Jake wrote six consecutive numbers on six white pieces of paper, one number on each piece. He stuck these bits of paper onto the top and bottom of three coins. Then he tossed these three coins three times. On the first toss, he saw the numbers 6, 7 and 8 and then coloured them red. On the second toss, the sum of the numbers he saw was 23 and on the third toss the sum was 17. What was the sum of the numbers on the remaining three white pieces of paper?
A 18 B 19 C 23 D 24 E 30
A
The sum of the three numbers visible on the first toss was 6 + 7 + 8 = 21. Therefore, since the sum of the numbers visible on the second toss was 23, which is greater than 21, and the sum of the numbers visible on the third toss was 17, which is less than 21, Jake’s list of consecutive numbers included at least one number greater than 8 and at least one number less than 6. Hence the two possibilities for his six consecutive numbers are 4, 5, 6, 7, 8, 9 and 5, 6, 7, 8, 9, 10. However, for the second list, the minimum total of three different numbers is 5 + 6 + 7 = 18 which is more than the observed third total. Therefore the six numbers Jake used were 4, 5, 6, 7, 8 and 9 and hence the sum of the three numbers on the three remaining white pieces of paper was 4 + 5 + 9 = 18.
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