9,999 different combos well 10,00 if you add 0 in with it
Let first COI=n=3 second COI=n+2=5 third COI=n+4=7 (n)(n+2)=n+4+8 check: n^2+2n=n+12 n^2+n-12=0 -n-12 -n-12 -4^2+(-4)-12=0 n^2+2n-n-12=0 16+(-4)-12=0 n^2+n-12=0 12-12=0 (n+4)(n-3)=0 0=0 n+4=0 n-3= 0 n^2+n-12=0 -4 -4 +3 +3 3^2+3-12=0 n=-4 n=3 9+3-12=0 reject 12-12=0 because is negative and it needs to be positive 0=0
x2 - 18x + 72 = 0 (x - 6)(x - 12) = 0 x ∈ {6, 12}
3(0)^2+7(x)-12 = -12
2y - 4x = -12 Algebraically rearrange to 2y = 4x - 12 Divide both sides by '2' y = 2x - 6 So at the 'y' intercept, x = 0 Substituting x = 0 We have y = 2(0) - 6 Anything multiplied '0' is equal; to '0' Hence y = 0 - 6 y = -6 is the value of the y-intercept.
#include <stdio.h> int main (void) { puts ("1 22 333 4444 55555"); return 0; }
772
printf ("12345 1234 123 12 1\n"); ... or, did you mean to do it with loops ? ... int i, j; for (i=5; i>0; i--) { for (j=1; j<=i; j++) printf ("%d", j); printf (" "); } printf ("\n");
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