Brain-Teaser 652: Stumped!
From The Sunday Times, 6th January 1974 [link] [link]
The scoreboard at our local cricket ground shows the total number of runs scored, the number of wickets fallen and the scores of the two batsmen in at the time.
At the match last Sunday we declared at 500 for 8 wickets, when our first man in had just reached 100 runs (I was the other opener and was already out for 20). The seventh batsman had also scored a century, and there were no ‘extras’ at all in the match (i.e. all the runs were scored by the batsmen).
Some time after my dismissal I noticed that the scoreboard showed eight similar digits (e.g. 999 for 9, both men in having scored 99 each) and later the board showed eight consecutive digits in ascending order, though with 0 following 9 (e.g. 678 for 9; batsmen with 01 and 23).
What were the two scores I had seen?
This puzzle is included in the book The Sunday Times Book of Brain-Teasers: Book 2 (1981). The puzzle text above is taken from the book.
[teaser652]
S2T2 
Jim Randell 8:56 am on 1 February 2022 Permalink |
The innings was declared 500 for 8, so the options are:
For scores greater between 20 and 500 we have:
The last of these is not possible. If only one batsman has been dismissed, it is the setter for 20, so the total score would be 20 + 11 + 11 = 42, not 111.
The second time the scoreboard could be showing (assuming the “in” scores can have a leading zero):
Again the last of these is not possible, as the sum of the scores of the “in” batsmen is more than the total score.
So we have the following potential observations:
But some of these transitions are not possible.
Which leaves only the following transitions:
Consider the “333 for 3” → “456 for 7” transition:
Initially the score is “333 for 3”, and the “in” batsmen have both scored 33. So the 3 dismissed batsmen must have scored 267 between them. We know one of these is the setter, who scored 20. Which leaves 247 to be scored between the remaining 2 dismissed batsmen. We’ll distribute the runs as 123 and 124 run to batsmen 3 and 4:
By the time we get to “456 for 7”, the first batsman must have increased his score to 89, and the remaining “in” batsman has a score of 1 (let’s say this is batsman 9):
But these already sum to 490, which is more than 456, so this situation is not possible.
Consider the “222 for 2” → “456 for 7” transition:
Initially the score is “222 for 2”, and the “in” batsmen have both scored 22. So the 2 dismissed batsmen must have scored 178 between them, and we know one of them is the setter, who scored 20. So the other must have scored 158 (let’s say that was the 3rd batsman):
By the time we get to “456 for 7” the 1st batsman must have increased his score to 89, and the remaining “in” batsman has a score of 1 (let’s say this is batsman 9):
This leaves 66 runs to distribute between 4, 5, 6, 7, 8. Which is possible.
So this is the only possible situation.
Solution: The two scores are: “222 for 2, (22 and 22)” and: “456 for 7, (89 and 01)”.
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