# maths question again (try this)

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#1
sum of the digits of

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

sorry edited by accident
0
15 years ago
#2
-111111111111111111111111111111

MdSalih
0
#3
that;'s not the sum of digits though :-)
0
15 years ago
#4
The number is -11...1(14 digits) x 10^16
So the sum of digits is 14
0
#5
well done harder question
what three digit number divides (2^32 + 1)?
0
15 years ago
#6
(Original post by mik1w)
well done harder question
what three digit number divides (2^32 + 1)?
641 divides F_5

(c) Mr Euler
0
15 years ago
#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?

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
0
15 years ago
#8
(Original post by MdSalih)
-111111111111111111111111111111

MdSalih
Just wondering why imperial is your first choice, not Cambridge?
0
15 years ago
#9
i am wondering as well
0
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