*a + 9*b
is divisible by 3, so the total will be divisible by 3 if and only if
a + b + c is.
This explanation also works to prove the divisibility by 9 test.
Divisibility by 4
A number is evenly divisible by 4 if the number formed by the last
two individual digits is evenly divisible by 4.
Why?
Consider a three digit number (100*a + 10*b + c)
Factor the first digit: 4*25*a + 10*b + c
Since we know 4*25*a is divisible by 4, we only need to consider 10*b + c,
the last two digits, for divisibility by 4.
Divisibility by 5
A number is evenly divisible by 5 if its ones digit is a 5 or a 0.
Why?
Consider the two digit number (10*a + b)
Factor the first digit: 5*2*a + b
Since 5*2*a is evenly divisible by 5, then only b needs to be evenly divisible
by 5, which means b must be 5 or 0.
Divisibility by 6
A number is evenly divisible by 6 if it is evenly divisible by both 2 and 3.
Why?
Since 2 and 3 are the prime factors of 6, any number with these two prime
factors is also divisible by 6.
Divisibility by 7
Is a number evenly divisible by 7? Take the ones digit off of n, double it
and subtract the doubled number from the remaining number. If the
result is evenly divisible by 7 (e.g. 14, 7, 0, -7, etc.), then
the number is divisible by seven. This process may need to be repeated
several times.
Example: 3101. Take off ones: 310. subtract from this ones doubled:
310-2: 308. take off ones: 30. Subtract the ones double, which is 16: 14.
14 is divisible by 7, so 3101 is too.
Why?
When I get a little time, I'll write this up.
Divisibility by 8
A number is evenly divisible by 8 if the number formed by the last
three digits is evenly divisible by 8.
Example: The last three digits of the number 3624 are
624, which is evenly divisible by 8 so 3624 is evenly divisible by 8.
Still a little tough for this one.
Why?
Consider the four digit number (1000*a + 100*b + 10*c + d)
Factor the first digit: 8*125*a + 100*b + 10*c + d
Since we know 8*125*a is divisible by 8, we only need to consider
100*b + 10*c + d, the last three digits, for divisibility by 8.
Divisibility by 9
A number is evenly divisible by 9 if the sum of the individual digits is evenly
divisible by 9.
Why? See the explanation for divisibility by 3.
Divisibility by 10
A number is evenly divisible by 10 if its ones digit is a 0.
Why?
Consider the two digit number (10*a + b)
Since we know 10*a is evenly divisible by 10, and no non-zero single digit b
can be evenly divisible by 10, only if b is zero (causing it to "drop out" of
the equation) can 10*a + b be evenly divisible by 10.
Divisibility by 11
A number is evenly divisible by 11 the alternating sum of its digits are
divisible by 11 (including zero).
Example: 82,762,262 is divisible by 11 because:
8-2+7-6+2-2+6-2 = 11, which is divisible by 11.
The rule can also be applied this way: add the total of the odd digits (1st, 3rd, 5th, ...) and subtract the total of the even digits (2nd, 4th, ...)
(8+7+2+6) - (2+6+2+2) = 23 - 12 = 11
Why?
When I get time, I'll write this one up.
Divisibility by 12
A number is evenly divisible by 12 if it is evenly divisible by both 3 and 4.
Why?
Since 3 and 4 are the prime factors of 12 any number with these two prime
factors is also divisible by 12.
Divisibility by 13
A number is evenly divisible by 13 if, when you
delete the last digit from the given number, then subtract nine times the
deleted digit from the remaining number, what is left is divisible by 13,
then so is the original number. This process may need to be repeated
several times.
Example: 1313 is divisible by 13 because:
131-27 = 104, 10-36 = -26. -26 is divisible by 13.
Why?
When I get time, I'll write this one up.
Divisibility by 14
A number is evenly divisible by 14 if it is evenly divisible by 2 and 7. Why?
2 and 7 are the prime factors of 14.
Divisibility by 15
A number is evenly divisible by 15 if it is evenly divisible by 3 and 5. Why?
3 and 5 are the prime factors of 15.
Divisibility by n
From here you can easily derive your own rules using prime factors,
extensions of powers of two (ie, a number is evenly divisible by 16 if
its last four digits are evenly divisible by 16) or a modulus
method (ie, for divisibility by 7 and 13) which I have yet to write up.
(Of interest: any number whose digits add up to a power of 3 in base 10
are divisible by that power of three...)
Divisibility rules in bases other than base 10
There is a cool pair of general proofs using modular arithmetic that say:
1) in base b, a number is divisible by b-1 if and only if its digits added
together in base b are divisible by b-1.
2) in base b, a number is divisible by b+1 if and only if the alternating sum
of its digits in base b is divisible by b+1.
The proof basically extends the proof of divisibility by 9 and 11 into any base you care to use.
Examples:
511 in base 8 is evenly divisible by 7 in base 8 because 5+1+1 = 7 in base 8.
515 in base 8 is evenly divisible by 11 in base 8 because 5-1+5 = 11 in base 8. (11 in base 8 is written 9 in base 10.)
A12711 in base 12 is evenly divisible by B in base 12 because A+1+2+7+1+1 = 1A in base 12.(B is 11 decimal, 1A is 22 decimal. (not really helpful for you? Sorry!))
