Friday, August 03, 2007
Harder divisibility test
We can see by (a) above that most divisibility tests are now covered. The exception being prime numbers such as 7, 13, 17 ... So, here's the trick for divisibility by prime numbers (or multiples of). First find the first multiple of the number (let's call it N) that ends in 9 or 1 (e.g. 9, 11, 19, 21, 29...). Second, round up or down the number to the nearest multiple of ten (10, 20, 30...). Now your multiplying constant (let's call it K) becomes the tens digit (1, 2, 3...). If you rounded up it is positive (for add), if you rounded down it is negative (for subtract). Now take the number you wish to test for divisibility and multiply the units digit by K. This value is then added to the tens column or subtracted if K is negative. If the number that results is divisible by the number N then so is the original number! Of course, by logical extension, this process can be repeated until you recognize a number that is divisible by N.
Let's see if we can use this method to find a divisibility test for 7. First, multiply 7 until we have a number ending in 9 or 1. OK. 7, 14, 21. Now, we need to round this down to 20, so K = -2. Lets' test it. Try 3199. 319 - 2x9 = 301. 30 - 2 = 28. 28 is divisible by 7 so is 3199. Note, we could have gone further with 28. As 2-16 = -14. And 1-8 = -7. The factor K = -2 is also a test for divisibility by 21. The difference, being that we need to check if the result is divisible by 21 rather than 7! Let's try 9576. 957 - 2x6 = 945. 94 - 2x5 = 84. 8 - 2x4 = 0. The number 9576 is divisible by 21 (and 7 and 3 in fact). A couple of very interesting observations can be made. Firstly, all two digit numbers where the tens digit is twice the units digit (21, 42, 63 and 84) are all divisible by 21 (and by default 7 and 3). You should also notice the similarity of this test with the second test for divisibility by 3 above. In fact, this test can also be used to test for divisiblity by 3!
Alternatively, when testing for divisibility of 7, since 7x7 = 49, we can use K=+5. Let's try this with 294. 29 + 5x4 = 49. 49 is divisible by 7 and so is 294! Interestingly, we can't go any further with 49, since 4 + 5x9 = 49!
Note, despite what some books may tell you, tests for divisiblity by seven are not very 'practical'. I would recommend simply dividing the number by seven (you can discard the result if you're only interested in the remainder) and if the final remainder is zero then the number is divisible by seven! If you are using pen and paper then the number can be simplified first by crossing out 7's or any 2 digit multiple of 7 (make them 0), changing 8's to 1 and 9's to 2 (i.e. the remainders when divided by 7) and, since 7 divides into 1001, 10010, ... exactly, subtract one digit from another which are separated by 2 digits. Remember any leading or trailing zeros can be removed. e.g. Try 59633, cross off 63 = 59003, make 9 a 2 = 52003, subtract 3 from 2 = 49000, 49 is divisible by 7 so is 59633. Note, after obtaining 59003, the result could be obtained easier by taking the 3 from the 9 = 56000, 56 is divisible by 7! Also, since 1001 is a multiple of 7, K= -100. Which is the same as subtracting it from the 4th digit!
How about a divisibility test for 13? First 13, 26, 39. Round up to 40, therefore K = +4. Testing it, let's try 585. 58 + 4x5 = 78. 7 + 4x8 = 39. 39 is divisible by 13 and so is 585. What's real interesting about this method is that if testing for divisibility by 3 or 9, the value of K = +1. In other words, you repeatedly add the units digit to the tens column. If you continue to do this you end up with the digital root of the number! The value of K for other prime numbers is; K= -1 for 11. K= -5 for 17. K = +2 for 19. K= +7 for 23. Note, it's pretty easy to work out the values for K as required. i.e. There's no need to memorize them!
Finally, you may wish to find a negative value for K rather than use a positive value (the number reduces quicker and therefore the answer is found quicker). Simply subtract the number N from K. e.g. for N=3 and K=+1, the alternate value for K is 1-3 = -2. Try it and see if it works! To convert a negative K to a positive we can use N + K. e.g. for N=7 and K= -2, the alternate value of K is +5. For N=9, we can also use K= -8. For N=11, K= +10. etc. Note, with K= +10 for divisibility by 11, we can simply take the units digit and add it to the thousands column, if the resulting number is divisible by 11 so is the original number! Let's try 517, (5+7)1 = 121 we could stop here, but if we do it again (1+1)2 = 22. Clearly divisible by 11. Note, you can't go any further with this method (K= +10) once you have a 2 digit number. Of course, you could always continue with K= -1.
Let's see if we can use this method to find a divisibility test for 7. First, multiply 7 until we have a number ending in 9 or 1. OK. 7, 14, 21. Now, we need to round this down to 20, so K = -2. Lets' test it. Try 3199. 319 - 2x9 = 301. 30 - 2 = 28. 28 is divisible by 7 so is 3199. Note, we could have gone further with 28. As 2-16 = -14. And 1-8 = -7. The factor K = -2 is also a test for divisibility by 21. The difference, being that we need to check if the result is divisible by 21 rather than 7! Let's try 9576. 957 - 2x6 = 945. 94 - 2x5 = 84. 8 - 2x4 = 0. The number 9576 is divisible by 21 (and 7 and 3 in fact). A couple of very interesting observations can be made. Firstly, all two digit numbers where the tens digit is twice the units digit (21, 42, 63 and 84) are all divisible by 21 (and by default 7 and 3). You should also notice the similarity of this test with the second test for divisibility by 3 above. In fact, this test can also be used to test for divisiblity by 3!
Alternatively, when testing for divisibility of 7, since 7x7 = 49, we can use K=+5. Let's try this with 294. 29 + 5x4 = 49. 49 is divisible by 7 and so is 294! Interestingly, we can't go any further with 49, since 4 + 5x9 = 49!
Note, despite what some books may tell you, tests for divisiblity by seven are not very 'practical'. I would recommend simply dividing the number by seven (you can discard the result if you're only interested in the remainder) and if the final remainder is zero then the number is divisible by seven! If you are using pen and paper then the number can be simplified first by crossing out 7's or any 2 digit multiple of 7 (make them 0), changing 8's to 1 and 9's to 2 (i.e. the remainders when divided by 7) and, since 7 divides into 1001, 10010, ... exactly, subtract one digit from another which are separated by 2 digits. Remember any leading or trailing zeros can be removed. e.g. Try 59633, cross off 63 = 59003, make 9 a 2 = 52003, subtract 3 from 2 = 49000, 49 is divisible by 7 so is 59633. Note, after obtaining 59003, the result could be obtained easier by taking the 3 from the 9 = 56000, 56 is divisible by 7! Also, since 1001 is a multiple of 7, K= -100. Which is the same as subtracting it from the 4th digit!
How about a divisibility test for 13? First 13, 26, 39. Round up to 40, therefore K = +4. Testing it, let's try 585. 58 + 4x5 = 78. 7 + 4x8 = 39. 39 is divisible by 13 and so is 585. What's real interesting about this method is that if testing for divisibility by 3 or 9, the value of K = +1. In other words, you repeatedly add the units digit to the tens column. If you continue to do this you end up with the digital root of the number! The value of K for other prime numbers is; K= -1 for 11. K= -5 for 17. K = +2 for 19. K= +7 for 23. Note, it's pretty easy to work out the values for K as required. i.e. There's no need to memorize them!
Finally, you may wish to find a negative value for K rather than use a positive value (the number reduces quicker and therefore the answer is found quicker). Simply subtract the number N from K. e.g. for N=3 and K=+1, the alternate value for K is 1-3 = -2. Try it and see if it works! To convert a negative K to a positive we can use N + K. e.g. for N=7 and K= -2, the alternate value of K is +5. For N=9, we can also use K= -8. For N=11, K= +10. etc. Note, with K= +10 for divisibility by 11, we can simply take the units digit and add it to the thousands column, if the resulting number is divisible by 11 so is the original number! Let's try 517, (5+7)1 = 121 we could stop here, but if we do it again (1+1)2 = 22. Clearly divisible by 11. Note, you can't go any further with this method (K= +10) once you have a 2 digit number. Of course, you could always continue with K= -1.
// posted by Kritidas @ 2:35 AM 
