[SOC] A Little Math Problem

[Fred Jensen]

OK, just a little fun exercise.

Responses with comment so far:

 > Would it be a minimum factor or a max factor?

Doesn't matter, any factor[s] will do.  Problem is to factor it, no 
reference to prime factors.  Part of this test is reading the problem 
very carefully.  However, it is sort of SOC-ish not to do that, kinda 
like immediately adjusting the insides of your radio to fix "no power 
out" without first noting the power connector has been pulled out.

 > 12,345,678,987,654,321 divided by 1 = 12,345,678,987,654,321

True, but then all integers are divisible by 1 [and themselves], and 
"number dudes" as it was put below call these "trivial factors" and 
don't count them.  Sorry.

 > Well, Fred, to a biologist it is divisible by 123456789 within
 > significant digits, but I'm only counting fingers so that is
 > 8 digits and two thumbs.

Unfortunately, the domain of factoring is the natural numbers [positive 
integers], and all the digits are significant.  Not a lot of calculators 
will compute with integers that big.

 > The Square Root returns 111,111,111.

Indeed it does, 12,345,678,987,654,321 is a perfect square so one 
potential right answer is "one hundred and eleven million, one hundred 
and eleven thousand, one hundred and eleven, twice."

 > However, "number dudes" would continue until all prime factors
 > are listed (a prime is divisible only by itself and 1).
 > What does any of this have to with SOC, you ask?
 > You let someone else do the work! (Google
 > "factor 12345678987654321" )
 > 12345678987654321 =  3 x 3 x 3 x 3 x 37 x 37 x 333667 x 333667

And "number dudes" would accept this as the "'proper' right answer," the 
prime factors are 3 [4 times], 37 [twice], and 333667 [twice].  And 
answering such a problem as this using Google might be a little SOC-ish 
in itself.  Talk about getting someone else to do the work! :-)

How could a SOC-member have done this pG [pre-Google, say in 1960 when 
ham radio -- and second class operators -- certainly existed but the 
Google guys' parents were likely still in HS, and calculators worked 
with gears, levers, and number wheels, and none of them had 17 digits]? 
  One way is to treat it as more of an inductive arithmetic 
visualization problem -- noting the regularity of 12345678987654321 and 
starting from the least significant digit, think:

1 = 1
2 = 1+1
3 = 1+1+1
4 = 1+1+1+1
.
.

Writing this down in the form we normally use to add numbers, starting 
from the least significant digit [you need to set your mail client to a 
monospace font for this part]:

           111111111
+        111111111
+       111111111
+      111111111
+     111111111
+    111111111
+   111111111
+  111111111
+ 111111111
   -----------------
   12345678987654321

This looks suspiciously like what you'd get if you worked out a simple 
multiplication problem with pencil and paper [remember pencil and paper?]:

           111111111
         X 111111111
         -----------
           111111111
+        111111111
+       111111111
+      111111111
+     111111111
+    111111111
+   111111111
+  111111111
+ 111111111
   -----------------
   12345678987654321

so our number 12,345,678,987,654,321 is a perfect square and we have 
answered the original problem.

"Number dudes" from that era, who misread the problem and think they're 
looking for all the prime factors, would know they were home free at 
this point because they can factor 111,111,111 just by brute force 
division by known primes.  The first odd prime [3] works and leaves 
37,037,037 [2 is technically a prime, the only even prime, but 
111,111,111 is odd so we know immediately it isn't divisible by 2].

1960's SOC-member could then divide that successively by 5, 7, 11, 13, 
17, 19, 23, 29, and 31 [none of which divide it evenly], but I'd be 
inclined to try 37 first just for grins given the pattern of the digits. 
  37 works and yields a remainder of 1,001,001 [note that 37,037,037 is 
itself divisible by 3, but 37 just calls out to be tried first, and 
we'll get the second 3 later anyway].

Now, dividing 1,001,001 by known primes, first odd one [3] yielding 
333667, which got us the second 3 we ignored above.  Tables of primes 
beyond a million existed in 1960, so taking the book off the shelf 
[remember books ... on shelves?], I find that it is indeed a prime.

So, starting with 111,111,111, I've got two 3's, a 37, and a 333,667 for 
its prime factors.  There are two instances of 111,111,111 in the 
problem, so I end up with four 3's, two 37's, and two 333,667's with but 
a three paper-pencil long division exercises.

I posed this problem to our 21 year old grandson, a high school 
graduate.  He Googled the number, found a link, and gave me the answer. 
  Google really is remarkable, no?  More remarkable, he did it on his 
cell phone.  If Chris ever gets a ham license, I will invite him to join 
SOC, I think he is a prime candidate.  OK, that was bad ;-(

73,

Fred K6DGW
#633 [prime factors = 3 and 211]