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
THE CORRECT COMBINATION IS 65292. SINCE THE THIRD DIGIT IS THREE LESS THAN THE SECOND, AND THE FOURTH IS FOUR GREATER THAN THE SECOND, THERE ARE ONLY THREE POSSIBLE COMBINATIONS FOR THE SECOND, THIRD AND FOURTH DIGITS. THESE ARE -307-, -418-, AND -529-. WITH THE FIRST DIGIT THREE TIMES THE FIFTH, THE ONLY POSSIBLE COMBINATIONS FOR THE FIRST AND FIFTH DIGITS ARE 0 0,3 1, 6 2, AND 9 3. THE SOLUTION ARISES FROM COMBINING THESE TWO SETS OF POSSIBILITIES, WITH THE ADDED CRITERIA THAT THERE ARE THREE COMBINATIONS OF TWO DIGITS THAT THAT EACH SUM TO 11 THE BUTLER DID IT. MR. ELLIS KNEW THE BUTLER WAS LYING BECAUSE PAGES 35 AND 36 IN A BOOK ARE ALWAYS PRINTED ON OPPOSITE SIDES OF THE SAME PIECE OF PAPER 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 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