Easy Numbers

Let the three integers be a, b, c.

N = abc = 9(a + b + c) and c = a + b.

N = ab(a + b) = 9(a + b + a + b) = 18 (a + b).

Since a and b are positive, ab = 18 so {a, b} is one of ( 1,18) (2, 9) (3, 6).

c= a+b is one of 19, 11, 9.

N is one of 342, 198, 162.

Sum of all values is 702.

A=B+C

ABC = 9(A+B+C)

(B+C)BC = 9(B+C+B+C)

(B+C)BC = 9(2B+2C)

(B+C)BC = 9 x 2(B+C)

(B+C)BC = 18(B+C)

BC = 18

(A,B,C) = (19,1,18) OR (11,9,2) OR (9,3,6)

THE ANSWER IS SUM OF N = (19 x 1 x 8) + (11 x 9 x 2) + (9 x 3 x 6) = 342 + 198 +162

= 702

simple trial and error program in java......the output will provide many similar values......avoid others and take one value at once........my for loop's limit is till 1000 coz brilliant.org doesn't support values after 999 ............HERE'S the PROGRAM:-

class number { void main() { for(int i=0;i<=1000;i++) { for(int j=0;j<=1000;j++) { for(int k=0;k<=1000;k++) { int sum=i+j+k; int product=i j k; if((i==j+k)||(j==i+k)||(k==i+j)) { if(product==9*sum) System.out.println(product); } } } } } }