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#1 |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
588010 Posts |
I am currently solving this crossnumber and finding it very difficult. I also thought that this puzzle would interest many on the forum.
https://issuu.com/chalkdust/docs/chalkdust-issue-03/47 |
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#2 | |
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"Forget I exist"
Jul 2009
Dumbassville
838410 Posts |
Quote:
44A gives a hint that 29D ends with an even digit also since 44A and 29D meet at the start of 44A ( the last part of 29D) and 44A is half of 29D we also know that that even digit is half of the start digit of 29D so we are looking for multiples of 7 that are even starting with either 4 or 8 and ending with either 2 or 4 respectively. which makes 44A end with 1 or 2 respectively. |
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#3 |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
101010000000012 Posts |
The old ones arre the best. Because seven ate nine.
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#4 |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
23·3·5·72 Posts |
Have solved 11D, 15A, 23D, 30A, 34D, 39A, 42A, 43D so far,
Down to two possibilities for 31D. 39D is bugging me |
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#5 |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
okay well it's 2 digits and doesn't have a sum to add to 1
let's see we can eliminate square numbers because they are n*n and n/n = 1 that eliminates 6 numbers out of 90. and 11 is taken out because it's 1 less than double a perfect number ( a number that can be defined as a number for which the sum of the reciprocal of the divisors sums to 2, the reciprocal of 1 is 1) so that also eliminates 55. so we have now taken out 8 numbers out of 90. and since xilman gave you 39 A we can cut to the chase and know that it starts with a 7. secondary edit: and the clue for 24 A shows us the sum of digits will be a two digit number and that doesn't work for secondary digits of 0,1,2, so the second digit must be greater than 2. Last fiddled with by science_man_88 on 2016-08-18 at 14:47 |
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#6 | |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
22×7×167 Posts |
Quote:
I can find a solution for every 7x number. Seems 23 is the highest 2 digit number though. |
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#7 |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
could be 678 in theory I guess or 786 or 78 <digit> or <digit>78 in which case 23 could work I'm trying to figure it out with PARI ( yeah I know I'm cheating).
Last fiddled with by science_man_88 on 2016-08-19 at 23:06 |
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#8 | |
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Sep 2014
29 Posts |
Quote:
By the way, I have attached a pdf-file which should contain the solution of the puzzle. Last fiddled with by Anonuser on 2016-08-19 at 23:22 |
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#9 | ||
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"Forget I exist"
Jul 2009
Dumbassville
203008 Posts |
Quote:
Quote:
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#10 | |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
588010 Posts |
Quote:
I made a mistake. Once I have fixed that I will post my full progress in solving the puzzle. There are some interesting programming challenges in this puzzle. |
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#11 |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
sadly I can't help you much in one sense I made my own test to see what I knew in pari ( by asking yes or no to a question like do you know about or do you know how to use ... though I was wishy-washy on some based my answers I know how to use to some extent 184/832 PARI/gp functions so under 23% is what I know.
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