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# Is X divisible by Y ?

MikeFromHC
Even today with calculators and computers there are times when it is helpful to see if one number will divide into another.

There are simple rules for 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12 and any number which is a form of 2^n.

Seven is the exception and while there are some rules it is easier to just divide the thing.

I'll leave one to you

Two
Two itself is trivial but follows the rule. If the last n digits are divisible by 2^n, the number is divisible by 2^n2 is 2^1, 4 is 2^2, 8 is 2^3...
All are forms of 2^n

If the last digit is even, the number can be divided by 2
If the last two digits are divisible by 4 (2^2) then the number can be divided by 4
If the last three digits are divisible by 8 (2^3) then the number can be divided by 8
etc.

Three
Sum the digits until you have a single digit or see the answer.
If the sum is divisible by three the number is also. (See below for a bit more)

12345 = 1+2+3+4+5= 15 = 1+5 = 6
The number can be divided by 3.

Four
A form of 2^n If the last two digits are divisible then the number is.

Five
If it ends in five or zero it is divisible

Six
If it is even and divisible by three it is divisible by six

Seven
Just do the math.

Eight
A form of 2^n If the last three digits are divisible then the number is.

Nine
Same as three, sum the digits and if the result is 9 it is divisible.(See below for a bit more)

Ten
If it ends in zero it is divisible.
Note that this also follows the 10^n rule.

Eleven
First digit minus second digit plus third digit minus...
If the answer is eleven or zero the number is devisable by eleven.
65*11 equals 715
7-1+5 = 11

99
9-9 = 0

Twelve
If divisible by three and four it can be divided by twelve.

Below The rule for nine and three is known as "casting out" and especially for nine has many other uses.
Summing the digits is not always needed as you can "cast out" sets of numbers.
For three
12345 = 1+2+3+4+5
Note that 1+2 = 3 so you can just omit them, the 3 itself, and just look at 45.
For nine
1234512 = 1+2+3+4+5 + 1 + 2
4+5 = 9 so I am left with 12312 which sums to 9.

NOTE If the sum of the digits is not equal to 9 then that number is the remainder.

Final comment. If anyone is interested in how useful casting out nines is, let me know.
hw3patch
Interesting, I never knew about that method of checking for divisibility by 11 or about the 2^n rule.

Concerning the 2^n rule, by the time n gets above 3 or 4, it becomes pretty cumbersome to apply. For divisibility by 16, checking if a three or four digit number is divisible by 16 requires pencil and paper (for me, at least, unless I get a number that's convenient to use, such as 688*). Same for 32 and other large numbers.

* 64 = 16 * 4, so 16 *40 = 640. Now I just need to check if 48 is divisible by 16. Now that I think about it, three digit numbers are pretty manageable. Between 100 and 1000, there are frequent checkpoints which make the process easier- by subtracting a multiple of 16*10 (or 16*5 to make it even easier).
160
320
480
640
800
960
So a number that at first seems daunting, like 588, only requires you to find out if 52 is divisible by 16 (640-588=52). Once a four digit number needs to be checked for divisibility by 16, overhead becomes immense. Say we have a random number, say 6980. At a glance, you can tell that it's divisible by two and four, so it would not be a waste to continue to pursue this. For every 1000 the number contains, we can subtract those thousands off, but each time we subtract off 1000, we need to add 40, since 1000 is not a multiple of 16, but 960 is. In other words, we're just subtracting 960; it's easier to subtract 1000 and add 40 in your head (performing the operation in two steps) than it is to subtract 960 (in one step), though. So we now need to check if 6980 - 7000 + 40*7 = -20 + 280 = 260 is divisible by 16, which it isn't.

Once n reaches 5 (divisibility by 32), the overt overhead will start to overwhelm most people. The 2^n rule is still really interesting though. 10.000 (10^4) is divisible by 16, 100.000 (10^5) is divisible by 32, 1.000.000 (10^6) is divisible by 64. That is, 10^r is divisible by 2^r. The reason for that is:
2^r * 5^r = (2*5)^r
ocalhoun
 Quote: Seven Just do the math.

You fail to deliver on your promise there a little.
likeabreeze
 hw3patch wrote: Interesting, I never knew about that method of checking for divisibility by 11

assume the last three digit of a number is a,b,c.
a b c
* 1 1
--------------------------------------
a b c
a b c
--------------------------------------
---------- (b+c) c if b+c<10;
-----(a+b+1) (b+c-10) c if b+c>=10;
----------------------------------------
so , you can see why
 Quote: First digit minus second digit plus third digit minus... If the answer is eleven or zero the number is devisable by eleven.
Kenji_Kensuki
Wow! I only knew the ones for 2,4,5, and 10. They REALLY need to teach this in the more basic math classes. With any luck, maybe they will.

Of course, in computers, you can just do X%Y.
asnani04
all these rules are quite useful in daily mathematics, and since they are taught in class 4, they become a habit by the time one reaches class 7 or 8. The rule of co-primes is very useful too.

By that rule, a number is divisible by a number Y, if it is divisible by two of Y's co-prime factors. For example, a number is divisible by 18 if 2 and 9 divide that number. A number is divisible by 15 if 3 and 5 divide it. This result is very useful in certain proofs of Number Theory.