Talk:Rare numbers: Difference between revisions

===20 mins not 20 years===

If Shyam Sunder Gupta has really spent 20 years on this he should have stayed in bed. Let me spend 20mins on it. The following took 9mins so any questions and I have 11mins to spare.

<pre>

Let me consider n-r=l for a 2 digit number ng n<g. Then l=(10g+n)-(g+10n)=9(g-n) where n is 0..8 and g is 1..9.

l is one of 9 18 27 36 45 54 63 72 81. l must be a perfect square so only 9 36 and 81 are of interest.

9 -> ng=89 78 67 56 45 34 23 12 01

36-> ng=59 48 37 26 15 04

81-> ng=09

For each of these candidate ng I must determine if ng+gn is a perfect square.

09+90 99 n

59+95 154 n

48+84 132 n

37+73 110 n

26+62 114 n

15+51 66 n

04+40 40 n

89+98 187 n

78+87 165 n

67+76 143 n

56+65 121 y

45+54 99 n

34+43 77 n

23+32 55 n

12+21 33 n

01+10 11 n

From which I see that 65 is the only Rare 2 digit number.

I love an odd number of digits. Let me call the 3 digit number nxg. l=(100g+10x+n)-(g+10x+100n).

x disappears and I am left with 99(g-n). None of 99 198 297 396 495 594 693 792 or 894 are perfect squares.

So there are no Rare 3 digit numbers.

At 4 I begin to think about using a computer. Consider nige. l=(1000e+100g+10i+n)-(e+10g+100i+1000n).

I need a table for l as above for 9(111(e-n)+10(g-i)) where n<=e and if n=e then i<g.

Before turning the computer on I'll add that for 5 digits nixge l=(10000e+1000g+100x+10i+n)-(e+10g+100x+1000i+10000n).

x dissappears leaving 99(111(e-n)+10(g-i))

</pre>--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 13:34, 12 September 2019 (UTC)