Michael J. Swart

IBM holds a monthly math/programming challenge called “Ponder This”.
This month’s problem (July 2017) is an interesting one because it can (mostly) be solved using SQL and highschool math:

“Find a way to assign a unique digit to each letter such that the 863,172nd digit after the decimal point of sin(PONDER) and of sin(THIS) are not zeros, but more than a fifth of the first million digits of sin(PONDER)*sin(THIS) are zeros.

“The parameters are in degrees, and we count the digits from 1, so for example, sin(100) = 0.9848077530122… so the fifth digit after the decimal point of sin(100) is zero.”

If you’re up to the challenge, use any technique you like, but you’ve only got a week. Next week (August 1st) after IBM publishes their solution, I’ll publish the SQL based solution that I used.

Update: My Solution

First I realized that PONDER and THIS represent integers greater than 360 degrees. That’s okay. All I need to do is find two numbers A and B between 1 and 360 with the desired properties and then take that solution and add 360 to either number until I get a six digit number and a four digit number with no shared digits.

So my guess is that SIN(A) * SIN(B) is a regular number.
I can look for that using SQL:

with Nums as
(
    select top 360 row_number() over (order by (select 1)) as n
    from sys.syscolumns
)
select A.n as a, 
       B.n as b, 
       sinA, 
       sinB,
       sinA * sinB as sinA_x_sinB
from Nums A
cross join Nums B
cross apply 
      (
         select SIN(RADIANS(A.n + 0.0)) as sinA,
                SIN(RADIANS(B.n + 0.0)) as sinB
      ) calc
where right(format(sinA * sinB, '###.00000000000'), 4) = '0000'
  and right(format(sinA, '###.00000000000'), 4) <> '0000'
  and right(format(sinB, '###.00000000000'), 4) <> '0000'
  and A.n % 10 <> B.n % 10
order by A.n, B.n

Notice the one extra filter I snuck in there: A.n % 10 <> B.n % 10. If A and B share the last digit, then PONDER and THIS would share the last digit and we can’t have that.

I’m going to focus on one solution, A = 54, and B = 18.
Can I add 360 to each of these to get an 6 digit and a 4 digit solution for PONDER and THIS? Yes.

select TOP (27) 54 + 360 * row_number() over (order by (select 1)) as id
into #nums4
from sys.messages a;
delete #nums4 where id < 1000;
 
select TOP (2777) 18 + 360 * row_number() over (order by (select 1)) as id
into #nums6
from sys.messages a;
delete #nums6 where id < 100000;
 
select #nums4.id, #nums6.id 
from #nums4
cross join #nums6
where dbo.HasDuplicate(10000*#nums6.id+#nums4.id) = 0

Where HasDuplicate is a CLR function I wrote that returns whether a number has repeat digits.
This gives me a few solutions, for example PONDER = 102978 and THIS = 3654.

I get a kick out of these puzzles and as an added bonus, I get to exercise my SQL skills.

Update 2

Stephen pointed out in the comments that I didn’t even check to see whether the 863,172nd digit of SIN(PONDER) or SIN(THIS) is nonzero. How did I do that? Well, I didn’t use SQL. I used google. Searching for “What’s the exact value of SIN(18)” gives me this page Exact Value of sin 18°. The internet’s amazing. So I trust that the exact value is sin(18) = ( -1 + sqrt(5) ) / 4 which has the same 863,172nd digit as sqrt(5) / 4. The same reasoning works for SIN(54) because sin(54) = ( 1 + sqrt(5) ) / 4.

Rather than figure out how to calculate the 863,172nd digit of sqrt(5) / 4, I searched for and found the first million digits of sqrt(5). The internet’s amazing. I know that when you divide a number by four, any digit in that number usually has a limited scope of influence on the digits in the answer. I divided the digits around the 863172 place by 4 to convince myself that the 863172nd digit of sqrt(5) / 4 is not 0.