Divisibility by three and remainders

I'm sure that anyone can recognise that 3 divides into 9 exactly. So, if we divide the digital root by 3 we will get the remainder. If that remainder is zero then the number is divisible by 3, otherwise we will get 1 or 2. Let's try the number 1416. 1+4+1+6=12. 1+2=3. 3/3=1 with no remainder! Again, there's a simplier way of working this out. Obtaining what I call the 3's digital root. In this case you cross out any digit that is 3, 6 or 9. And any combination of digits that sum to 3, 6, 9, 12 etc. Let's try the number 3485452612. Cross off 3, 6. Also 4+8, 5+4, 1+2. You are left with 5+2. 7/3 = 2 with 1 remainder. Simple. Or, if you prefer, cross off 3, 6 or 9. Replace both 4 and 7 with 1's and replace both 5 and 8 with 2's (the remainders if divided by 3). Then cross off any pairs of 1 and 2, then simply add up any 1's or 2's remaining - divide that by 3 and bingo!

Finally, out of interest, there is another (more complex) method of testing for divisibility by 3 - AND for finding the remainder! First add all the ODD positioned digits (let's call it A), then add all the EVEN positioned digits (B). If A-2B is divisible by 3 so is the original number! If it's the remainder you're after, the result must be between -9 and +9 (a single digit), otherwise do it again and again if necessary. Once you have a single digit, divide it by 3 to obtain a remainder. If it's negative, add 3 giving a value of positive 1 or 2. Let's try 82134682. A=2+6+3+2=13, B=8+4+1+8=21, A-2B=13-42= -29, -(9-4)= -5, -5/3 = -1 and -2 remaining, -2+3=1. The number has a remainder of 1 when divided by 3! Let's try another, 213468. A=1+4+8=13, B=2+3+6=11, A-2B= 13-22= -9. This number is exactly divisible by 3. One interesting observation about this method is that it implies that any two digit number where one digit is twice the other digit will ALWAYS be divisible by three. Let's see - 12, 21, 24, 42, 36, 63, 48 and 84 - all divisible by three! It also implies that any three digit number where the digits are the same will also be divisible by three. 111, 222, 333 ... and 999 are all divisible by three!

Divisibility by nine and remainders

You may wonder why divisibility tests are important. If you wish to be quick and efficient in maths then recognizing divisibility is important when simplifying fractions, simplifying or solving algebraic equations and for pattern recognition...plus more! I think most people know that any number divisible by 2 will end in 0,2,4,6,8 (even numbers), divisible by 5 will end in 0 or 5 and divisible by 10 will end in 0. And, with a little bit of thought, you'll probably be able to work out the remainder of any number divided by 2, 5 or 10. But what about other numbers? Digital roots, once again, allow us to quickly work out, not only, if a number is divisible by 9 but also the remainder - if necessary. Let's try the number 3644. 3+6+4+4=17, 1+7=8. This number is NOT exactly divisible by 9, it would have a remainder of 8. In fact I can also tell you that 3645 is the next number divisible by 9 because 8+1=9. And that 3636 is the prior number divisible by 9 (just subtract 8). You should also note that any number divisible by 9 has a digital root of 9 - the remainder is 0 not 9 in this case. Now, for the 'true digital root' or 9's digital root. This method makes it quicker and easier to work out the remainder or divisibility of a number by 9, particularly if you have a pen and paper handy and can write down the number and cross off digits. The rule is simple - 'any digit which is nine is crossed off and replaced by zero'. You can even extend this further by saying any combination of digits that sum to 9, 18, 27 etc. can also be crossed off! I can now write a number like 1234567892 on paper and within a few seconds tell you that this number has a remainder of 2 if divided by 9. How? First cross off the 9, then 1+8, 2+7, 3+6, 4+5 and all you have left is 2. Or, if you prefer, as soon as the sum equals or exceeds 9 - subtract 9 (or add the two digits) and continue. Remember, with this method, you can never be left with 9 because it is simply crossed off and replaced with zero - automatically giving you the correct remainder!

In fact, you should never have to add ANY sums bigger than 8+8=16, because 16-9 or 1+6=7. You can simply continue with 7 added to the next digit etc. For example with 567862. 5+6=11, 1+1=2, 2+7=9 (becomes 0), 8+6=14, 1+4=5, 5+2=7. Or maybe it's easier with, 5+6+7=18 (becomes 0), 8+6+2=16, 1+6=7. You decide!

Divisibility tests

A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8).

A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.

A number is divisible by 4, if last 2 numbers are divisible by 4. Likewise the number is divisible by 8, if its last 3 digits are divisible by 8. So this holds for powers of 2.

A number is divisible by 5 if the last digit is 5 or 0.

Most people know (only) those 3 rules. Here are the rules for divisibility by the PRIMES up to 50. Why only primes and not also composite numbers? A number is divisible by a composite if it is also divisible by all the prime factors (e.g. is divisible by 21 if divisible by 3 AND by 7). Small numbers are used in these worked examples, so you could have used a pocket calculator. But my rules apply to any number of digits, whereas you cannot test a 30 or more digit number on your pocket calculator otherwise.
Lets assume L is the last digit and A is the remaining truncating number....

Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.
There are similar rules for the remaining primes under 40, i.e. 11,13, 17,19,23,29,31,37,41,43 and 47.
(A-2L) / 7


Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.
(A-L) / 11
     Another approach is to sum up numbers in 2 parts - 1st part containing all odd positioned numbers and other part containing all numbers positioned at  even positions. Find the difference and if the difference is divided by 11, i.e. if it is 11 or 0 it means number is divided by 11. eg. 19151 ,
odd position sum = 3, even position sum = 14; Difference = 11, that means no. is divisible by 11.

Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

(A+4L) / 13

Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
(A-5L) / 17




  Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
(A+2L) / 19

I think what I have above written is sufficient..but if you want more rules please carry on... :)

Test for divisibility by 23. Add seven times the last digit to the remaining leading truncated number. If the result is divisible by 23, then so was the first number. Apply this rule over and over again as necessary.
Example: 17043-->1704+7*3=1725-->172+7*5=207-->20+7*7=69 which is 3*23, so 17043 is also divisible by 23.
(A+7L) / 23




Test for divisibility by 29. Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.
Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.
(A+3L) / 29


Test for divisibility by 31. Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.
Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.
(A-3L) / 31

Test for divisibility by 37. This is (slightly) more difficult, since it perforce uses a double-digit multiplier, namely eleven. People can usually do single digit multiples of 11, so we can use the same technique still. Subtract eleven times the last digit from the remaining leading truncated number. If the result is divisible by 37, then so was the first number. Apply this rule over and over again as necessary.
Example: 23384-->2338-11*4=2294-->229-11*4=185 which is five times 37, so 23384 is also divisible by 37.
(A-11L) / 37


Test for divisibility by 41. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary.
Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.
(A+2L) / 41


Test for divisibility by 43. Now it starts to get really difficult for most people, because the multiplier to be used is 13, and most people cannot recognise even single digit multiples of 13 at sight. You may want to make a little list of 13*N first. Nevertheless, for the sake of completeness, we will use the same method. Add thirteen times the last digit to the remaining leading truncated number. If the result is divisible by 43, then so was the first number. Apply this rule over and over again as necessary.
Example: 3182-->318+13*2=344-->34+13*4=86 which is recognisably twice 43, and so 3182 is also divisible by 43.
(A+13L) / 43


Test for divisibility by 47. This too is difficult for most people, because the multiplier to be used is 14, and most people cannot recognise even single digit multiples of 14 at sight. You may want to make a little list of 14*N first. Nevertheless, for the sake of completeness, we will use the same method. Subtract fourteen times the last digit from the remaining leading truncated number. If the result is divisible by 47, then so was the first number. Apply this rule over and over again as necessary.
Example: 34827-->3482-14*7=3384-->338-14*4=282-->28-14*2=0 , remainder is zero and so 34827 is divisible by 47.
(A+2L) / 19


I've stopped here at the last prime below 50, for arbitrary but pragmatic reasons as explained above.
Lets summarize A+mL divisibilities :

Number	Coeff of L (m)
7	-2
11	-1
13	4
17	-5
19	2
23	7
29	3
31	-3
37	-11
41	2
43	13
47	-14

Other blogreaders (sadly even people from .edu domains, who should be able to do the elementary algebra themselves) have asked why I sometimes say ADD and for other primes say SUBTRACT, and ask where the apparently arbitrary factors come from. So let us do some algebra to show the method in my madness.
We have displayed the recursive divisibility test of number N as f-M*r where f are the front digits of N, r is the rear digit of N and M is some multiplier. And we want to see if N is divisible by some prime P. We need a method to work out the values of M. What you do is to calculate (mentally) the smallest multiple of P which ends in a 9 or a 1. If it's a 9 we are going to ADD, if it's a 1 we are going to SUBTRACT later. Then we will use the leading digit(s) of the multiple as our multiplier M.
Example for P=17 : three times 17 is 51 which is the smallest multiple of 17 that ends in a 1 or 9. Since it's a 1 we are going to SUBTRACT later. The leading digit is a 5, so we are going to SUBTRACT five times the remainder r. The algorithm was stated above. Now let's do the algebraic proof. Writing N=10f+r, we can multiply by -5 (as shown in the example for 17), getting -5N=-50f-5r. Now we add 51f to both sides (because 51 was the smallest multiple of P=17 to end in a 1 or a 9), giving one f (which we want), so 51f-5N=f-5r. Now if N is divisible by P (here P=17), we can substitute to get 51f-5*17*x=f-5r and rearrange the left side as 17*(3f-5x)=f-5r and therefore f-5r is a multiple of P=17 also. Q.E.D.

Special numbers

A number of 9 digits has the following properties:

• The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.

• Each digit in the number is different i.e. no digits are repeated.

• The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.

Find the number.

Answer
The answer is 381654729 One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.
At odd points we have  3 1 5 7 9
At even points we have 8 6 4 2

Armstrong number
Armstrong numbers are the sum of their own digits to the power of the number of digits. example:153 = 1³ + 5³ + 3³[as sum of the digits are 3], 1⁴ + 6⁴ + 3⁴ + 4⁴ = 1634.