bonanova Posted January 12, 2009 Report Share Posted January 12, 2009 You've had the warm-up. Now this. This may be the most difficult puzzle of its type: no starting numbers are given. [The zeros simply remind that you're beyond the last digit of the dividend.] The quotient of course has a decimal point [not indicated] but it's easy to see where it goes. The last nine digits repeat. The calculation is not shown beyond the point [2nd asterisk] where the process repeats [from first asterisk] Enjoy! ..........._________.........________ ........XXXXXXXXXXXX <- last 9 digits repeat XXXXXX /XXXXXXX ........XXXXXX ........XXXXXXX ........_XXXXXX .........XXXXXX0 .........XXXXXXX ..........XXXXXX0 <- * Same digits as below .........._XXXXXX ...........XXXXXX0 ...........XXXXXXX ............XXXXXX0 ............XXXXXXX .............XXXXXX0 .............XXXXXXX ..............XXXXXX0 ..............XXXXXXX ...................XX0000 ...................XXXXXX ...................XXXXXX0 <- * Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 12, 2009 Report Share Posted January 12, 2009 Unfortunately, my heart condition excludes me from this puzzle. *whew!* Good luck to the rest of you. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 15, 2009 Author Report Share Posted January 15, 2009 Would a hint be useful? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 15, 2009 Report Share Posted January 15, 2009 I haven't had time to start on it yet, but I do plan on attempting this some time in the next few days. Quote Link to comment Share on other sites More sharing options...
0 Prime Posted January 16, 2009 Report Share Posted January 16, 2009 I am not all that familiar with long division, but here is some stuff that seems more or less obvious to start with... ..........._________.........________ ........XX.XXXXXX000X <- last 9 digits repeat XXXXXX /XXXXXXX ........XXXXXX ........1XXXXXX ........_XXXXXX .........XXXXXX0 .........XXXXXXX ..........1XXXXX0 <- * Same digits as below .........._XXXXXX ...........XXXXXX0 ...........XXXXXXX ............XXXXXX0 ............XXXXXXX .............XXXXXX0 .............XXXXXXX ..............XXXXXX0 ..............XXXXXXX ...................XX0000 ...................XXXXXX ...................XXXXXX0 <- * From the three lines at the bottom, I can tell the Divisor (let's call it R), is less or equal 880,000. Also, R > 111,000 ( since it can yield a 7-digit number when multiplied by one-digit number.) The two digits in the whole part of the quotient must be less than 9 each (the divisor multiplied by those is a 6-digit number). Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 16, 2009 Author Report Share Posted January 16, 2009 These are all correct. And of course you can add the leading number in the bottom row of repeating digits. Quote Link to comment Share on other sites More sharing options...
0 Prime Posted January 16, 2009 Report Share Posted January 16, 2009 Upon further consideration, we could figure out the divisor, which might help solving the puzzle. Let’s concentrate on the periodical part of the quotient. It comes from the 4th step (marked by asterisk). There we divide a six-digit number 1xx,xxx by another 6-digit number (the original divisor) and obtain 0.[xxxxxxxxx]… That infinite decimal fraction may be expressed as a geometric sum: xxxxxxxxx/109 + xxxxxxxxx/1018 + xxxxxxxxx/1027 + … Using the formula for geometric sum we find it is equal to xxxxxxxxx/999,999,999. But in the long division we have obtained that fraction by dividing one six-digit number by another (in the step 4). Therefore, the divisor must be a six-digit factor of 999,999,999. Using Prime’s Template for breaking a number into its prime factors, we find: 999,999,999 = 34*37*333,667. The only 6-digit factor possible is 333,667. (The product of all other factors is only 2997, whereas multiplying 333,667 by the smallest of other factors 3, yields 1,001,001 -- a 7-digit number.) So we have the only possible divisor and can resolve few of the other digits easily: ..........._________.........________ ........X2.X2XXXX000X <- last 9 digits repeat 333667 /XXXXXXX ........XXXXXX ........100XXXX ........_667334 .........33XXX60 .........XXXXXXX ..........100XXX0 <- * Same digits as below .........._667334 ...........33XXX60 ...........XXXXXXX ............XXXXXX0 ............XXXXXXX .............XXXXXX0 .............XXXXXXX ..............XXXXXX0 ..............XXXXXXX ...................XX0000 ...................XXXXXX ...................100XXX0 <- * Quote Link to comment Share on other sites More sharing options...
0 Prime Posted January 16, 2009 Report Share Posted January 16, 2009 Unfortunately, the divisor 333,667 does not fit into the pattern. Upon re-examination, the first member of geometric progression should have been xxx,xxx,xxx*10-10, rather than …*10-9. That opens another possibility for the divisor. So the sum of the geometric progression is actually xxx,xxx,xxx/9,999,999,990. The 9,999,999,990 breaks down into prime factors as 2*34*5*37*333,667 and allows another 6-digit factor, namely 333,667*2=667,334. If that was the divisor, we get: ..........._________.........________ ........11.X1XXXX0001 <- last 9 digits repeat 667334 /XXXXXXX ........667334 ........1XXXXXX ........_667334 .........XXXXXX0 .........XXXXXXX ..........1XXXXX0 <- * Same digits as below .........._667334 ...........XXXXX60 ...........XXXXXXX ............XXXXXX0 ............XXXXXXX .............XXXXXX0 .............XXXXXXX ..............XXXXXX0 ..............XXXXXXX ...................XX0000 ...................667334 ...................1X26660 <- * Quote Link to comment Share on other sites More sharing options...
0 Prime Posted January 18, 2009 Report Share Posted January 18, 2009 The divisor 667334 fits into the pattern just fine and it seems like the only solution. However, I have a retraction to make. I reasoned earlier that 333,667 and 667,334 were the only two possibilities for the divisor. That reasoning was wrong, even though it yielded the solution. Really, there are more possibilities that had to be excluded. So my solution is based on the ratio implied by the 9-digit periodical part of the fraction. Namely, the the remainder after the 3rd step R3 to the original 6-digit divisor, D, is the same ratio as the 9-digit periodical part of the fraction, P, to the number 999,999,999. R3/D = P/999,999,999. It is also useful to break that big number into its prime factors: N = 999,999,999 = 34*37*333,667. From the above ratio I reasoned, the only way to get a ratio R/D where R and D are whole numbers, is to divide N by (34*37) factors, obtaining a whole-number D=333,667. But really a whole number ratio like P/N with numbers smaller than P and N could be obtained by dividing each of those numbers by their largest common factor, F, and then multiplying each by any whole number W, as fits. That is (P*W/F) / (N*W/F) = P/N. The only two possibilities for the D, where common factor does not include 333,667 are D=333,667 and D=667,334, where the former does not fit the pattern and the latter does. For other possibilities, the common factor must include 333,667 and therefore, 9-digit periodical part P must be divisible by 333,667. The only two numbers in the 9-digit range which are divisible by 333,667 and end with "000X" (as we know it should) are 336,670,003 and 673,340,006. Both of which quite obviously don't fit the pattern. So what we have: ..........._________.........________ ........11.X1XXXX0001 <- last 9 digits repeat 667334 /XXXXXXX ........667334 ........1XXXXXX ........_667334 .........XXXXXX0 .........XXXXXXX ..........1X26660 <- * Same digits as below (R3) .........._667334 ...........XXXXX60 ...........XXXXXXX ............XXXXXX0 ............XXXXXXX .............XXXXXX0 .............XXXXXXX ..............XXXXXX0 ..............XXXXXXX ...................XX0000 ...................667334 ...................1X26660 <- * Where R3=1X2666. From here we can resolve some more digits. And turn to that ratio R3/D = P/N, or 1X2,666/667,334 = P/999,999,999. After a little bit of trial and elimination we find the only candidates for R3 and P: R=112,666 and P=168,830,001. The rest resolves relatively easy. The fraction is 7,752,341 / 667,334 = 11.6[168830001]... That was a tough assignment. What's the time limit for solving a problem like that? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 19, 2009 Author Report Share Posted January 19, 2009 This one has no time limit... Four gold stars. Great job. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
You've had the warm-up. Now this.
This may be the most difficult puzzle of its type: no starting numbers are given.
[The zeros simply remind that you're beyond the last digit of the dividend.]
The quotient of course has a decimal point [not indicated] but it's easy to see where it goes.
The last nine digits repeat.
The calculation is not shown beyond the point [2nd asterisk] where the process repeats [from first asterisk]
Enjoy!
..........._________.........________
........XXXXXXXXXXXX <- last 9 digits repeat
XXXXXX /XXXXXXX
........XXXXXX
........XXXXXXX
........_XXXXXX
.........XXXXXX0
.........XXXXXXX
..........XXXXXX0 <- * Same digits as below
.........._XXXXXX
...........XXXXXX0
...........XXXXXXX
............XXXXXX0
............XXXXXXX
.............XXXXXX0
.............XXXXXXX
..............XXXXXX0
..............XXXXXXX
...................XX0000
...................XXXXXX
...................XXXXXX0 <- *
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