 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Shiny Numbers (Posted on 2013-06-20) A positive integer is called shiny if it can be written as the sum of two not necessarily distinct integers a and b which have the same sum of their digits. For instance, 2012 is shiny, because 2012 = 2005 + 7, and both 2005 and 7 have the same sum of their digits. Find all positive integers which are not shiny (the dark integers).

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Some thoughts Comment 6 of 6 | Let the number being investigated be n.
Let the sum of digits of n be SOD(n)
Let a=(n-1)/2
Let b=(n+1)/2
I   SOD(a)+SOD(b)=SOD(n)+9r, r in Z
On each of k steps, add 1 to a and deduct 1 from b
II   SOD(a+k)+SOD(b-k)=SOD(n)+9s, s in Z
Let k=5
Now (a+5)=(b-5)+9;
And SOD(a+k)+SOD(b-k)=SOD(n) or SOD(n)+/-9s (II)
This is necessarily a solution unless:
(i)  (b-5) is negative (i.e.1,3,5,7); or
(ii) The number of zeroes in (a+5) or (b-5) has changed (the converse is not implied) (III)

Next, add 9 to  (a+5), and deduct 9 from (b-5) .
All other things being equal, there will be no change in the values of SOD(a+14) and SOD(b-14), unless the number of zeroes in one of them has changed (IV)
If so, either:
(i) SOD(a+14) = SOD(b-14), or
(ii) SOD(a+14) = SOD(a+5), and SOD(b-14) = SOD(b-5), in which case either:
(ii)(a) SOD(a+23) = SOD(b-23), or
(ii)(b) The number of zeroes in SOD(a+23) or SOD(b-23) has changed (the converse is not implied).

Obviously this process can be repeated until a solution is found or (b-9k-5)=0

e.g. 797:
k=0, {(a+5), (b-5), SOD(a+5), SOD(b-5)} = {403,394, 7,16} (III)
k=1, {(a+5), (b-5), SOD(a+14), SOD(b-14)}= {412,385,7,16}
k=2, {(a+23), (b-23),SOD(a+23), SOD(b-23} = {421,376,7,16}
k=3, {(a+32), (b-32),SOD(a+32), SOD(b-32} = {430,367,7,16} (IV)
k=4, {(a+41), (b-41),SOD(a+41), SOD(b-41} = {439,358,16,16} (IV(ii)(a))

e.g. 799:
k=0, {(a+5), (b-5), SOD(a+5), SOD(b-5)} ={404,395,8,17} (III)
...
k=41, {(a+41), (b-41), SOD(a+5), SOD(b-5)} ={440,359,8,17} (IV)
but:
k=50, {(a+50), (b-50), SOD(a+50), SOD(b-50)} ={449,350,17,8} (IV(ii)(b))
k=95, {(a+95), (b-95), SOD(a+95), SOD(b-95)} ={494,305,17,8} (IV)
But:
k=104, {(a+104), (b-104), SOD(a+104), SOD(b-104)} ={503,296,8,17} (IV(ii)(b))
... And so on down to (a+9k+5)=799, (b-k-5)=0, with no solutions.

It seems clear enough from the above that with the exception of {1,3,5,7,9} all digits but the first must be a 9, for a number to be 'dark'. Let's assume that the number of 9's is even, e.g. 99, 9999, ... etc., prefixed by no other number;
(98/2+5) = 54, (100/2-5 = 45)
(9998/2+50) = 5049, (10000/2-50 = 4950)
(999998/2+500) = 500499, (1000000/2-500 = 499500)
Generally, SOD (10^(2n)-2)/2+5*10^(n-1) = SOD (10^(2n))/2-5*10^(n-1); if the number of 9's is even, then the number is shiny. (V)

Let's assume that the even string of 9's is prefixed by an even integer: e.g. 299 splits into 149 and 150; 100 is much larger than 5, so the addition of the prefix makes no difference.

Edited on June 22, 2013, 7:12 am
 Posted by broll on 2013-06-22 07:10:02 Please log in:

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