Viewing a response to: @rycharde/brainsteem-mathematics-challenges-fibonacci-2018
All right. If we start with G(0) as first therm it would be G(9) which would be after the formula I pointed out in another comment 2018=21a+34b again 2018 mod 55 is 38 2018-38=1980 1980/55=36 so the a has to be 36 or less. 2018-21a=34b 1009-21a/2=17b a/2=c-> cmax=18 (1009-21c)/17=b for a+b minimal a has to be as small as posible since b has a bigger impact (2ax=34 x has to be bigger than 1.5) So you can either make a table or test for each number for c beginning at 1 and stopping as soon as b is an integer. I dicided to go with the last option. I won't post the boring calculations. It turns out it works for c=10 and so b is 47 and a is 20. I checked even further and it turns out there is no other possible pair smaler than c=18 and if c larger 18 a smaller b would not apply.
author | sammy111100 |
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rycharde | 0 | 38,598,294,562 | 51% | ||
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Thanks, upvoted two of you!
author | rycharde |
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Thanks for the problem. I wish they'd give us similarily hard problems in school. Hope to see another one from you somtimes.
author | sammy111100 |
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Most countries have national mathematics competitions at various levels, but it needs the school to actually supervise the tests. So you need to ask your head of maths to apply and get those interested students to practise such questions. :-)
author | rycharde |
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