题目信息
Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10 ?
A:Three
B:Four
C:Five
D:Six
E:Seven
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正确答案：
E:Seven
Arithmetic Elementary combinatorics
To simplify the discussion, we will refer to the drawing of the slip of paper with the integer n written on it as “drawing the integer n.” The number of integers that must be drawn is at least seven, because if the six integers 0 through 5 were drawn, then no two of the integers drawn will have a sum of 10. In fact, it is easy to see that the sum of any two of these six integers is less than 10.
0, 1, 2, 3, 4, 5
Of the answer choices, only seven is not eliminated.
Although it is not necessary to show that seven is the least number of integers that must be drawn to ensure there exists a pair of the drawn integers that has a sum of 10, we provide a proof that seven is the least such number. Thus, we will show that if seven integers were drawn, then there exists a pair of the drawn integers that has a sum of 10. Since the integer 0 is such that none of the other integers can be paired with 0 to give a sum of 10, and similarly for the integer 5, it will suffice to show that if five integers were drawn from the eight integers 1, 2, 3, 4, 6, 7, 8, and 9, then there exists a pair of the drawn integers that has a sum of 10. Note that each of these eight integers differs from 5 by one of the numbers 1, 2, 3, or 4, as shown below.
With these preliminaries out of the way, assume that five integers have been drawn from these eight integers. Of the five integers that have been drawn, at least two must differ from 5 by the same number, say k, and since these two integers must be different, it follows that one of these two integers is 5 + k and the other is 5 − k, and hence these two integers have a sum of 10.
The correct answer is E.
To simplify the discussion, we will refer to the drawing of the slip of paper with the integer n written on it as “drawing the integer n.” The number of integers that must be drawn is at least seven, because if the six integers 0 through 5 were drawn, then no two of the integers drawn will have a sum of 10. In fact, it is easy to see that the sum of any two of these six integers is less than 10.
0, 1, 2, 3, 4, 5
Of the answer choices, only seven is not eliminated.
Although it is not necessary to show that seven is the least number of integers that must be drawn to ensure there exists a pair of the drawn integers that has a sum of 10, we provide a proof that seven is the least such number. Thus, we will show that if seven integers were drawn, then there exists a pair of the drawn integers that has a sum of 10. Since the integer 0 is such that none of the other integers can be paired with 0 to give a sum of 10, and similarly for the integer 5, it will suffice to show that if five integers were drawn from the eight integers 1, 2, 3, 4, 6, 7, 8, and 9, then there exists a pair of the drawn integers that has a sum of 10. Note that each of these eight integers differs from 5 by one of the numbers 1, 2, 3, or 4, as shown below.
| 1 = 5 − 4 2 = 5 − 3 3 = 5 − 2 4 = 5 − 1 |
6 = 5 + 1 7 = 5 + 2 8 = 5 + 3 9 = 5 + 4 |
With these preliminaries out of the way, assume that five integers have been drawn from these eight integers. Of the five integers that have been drawn, at least two must differ from 5 by the same number, say k, and since these two integers must be different, it follows that one of these two integers is 5 + k and the other is 5 − k, and hence these two integers have a sum of 10.
The correct answer is E.
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