Out of the Mouth of a Donkey?

Got an e-mail a couple of days ago from an Indian Mathematician named Sajeev Singh. Sanjeev asked me to let people know about his work at Total Gadha where he posts notes about mathematical ideas with some comments etc. I took a look and found some really well done posts on some interesting ideas about math that is accessible to high-school and college age folks. Gadha, by the way, if my memory serves me in old age, means Donkey, but I'm not sure how that reflects in the local culture. One more reason to visit India some day.

One of the notes he posted took me back, yet again, to the question about the 6! or 720 six-digit numbers that can be made using only the digits 1,2,3,4,5, and 6 once each. I realized that with a single brief mental calculation you could confirm that a) none of them were prime, and b) none of them were perfect squares. Why NOT?? (think, then read)

The reason has to do with the digit root; what used to be called casting-out-nines(and some of us still do call it that). If you add the digits, 1+2+3+4+5+6 the total is 21, and if you add 2+1 you get 3, so the remainder when any of the 720 numbers is divided by nine is going to be three, try it. 243561, for example, when divided by nine has a quoatient of 27062 with a remainder of three. Any number that has a digit root of three is divisible by three. All 720 of the six-digit arrangements are divisible by three. But how do I know none of them is a perfect square???? Same answer, the digit root is 3. THe digit root of a square number is the square of the digit root of the square root. Let's look at some numbers and their squares:

1² = has digit root of one

2²=has digit root of four

3²=digit root of nine

4²=16, so digit root of seven

5²=25 so digit root of seven

6²=36 so digit root of nine

7²=49 so digit root of four

8²=64 so digit root of one

9²=81 so digit root is nine

Keep goin.. these are the only four numbers you get for the digit root of a square,... ever...square 37?? digit root is one... square 155?... digit root is four.. you do these in your head.. To find the digit root of the square of 23 you add the 2 and 3 to get five, so the digit root of 23 is five. That means the digit root of 23 squared is the digit root of 5 squared.... seven. 23 ² = 529. Add the digits you get 16 and add those digits you get...yep, 7. What does NOT ever happen is a digit root of 3; so none of the 720 numbers are square numbers. Can you prove than none of them are perfect CUBES????

Two things better than being reminded of something you once knew are to learn something you never knew, and even better, to be stimulated by a question to discover something new on your own.... and both those things happened after going to TotalGahda. Rather than throw all that at you and once though, I will give you a chance to experiment with one.

Think of a two digit number, now write it down three times xyxyxy and divide by 7. What is the remainder? Wait, divide it by 13.... NOw what is the remainder? I just discovered this morning that this works.. and proved a general version of it which I will include next time with a conjecture about a couple of extensions to all primes... not sure if or who might have done it before... and I can add that if you write the two digit number out eight times, it will be divisible by 17. It will also work if you write down any four digit number four times, or any eight digit number twice (wow, what a clue)

If you have trouble dividing big long numbers and finding the remainder, try using the Google search engine. Just enter something like "4512451245124512 mod 17 =" in the search box and the Google calculator takes over and gives you the remainder (modulus)

Ok, IF you get stuck on that one, you can try this, write down any THREE digit number, and copy it again...xyzxyz to make a number... now divide by seven again, and by 13???? The reason this one works may be a little easier(I've know it has been around in calculator books before).... see if you can figure it out. I'll be back in a couple of days with the additional material, and if you KNOW this has been done already, especially if you have a citation, Please advise.

A Problem About Elevens, and Some Methods of "Casting Out Sevens"

Expanding an Archive blog from 2008 with some new insights: 


Sometimes I enter the math contest they host at the Wild About Math blog site, and usually am among the people who get a correct answer, but haven't been the lucky name pulled from the hat yet. A recent problem about divisibility got me thinking about testing by divisibility by sevens again.
The problem was actually about divisibility by eleven, and asked
Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.

So.... How many of these 6-digit numbers are divisible by 11?

The answer, of course, is none.
Of course, is a danger word, like obviously, or trivially, in which we dismiss the idea that there is thinking involved.  If I were presenting this question to a class, and I have, I would say it differently. 

"None of these numbers are divisible by eleven, can you figure out why without test dividing any of them? "
 If you don't see it, don't worry, I'll spill the beans on how I would prove it down the page.  Just take a moment and try it yourself. 

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The divisibility rule for eleven most commonly known is to take the digits abcdef and add every other one from the back, and then subtract the alternate ones.  So f-e+d-c+b-a  For example, 123456 would be  6-5+4-3+2-1=3.  Now they can only be divisible by eleven if the result is 0, or a multiple of 11. 

So if we think about the numbers involved in all these numbers, they all have three odd numbers, and three even, so no matter how you split them, you'll have one group is odd, the other is even, so the difference is an odd number.  You can't get zero.  But can we get eleven?  Ummm, NO, because if we put all the big ones in one clump (6+5+4), and all the small ones in another (3+2+1), the difference is less than eleven.... so NONE of them are divisible by 11. 

 My interest was piqued by a response by Jonathan, of the jd2718 blog made these observations about the 6! or 720 possible numbers formed with the six digits as described:

As a consolation, all 720 of them are multiples of 3.

Half of them (360) are even (multiples of 2)

One in 6 (120) are multiples of 5.

Eight in thirty (192) are multiples of 4.

None are multiples of 9.

Fourteen of 120 (84) are multiples of 8.

Multiples of 7? New puzzle, good place to stop.


It is Not surprising that Jonathon covered all the other one digit divisors but did NOT test seven. Seven is the one that books on mental math simply say "divide by seven"; but thinking about it, it shouldn’t be too hard
I wrote recently about a mental test for divisibility by seven but it did have one flaw. It worked fine to tell you a number was divisible by seven, but if the number was not divisible by seven it did not give the correct modulus. Casting out nines will always give you the correct modulus. The sum of the digits of 2134 is 10 which has a digit root of 1, so if you divide 2134 by 9 you get a remainder of one. My rule for seven didn't give the true remainder.

The remainder when you divide a number by seven (or any other number) is frequently called the modulus. I've mentioned this before, it is just a way to divide integers into sets. Odd numbers are all equivalent to 1 mod 2 (they all leave a remainder of one when you divide by 2) and even numbers are all equivalent to zero. If a number is equivalent to zero mod (something) that means it is divisible by that (something). One of the things that makes them effective is the simple rule that if a+b=c then for any modular index n(the number you are dividing by) c mod n = a mod n + b mod n. For example 25 mod 7 = 4.  We could have found that by breaking it into parts.  Since 25 = 20 + 5 we find 20 mod 7 = 6, and 5 mod 7 = 5... and the  6+5 = 11 which is equivalent to 4 mod 7. So now we have the basics.

Every increase of one in the units increases the modulus 1, and every increase in the tens increases the modulus three, (for example 21 mod 7 = 0; 31 mod 7 = 3,and 41 mod 7 = 6). I figured out that the rest would be an increase of two in the hundreds, six in the thousands (think minus one) four in the ten-thousands (minus three?) and five in the hundred-thousands (minus two….) so the sequence applied to our test number, 546231… and what a coincidence, if you multiply 5(-2)+4(-3)+6(-1)+2(2) +3(3) + 1(1) …YOU GET -14, which is zero mod 7, the sequence of moduli used in each place makes it a multiple of seven…(546231 = 0 [mod 7]…

 but wait, if you used the A-2B method in the blog  above, we would notice that 231 by itself is a multiple of seven, since 23 - 2(1) =21… and 546 is also. [54 - 2(6)=42]..   You could also apply that approach step by step, 54623-2*1=54621; and 5462-2(1) = 5460.  The zero we can throw away because 546 tens is not divisible by 7 if 546 is not, so we proceed to 54-2(6) = 42 and hey, we have a divisor by 7.  In each case we just combined two moduli to reduce the size of the number. 

There is even a neat graphic approach that works the same way.  I came across a nice video of that at a site by Presh Talwalker called Mind Your Decisions.  Each step uses the same modulus increase approach as the method above, but challenge your students to figure out why it works (not an easy task). 

To check other numbers, you can just write them out multiplied times the correct modulus for that place value and probably check it in your head, but you would have to check each one… so making up a sequence, 234561(this covers everything up to six digits, but if you have a super memory, figure out another period), we would think 2×5=10 (the first digit times the first modulus) drops to 3, plus 3(second digit)x4(second modulus) is 15 which drops to modulus of one, now add 4(6) to get 25 and drop to mod 4, then add 2×5 to get 14, and we are at mod 0 [so 234500 is divisible by seven] and we know that 61 is NOT divisible by seven, so we can be assured that 234561 == 61 == 5 mod 7… ok, not EASY, but certainly could be done sans calculator…. [footnote, for those of you who have just gone through a pre-calc class and somewhere along the way they taught you about vectors, you probably imagined that you would never run across another dot product in your life, but you just did. IF you think of the sequence of modular values as one vector (5,4,6,2,3,1) and the six-digits of the number as a second vector, then the dot product is an integer that has the same modulus as the original six digit number....come on... that's pretty cool use of vectors!!!]
It is easy enough to write/remember the modulus up to ten-thousands (6231) to make this pretty useful as a factoring tool if you really needed the correct modulus. If the number is 4723 we just think 4(6)=24--==; 3 + 7x2---==; still 3, + 2(3)=9 --==;2 and then + 3(1)= 5 so 6231 divided by seven leaves a remainder of five.