S _ L _ T _ _ N
F _ R
4
N _ M B _ R S :
1
+
1
+
2
+
4 =
1
X
1
X
2
X
4 .
S _ L _ T _ _ N
F _ R
5
N _ M B _ R S :
1
+
1
+
1
+
2
+
5 =
1
X
1
X
1
X
2
X
4
1
+
1
+
1
+
3
+
3 =
1
X
1
X
1
X
2
X
4
1
+
1
+ 2
+
2
+
2 =
1
X
1
X
2
X
2
X
2 Clue
20 TIMES. (8, 18, 28, 38, 48, 58, 68, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 98) SOLUTION FOR 4 NUMBERS: 1 + 1 + 2 + 4= 1 X 1 X 2 X 4. SOLUTION FOR 5 NUMBERS: 1 + 1 + 1 + 2 + 5= 1 X 1 X 1 X 2 X 4 1 + 1 + 1 + 3 + 3= 1 X 1 X 1 X 2 X 4 1 + 1 +2 + 2 + 2= 1 X 1 X 2 X 2 X 2 THE NUMBERS CAN BE GROUPED BY PAIRS: 999,999,999 AND 0; 999,999,998 AND 1′ 999,999,997 AND 2; AND SO ON.... THERE ARE HALF A BILLION PAIRS, AND THE SUM OF THE DIGITS IN EACH PAIR IS 81. THE DIGITS IN THE UNPAIRED NUMBER, 1,000,000,000, ADD TO 1. THEN: (500,000,000 X 81) + 1= 40,500,000,001 THE SECRET CODE IS 7,4,6,5,8. LET THE NUMBERS = A, B, C, D, E BASED ON 1ST CLUE: E + C = 14 BASED ON 2ND CLUE: A = 2B - 1 BASED ON 3RD CLUE: D = B + 1 BASED ON 4TH CLUE: B + C = 10 BASED ON 5TH CLUE: A + B + C + D + E = 30 ALL NUMBERS CAN BE REPRESENTED BY B EXCEPT FOR E SO WE NEED TO REWRITE EQUATION 1. SUBSTITUTE C = 10 - B FROM EQUATION 4 INTO EQUATION 1. E + (10 - B) = 14 E = B + 4 SUBSTITUTE A,C,D,E IN EQUATION 5 (2B - 1) + B + (10 - B) + (B + 1) + (B + 4) = 30 SOLVING FOR B GIVES 4. USE THE OTHER EQUATIONS TO FIND THE VALUES OF A,C,D AND E