Math Riddle: 12/15/2017 01:07:14 
TBestLittleHelper
Level 47
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Great, so let's solve this in the most inefficient way possible.
000 000 0  fail 000 000 1  fail
hm, this will take a while. Maybe someone will be able to do it faster then me :P

Math Riddle: 12/15/2017 03:02:26 
Padraig
Level 50
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What are the possible numbers we're looking for? Do you mean the unique digits or the number, or numbers made up of those unique digits? It would matter how long that number made up of the unique digits was. I am satisfied that I know what the unique digits are.

Math Riddle: 12/15/2017 04:22:19 
knyte
Level 58
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We're looking for a number with 7 unique digits. The sum of the first seven digits is 40. The sum of the seven middle digits is 30, the product is 0. The sum of the five middle digits is 25, the product is 448. The sum of the last 7 digits is 32, the product is 6400. What are the possible numbers we're looking for? 8790178185541 Digits: 0 1 4 5 7 8 9 (7 uniques) First 7: 8 7 9 0 1 7 8 (sum: 40) Middle 7: 0 1 7 8 1 8 5 (sum: 30; product: 0) Middle 5: 1 7 8 1 8 (sum: 25; product: 448) Final 7: 8 1 8 5 5 4 1 (sum: 32; product: 6400) Just one of the possibilities (I'm going to leave most of the work undone and turn this in for partial credit so I don't get nerdsniped any further). To generate the others, you can rearrange some digits (like 879) in here. Middle 5 have to be 1, 1, 7, 8, 8 (other factorizations of 64 do not add up to 25). Then you need a 0 on one side and a 5 on the other, since the sum of the middle 7 is 30 (25+5) but the product is 0 (so one of the digits has to be 0). Since 6400 is a multiple of just 2's and 5's, the last 7 digits need to be 4, 8, 8, 5, 5, and 2 1's. So the 5 has to go on the right of the middle 5. The first 3 are relatively easier since you just need to throw in 3 numbers that equal 24 once you've moved the appropriate digits in the middle 5 out o the first 7. @Tabby: Are you sure of that? I think the restrictions placed here might be sufficient to reduce this to a finite subset of integers.
Edited 12/15/2017 04:31:54

Math Riddle: 12/15/2017 05:54:14 
Hostile
Level 58
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First > 40 Middle > 30, 0 ; 25, 448 Last > 32, 6400
First > 9,8,7,6,5,4,1 = 40 > 9,8,7,6,5,3,2 = 40
Second > 0,X,X,X,X,X,y > sum of seven  sum of five = 30  25 so 5. > 0,X,X,X,X,X,5 > Factors(448) => 2*2*2*2*2*2*7 > 0,7,X,X,X,X,5 > 0,7,2,4,8,1,5
Last > 32;6400 > Factors(6400) => (5*2)*(5*2)*(2*4*8) > 4 short ??? > 5,5,4,8,8,1,1
nonunique digits (?)

Math Riddle: 12/15/2017 07:52:33 
Padraig
Level 50
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If we allow the first, middle, and last to overlap I believe the shortest solution would be 17 digits long.
For all solutions the number of sets of unique digits that would satisfy the conditions is 4, just as Bane hinted at.
8,7,5,4,1,0 & a seventh digit which could be 9,6,3, or 2.
{9,8,7,5,4,1,0} {8,7,5,4,3,1,0} {8,7,6,5,4,1,0} {8,7,5,4,2,1,0}
So it seems to me.
Edited because at 2:30 AM I should have been sleeping  heck!
Edited 12/15/2017 17:48:18

 downvoted post by Hot Brick
Math Riddle: 12/15/2017 14:39:20 
Hot Brick
Level 17
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I'm feeling too smart for this.

Math Riddle: 12/16/2017 09:54:46 
Derfellios
Level 59
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Farah dont cheat. For everyone who has never seen this. This is a problem from the aivd (Dutch secret service) christmas puzzle. Farah, try to connect it with what remains from ABDEJRSTU FKNPS DGHIKLRT BDGHIJNPRST BGINPRSTU ABERU DEKLNRSVY

Math Riddle: 12/16/2017 10:21:03 
Derfellios
Level 59
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I have not tried to solve that one. But good luck with the other questions!
Edited 12/16/2017 10:55:51

Math Riddle: 12/16/2017 10:28:45 
Derfellios
Level 59
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This edition is really difficult and long and less points for each question. We have about 10 points so far but I dont think I can reach the 70 we had last year

Math Riddle: 12/16/2017 13:49:35 
Hot Brick
Level 17
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I solved it days ago derfellios ;) Also, this isn't the exact same puzzle as the AIVD gave us. You work in the AIVD?

Math Riddle: 12/16/2017 15:43:31 
kyte
Level 24
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Man, Dutch intelligence must be seriously underfunded if they have to crowdsource solutions from Warzone.

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