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# Really hard maths question

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1. this is the question:

Sir Issac Newton died in the year x^3+1 and was 7x years old when he died( He had this birthday in the year that he died). In which year was he born?

(4 marks)
2. Don't tell me this is a real question
3. Maybe ask the guys in the A-Level forum? They would find this easy. I've tried, but the furthest I can go is x^3+1-7x = year he was born
4. (Original post by Gabzinc)
Don't tell me this is a real question
trust me it is a real question, its so difficult

(Original post by 13 1 20 8 42)
what?
5. 2027
6. can any of you guys solve this ? Zacken RDKGames SeanFM Ano123
7. (Original post by theBranicAc)
can any of you guys solve this ? Zacken RDKGames SeanFM Ano123
2027
8. (Original post by LionKing101)
2027
how did you work it out?
9. No it's 1876
10. This isn't that hard...
http://www.wolframalpha.com/input/?i...y+,+y+%3E+1000 gives you the answer. If you want to do it by hand, choose a reasonable search space based on the values of x^3, then use trial and error for different values of x to find the value of y. For example: I know that Newton was born in the middle of the 2nd millennium, and, obviously, 10^3 = 1000. Knowing that, I can plug in x = 11 into the equation (because 10^3 - 7*10 + 1 will be lower than 1000), get y = 1255 (which is too early), then try x = 12 and get the right answer. So, trial and error, but knowing where to start means you only have to try 2 values of x.
1642

Source: https://en.wikipedia.org/wiki/Isaac_Newton
12. I'm getting so many different answer yet I haven't see a proper algebraic solution
13. (Original post by beam3.14159)
This isn't that hard...
http://www.wolframalpha.com/input/?i...y+,+y+%3E+1000 gives you the answer. If you want to do it by hand, choose a reasonable search space based on the values of x^3, then use trial and error for different values of x to find the value of y. For example: I know that Newton was born in the middle of the 2nd millennium, and, obviously, 10^3 = 1000. Knowing that, I can plug in x = 11 into the equation (because 10^3 - 7*10 + 1 will be lower than 1000), get y = 1255 (which is too early), then try x = 12 and get the right answer. So, trial and error, but knowing where to start means you only have to try 2 values of x.
you only got the working out to the answer, after you saw the answer, but in a test you wont be able to do this
14. (Original post by theBranicAc)
I'm getting so many different answer yet I haven't see a proper algebraic solution
How do you define proper algebraic solution? I showed you how to find the date mathematically...
15. (Original post by beam3.14159)
How do you define proper algebraic solution? I showed you how to find the date mathematically...
yes but i don't understand your solution
16. appears to be looking for an integer solution.

if you've got a calculator you could estimate he died shortly after the great fire of london which was in 1666 (he was alive around that period)

cuberoot (1666) = 11.855

round up to 12
17. You should probably post this in the 'Maths Forum' section -It's easier to receive help that way http://www.thestudentroom.co.uk/forumdisplay.php?f=38
18. You can't solve this algebraically as a stand alone question because there isn't enough information. You have to use some general knowledge and also assume that it's an integer solution and then use the trial and error already discussed. I do wonder where you found such a question..
edit: Actually I don't think you have to assume it is an integer because x^3 must be an integer, and then x is either an integer or irrational, and if it is irrational then 7x is irrational which is stupid, but I doubt you are expected to realise this given the level of the exam I have been shown this question is on..
19. (Original post by 13 1 20 8 42)
You can't solve this algebraically as a stand alone question because there isn't enough information. You have to use some general knowledge and also assume that it's an integer solution and then use the trial and error already discussed. I do wonder where you found such a question..
Question 7 https://www.mcsoxford.org/resource.aspx?id=7599

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