set@bnl.UUCP (William Tatum) (08/18/84)
[the wiser of the wisemen will know the answer] here is a problem that gave me a little trouble.here it is for your enjoyment: what is the smallest whole number that,when divided by two leaves a remainder of 1;when divided by three leaves the remainder of two;and so on,and when divided by 10 leaves a remainder of nine????? comments? set
keith@seismo.UUCP (Keith Bostic) (08/19/84)
Bill, I think you're looking at this the wrong way: ====================================================================== tmp<159> @ c = 9 tmp<160> while 1 ? if ($c % 10 == 9 && $c % 9 == 8 && $c % 8 == 7 && $c % 7 == 6 && \ ? $c % 6 == 5 && $c % 5 == 4 && $c % 4 == 3 && $c % 3 == 2 && \ ? $c % 2 == 1) then ? echo "it's $c" ? break ? else @ c += 10 ? endif ? end it's 2519 tmp<161> See what I mean? Keith ARPA: keith@seismo UUCP: seismo!keith
mwg@mouton.UUCP (08/20/84)
++ Problem: Find N s.t. N/m yields a remainder of m-1 for All m in [2,10]. Solution: N+1 must be a multiple of all integers in [2..10], so N+1 = 2*2*2*3*3*5*7 -> N = 2519.
eklhad@ihnet.UUCP (K. A. Dahlke) (08/20/84)
The cheap way. main(){ register i,j; for(i=1;;++i){ for(j=2;j<11;++j) if(i%j!=j-1) break; if(j==11){ printf("%d\n",i); exit(0); } } } If you are not into computers, try the chinese remainder theorem. x%5=4 x%7=6 x%8=7 x%9=8 bc(1) or a calculator makes short work of this. -- Karl Dahlke ihnp4!ihnet!eklhad
mr@hou2h.UUCP (M.RINDSBERG) (08/21/84)
Here is a small program to find the number. Hard one ???????? #define MAXNUM 10000 main() { char a[MAXNUM]; int i, j; for(i = 0; i < MAXNUM; i++) a[i] = 0; for(j = 2; j <= 10; j++){ for(i = 0; i < MAXNUM; i++) if( i % j != j - 1) a[i] = 1; } for(i = 0; i < MAXNUM; i++) if(a[i] == 0){ printf("Number is %d\n",i); exit(0); } }
gary@mit-eddie.UUCP (Gary Samad) (09/01/84)
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Oops. Mine was an obvious correct answer but I didn't notice that
the smallest was required.
Gary Samad