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Question Number 191831 by Shlock last updated on 01/May/23

Answered by mehdee42 last updated on 02/May/23

suppose , n=<abcdefghij> ,is an <i.n> a+b+c+...+i+j≡^9 0⇒9∣n 9∣n , 11111∣n , (9,11111)=1⇒99999∣n let x=<abcde> & y=f<fghij> ⇒n=10^5 x+y≡^(99999) x+y ⇒x+y≡^(99999) 0 0<x+y<2×99999⇒x+y=99999 (a+f=b+g=...=e+j=9) there are 5!=120 ways to move <abcde>. also there are 2^5 =32 ways move for pair <(a,f),...,(e,j)> on thr other hand , a≠0. so the total number intersting : (9/(10))×32×120=3456

suppose,n=<abcdefghij>,isan<i.n>a+b+c+...+i+j≡90⇒9∣n9∣n,11111∣n,(9,11111)=1⇒99999∣nletx=<abcde>&y=f<fghij>⇒n=105x+y≡99999x+y⇒x+y≡9999900<x+y<2×99999⇒x+y=99999(a+f=b+g=...=e+j=9)thereare5!=120waystomove<abcde>.alsothereare25=32waysmoveforpair<(a,f),...,(e,j)>onthrotherhand,a≠0.sothetotalnumberintersting:910×32×120=3456

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