## Golf Is a Good Walk Ruined

*July 28, 2012 at 10:08 pm* *
5 comments *

The title of this post is a quote attributed to Mark Twain. I would amend it as follows:

Golf may be a good walk ruined — but it’s a good time to think about math.

Not too long ago, I attended a conference at the Asilomar Conference Center on the Monterey Bay Peninsula in California. The property is adjacent to beautiful 17-Mile Drive and historic Pebble Beach. With some free time on the last afternoon, I decided to treat myself to a round of golf.

“Hello, this is Edie. Thanks for calling Pebble Beach Golf Club. How can I help you?”

“Hi, Edie. I’m wondering if I can get on the course to play 9 holes today.”

“You sure can,” she said, “but the price is the same for 9 or 18 holes.”

“Oh, okay. And what’s the price?”

“$495,” Edie said matter-of-factly.

I did a quick mental calculation, and I realized that the price would be $55 per hole.

What I said: “On second thought, Edie, I’m not sure I’ll have enough time today. Let me call back if my schedule opens up.”

What I thought: “Are you f**kin’ kiddin’ me?”

So instead of playing Pebble Beach, I found an executive course in Cupertino where I played 9 holes, rented clubs, and bought six golf balls, a bag of tees, and a new golf glove — all for less than $50.

Not only does this incident represent an exercise in fiscal responsibility, it also brought to mind divisibility rules. For instance, I realized that 495 was divisible by 9 because 4 + 9 + 5 = 18.

The rule for divisibility by 9 is rather easy to implement: just add the digits. If the result is divisible by 9, then the original number is divisible by 9, too.

A couple of weeks ago, a colleague told me that she and some friends had discovered an elegant rule for divisibility by 7. However, she was unwilling to share the method with me — she said I’d have more fun if I discovered it on my own. (I simultaneously loved and loathed her for that.)

I knew a rule for divisibility by 7, I discovered an inelegant method, and I discovered another method that was less inelegant. I’ll share my three methods below, but if you know (or discover) a different method, please share.

**The Method I Already Knew**

The rule for divisibility by 7 that I had heard before, and that can be found easily via Google, works as follows:

- Remove the units digit, and double it. (If the original number was 10
*m*+*n*, where*m*and*n*are positive integers, you’d calculate 2*n*.) - Subtract the result of Step 1 from the remaining number. (That is, you’d find
*m*‑ 2*n*.) - Repeat Steps 1 and 2 until you can determine if the result is a multiple of 7. If it is, then the original number is a multiple of 7, too.

This is best shown with an example. Let’s say you want to know if 8,603 is divisible by 7. Then first find 860 ‑ 2(3) = 854. Since it may not be obvious that 854 is a multiple of 7, repeat the procedure: 85 ‑ 2(4) = 77. Because 77 is a multiple of 7, then 8,603 must be a multiple of 7, too.

The problem with this procedure is that it takes too long for big numbers.

** The Highly Inelegant Method That I Devised**

Take the number, and continually subtract known multiples of 7. For instance, again using 8,603, first remove 7,000 to leave 1,603. Then remove 1,400 to leave 203. Then remove 210 to leave ‑7, which is a negative multiple of 7. So that means 8,603 must be a multiple of 7, too.

Since any number of the form 7 × 10* ^{k}* is a multiple of 7, a time-saving step in this process is to reduce every digit in the original number by 7. For 8,603, that means reduce 8 to 1 to leave 1,603. That’s the same as subtracting 7,000. But for a bigger number like 973,865, that shortcut could be implemented three times to leave 203,165, which might help a little. Taking this even further, remove other multiples of 7; for instance, since 16 ‑ 14 = 2, you could further reduce the number to 203,025.

I rather like this method for its simplicity, but it also takes too long.

**The Less Inelegant Method That I Found **

The reason that the rule for divisibility by 9 works is that every power of 10 is 1 more than a multiple of 9. Consequently, when a number of the form *m* × 10* ^{k}* is divided by 9, the remainder will be

*m*. For instance, when 7,000 is divided by 9, the remainder is 7; when 300 is divided by 9, the remainder is 3; and, when 80 is divided by 9, the remainder is 8. So if you subtract a lot of 9’s from 7,380, you would be left with 7 + 3 + 8 = 18 as the remainder, and since that result is a multiple of 9, then 7,380 is a multiple of 9, too.

This same idea can be applied for divisibility by 7.

- When you divide a number of the form
*m*× 10^{6k}by 7, the remainder will be*m*. - When you divide a number of the form
*m*× 10^{6k + 1}by 7, the remainder will be 3*m*. - When you divide a number of the form
*m*× 10^{6k + 2}by 7, the remainder will be 2*m*. - When you divide a number of the form
*m*× 10^{6k + 3}by 7, the remainder will be 6*m*. - When you divide a number of the form
*m*× 10^{6k + 4}by 7, the remainder will be 4*m*. - When you divide a number of the form
*m*× 10^{6k + 5}by 7, the remainder will be 5*m*.

You can then determine if a number is divisible by 7 by multiplying the units digit by 1, the tens digit by 3, the hundreds digit by 2, and so on, following the sequence 1, 3, 2, 6, 4, 5, as given above. If the result is a multiple of 7, the original number is, too.

As an example, again consider 8,603. Then calculate

1(3) + 3(0) + 2(6) + 6(8) = 63.

Since the result is a multiple of 7, then 8,603 is a multiple of 7, too.

**What You Got?**

Those are my three methods. As I said above, the third is less inelegant that the first two, but I still wouldn’t call it elegant.

Have you got a rule for divisibility by 7 that’s better than any of these? Do tell!

Entry filed under: Uncategorized. Tags: divisbility, elegant, golf, Pebble Beach, rule.

1.Joshua Zucker | July 29, 2012 at 12:37 amI don’t have anything significantly better. Knowing that 1001 is divisible by 7 is a bit of a shortcut that can be added in to your last method(s).

Also, a typo. You wrote “(That is, you’d find 10m ‑ 2n.)” but you mean m – 2n there, I think.

2.venneblock | August 6, 2012 at 9:57 amThank you, sir. Post updated.

3.puntomaupunto.mau. | July 29, 2012 at 10:09 aminstead that multiplying for 1,3,2,6,4,5 I would use 1,3,2,-1,-3,-2. Smaller numbers and symmetry.

Otherwise, use a rule similar to that for 11 (sum and subtract digits from left to right) but taking

threedigits at a time. This is a criterion for 1001; you have so 7 and 13 at the same time.4.Silvio Moura Velho | September 30, 2012 at 10:54 amIf you want to know the first real rule for divisibility by 7 access:

primeasdivisor.blogspot.com

5.venneblock | October 1, 2012 at 8:12 pmThe full URL for Silvio’s post is http://primeasdivisor.blogspot.com/2012/09/the-most-difficult-elementary-problem.html.