+2446 | Divisibility Test Number | Process |
| 1 | Every integer is divisible by 1 |
| 2 | Check to see if the last digit is divisible by 2 |
| 3 | Recursively check if the sum of digits is divisible by 3 |
| 4 | Check to see if the last two digits are divisible by 4 |
| 5 | Check to see if the last digit is divisible by 5 |
| 6 | Check divisibility rules for 2 & 3 |
| 7 | Recursively check if subtracting twice the sum of the last digit from the rest of the number is divisible by 7 |
| 8 | Check if the last 3 digits are divisible by 8 |
| 9 | Recursively check if the sum of the digits is divisible by 9 |
| 10 | Check if the last digit is a 0 |
| 11 | Recursively check if subtracting the final digit from the rest is divisible by 11 |
| 12 | Check divisibility for 3 & 4 |
| 13 | Recursively add 4 times the final digit to the rest |
I understand the divisibility rules except 7, 11, and 13. What do you mean by the "rest?" Can you kindly explain further?
+2446 Of course! I will gladly delve deeper into this subject.
The explanation I provided is probably unsatisfactory and shoddy anyway. I believe that these rules are best demonstrated by example.
1. Divisibility by 7
Is 205226 divisible by 7? Well, let's use the process!
| 205226 | |
| 1. 20522−2∗6=20510 | I have no clue if this is indeed divisible by 7, so do this process again and again (hence recursion) |
| 2. 2051−2∗0=2051 | I still cannot tell, so I will do this again. |
| 3. 205−2∗1=203 | I still cannot tell. |
| 4. 20−2∗3=14 | I know that 14 is divisible by 7, so the original number is, too. |
How about 22604? Well, let's check it!
| 22604 | |
| 1. 2260−2∗4=2252 | Yet again, I cannot make a judgment. |
| 2. 225−2∗2=221 | Of course, we must keep going. |
| 3. 22−2∗1=20 | I know that this number is not divisible by 7, so the original number is not either. |
2. Divisibility by 11
Let's check if 43923 is divisible.
| 43923 | |
| 1. 4392−3=4389 | Let's do it again! |
| 2. 438−9=429 | One more time! |
| 42−9=33 | I know that 33 is divisible by 11, so the original number is, too. |
How about 123567?
| 123567 | |
| 1. 12356−7=12349 | This requires perserverance. Keep going! |
| 2. 1234−9=1225 | |
| 3. 122−5=117 | I know that 11∗11=121, so 117 is not divisible. |
3. Divisibility by 13
Is 19704 divisible? Let's find out!
| 19704 | |
| 1. 1970+4∗4=1986 | |
| 2. 198+4∗6=222 | |
| =22+4∗2=30 | 13∗3=36, so 30 is not divisible by 13 and nor is the given number. |
Is 9321 able to be divised?
| 9321 | |
| 1. 932+4∗1=936 | |
| 2. 93+4∗6=117 | |
| 3. 11+4∗7=39 | 39 is divisible by 13, so the original number is as well. |
+2446 I am glad you understand now! One unique thing about divisibility tests is that there are absolutely no exceptions to any of the rules listed.
+9675 Easier divisibility rule for 11:
if absolute value of ((1st digit + 3rd digit + 5th digit + ....) - (2nd digit + 4th digit + 6th digit + ...)) = 0 or anything divisible by 11,
then the number is divisible by 11.
Example:
Is 43956 divisible by 11?
(4 + 9 + 6) - (3 + 5) = 19 - 8 = 11 <--- divisible by 11.
So 43956 is divisible by 11.
Is 10293583762 divisible by 11?
(1 + 2 + 3 + 8 + 7 + 2) - (0 + 9 + 5 + 3 + 6) = 23 - 23 = 0
So 10293583762 is divisible by 11.
+130477 Here's an easy rule to decide divsibility by 11
Alternate signs on the digits and add.....if the result is 0 or 11, the number is divisible by 11
Example 43956 = +4 - 3 + 9 - 5 + 6 = 2 + 9 = 11
Note .....43956 = 11 * 3996
Another example 1331 = +1 - 3 + 3 - 1 = 0
And 1331 = 11^3
