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john !!
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#1
Report Thread starter 15 years ago
#1
sum of the digits of

444,444,444,444,445^2 - 555,555,555,555,555^2

sorry edited by accident
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MdSalih
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#2
Report 15 years ago
#2
-111111111111111111111111111111

MdSalih
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john !!
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#3
Report Thread starter 15 years ago
#3
that;'s not the sum of digits though :-)
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misshn
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#4
Report 15 years ago
#4
The number is -11...1(14 digits) x 10^16
So the sum of digits is 14
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john !!
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#5
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#5
well done

harder question
what three digit number divides (2^32 + 1)?
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RichE
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#6
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#6
(Original post by mik1w)
well done

harder question
what three digit number divides (2^32 + 1)?
641 divides F_5

(c) Mr Euler
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J.F.N
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#7
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#7
(Original post by mik1w)
well done

harder question
what three digit number divides (2^32 + 1)?
The more interesting question is how do we know that F_5 is divisible by 641?

Answer:
Let a=2^7, b=5, so that a - b^3 = 3, 1+ab-b^4=1+3b=2^4

Therefore,

2^(2^5) + 1 = (2a)^4 + 1 = (1 + ab - b^4)a^4 + 1 = (1+ab)a^4 + 1 - (a^4)(b^4), and this must be divisible by 1+ab=2^4 + 5^4=641
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trigtrig
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#8
Report 15 years ago
#8
(Original post by MdSalih)
-111111111111111111111111111111

MdSalih
Just wondering why imperial is your first choice, not Cambridge?
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Gentilhomme
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#9
Report 15 years ago
#9
i am wondering as well
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