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.

## 3 comments:

This was neat! You should have linked, way back then, so I could have seen.

In the meantime I do know two easy-ish tests for 7, and actually taught one of them to fairly amazed 14 year olds this year.

Jonathan

Oh, and I too am in the category of those who've gotten a pile of mmm's correct, but have never been chosen.

On the subject of sevens...

I once saw a circle of digits that behaved most fascinatingly. Do the following: draw a hexagon, and label the vertices in order clockwise (from any point) as 1, 4, 2, 8, 5, 7.

Now you have the number 142857 AND ITS MULTIPLES all expressed.

(Dashes used for spacing—see how the digits line up?)

A = 142857

2A = --285714

3A = -428571

4A = ----571428

5A = -----714285

6A = ---857142

...which can all be found on the circle [hexagon] by starting at any digit and going clockwise.

But the real kicker is when you see what 7A equals. Go on...I won't spoil it for you.

Then work out the decimal form of 1/7th and you're all set :)

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