# Maths Problem

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#61

(Original post by

That'll do, although i was thinking of a nicer method imho.

**theone**)That'll do, although i was thinking of a nicer method imho.

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#62

(Original post by

That'll do for you then it'll do for me, yay I got one right! Well sort of... please enlighten us theone but don't be all "haha I know everything and you suck" about it

**ZJuwelH**)That'll do for you then it'll do for me, yay I got one right! Well sort of... please enlighten us theone but don't be all "haha I know everything and you suck" about it

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#63

Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.

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(Original post by

Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.

**mikesgt2**)Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.

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#65

**mikesgt2**)

Why are we looking for a proof? It is bloody obvious that there are no solutions when there a 4 or more digits.

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#66

(Original post by

so what's your method?

**elpaw**)so what's your method?

Let n have x digits.

It is clear that s(n) =< 9x and n >= 10^(x-1)

So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)

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(Original post by

It is clear that s(n) =< 9x and n >= 10^(x-1)

**theone**)It is clear that s(n) =< 9x and n >= 10^(x-1)

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#68

(Original post by

Umm how is that clear?

**2776**)Umm how is that clear?

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#69

(Original post by

Let n be the integer and s(n) the sum of the digits of n.

Let n have x digits.

It is clear that s(n) =< 9x and n >= 10^(x-1)

So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)

**theone**)Let n be the integer and s(n) the sum of the digits of n.

Let n have x digits.

It is clear that s(n) =< 9x and n >= 10^(x-1)

So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)

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#70

(Original post by

Too small and concise for me, please explain to a future UEL graduate <vomits> how all that works...

**ZJuwelH**)Too small and concise for me, please explain to a future UEL graduate <vomits> how all that works...

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(Original post by

Because n has x digits. and s(n) is maximised with each digit being 9, so s(x) <= 9x. Also we know n has x digits and so must be greater than or equal to 10^(x-1).

**theone**)Because n has x digits. and s(n) is maximised with each digit being 9, so s(x) <= 9x. Also we know n has x digits and so must be greater than or equal to 10^(x-1).

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#72

(Original post by

umm how is it that n=> 10 ^(x-1) ?

**2776**)umm how is it that n=> 10 ^(x-1) ?

Now n is minimised with b=c=d=... = 0 and a=1 since a must be > 0.

So n is minimised with n = 10^x-1.

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(Original post by

So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)

**theone**)So if 10^x-1 > 99x (i.e. 11s(n)) then n > 11s(n) and the inequality holds with x >=4 (you can prove this by induction if u need convincing)

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(Original post by

Can u explain this bit please oh angel of maths

**2776**)Can u explain this bit please oh angel of maths

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#75

(Original post by

Can u explain this bit please oh angel of maths

**2776**)Can u explain this bit please oh angel of maths

10^x-1 is the minimum value of the number n with x digits. 99x is the maximum value of 11 times the sum of the digits. Therefore, the number must be greater than 11 times the sum of its digit as the minimum value of n is greater than the maximum value of 11 times the sum of the digits:

10^x-1 > 99x

This implies n>11S(n)

When x=4, then 10^x-1=10^3 and 99x=396.

Therefore, the basis case works.

Assume P(k) is true and prove P(k+1)

Therefore, 10^x-1>99x

If 10^x-10^(x-1)>99(x+1)-99x then P(k+1) will be true if P(k) is true:

10^x-10^(x-1)=[10^(x-1)](10-1)

=9(10^(x-1))

>9*99x

As x>=4, then 9*99x>=3564

99(x+1)-99x=99x+99-99x=99

=> 99(x+1)-99x<9*99x

Put back above, this gives 10^x-10^(x-1)>99(x+1)-99x

Therefore, if P(k) is true then P(k+1) is true. As the basis case P(4) it true, by induction P(x) is true where x is a positive integer greater than or equal to four.

Therefore, n>11S(n) when x>=4, so there are no solutions when the number n has more than three digits.

Nice proof theone and who have you got interviewing you at Trinity?

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