题目信息
If Whitney wrote the decimal representations for the first 300 positive integer multiples of 5 and did not write any other numbers, how many times would she have written the digit 5 ?
A:150
B:185
C:186
D:200
E:201
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正确答案：
E:201
Arithmetic Properties of integers
The number of times the digit 5 would be written is the number of times the digit 5 will appear in the units place plus the number of times the digit 5 will appear in the tens place plus the number of times the digit 5 will appear in the hundreds place.
Number of times the digit 5 will appear in the units place: This will be the number of terms in the sequence 5, 15, 25, 35, …, 1485, 1495. Adding 5 to each member of this sequence does not change the number of terms, and doing this gives the sequence 10, 20, 30, 40, …, 1490, 1500, which clearly has 150 terms (dividing the terms by 10 gives 1, 2, 3, 4, …, 149, 150). Thus, the digit 5 appears 150 times in the units place.
Number of times the digit 5 will appear in the tens place: This will be the number of terms in the sequence 50, 55, 150, 155, 250, 255, …, 1450, 1455. The digit 5 appears in the tens place twice for each consecutive change in the hundreds digit. Thus, the digit 5 appears 2(15) = 30 times in the tens place.
Number of times the digit 5 will appear in the hundreds place: This will be the number of terms in the sequence 500, 505, 510, 515, …, 590, 595, 1500. Thus, the digit 5 appears 20 + 1 = 21 times in the hundreds place.
Therefore, the number of times the digit 5 would be written is 150 + 30 + 21 = 201.
Tip The method used above to count the number of terms in the sequence 5, 15, 25, 35, …, 1,485, 1,495 can be applied to any arithmetic sequence, and it avoids the necessity of remembering certain formulas. For example, to determine the number of terms in the sequence 13, 19, 25, 31, 37, 43, …, 301, we first observe that consecutive differences are equal to 6, so we subtract from each term a number chosen so that the first term becomes (1)(6) = 6. Thus, we subtract 7 from each term and obtain the sequence 6, 12, 18, 24, 30, 36, …, 294, which has the same number of terms as the original sequence.The number of terms in this new sequence is now easy to find—divide each term of this new sequence by 6, and it will be clear that the number of terms is 49.
Alternatively, in the 2-digit multiples of 5, namely the multiples of 5 in the interval 5−95, there are twelve occurrences of the digit 5. The same number of occurrences of the digit 5 appear in the multiples of 5 in each of the intervals 100−195, 200−295, 300−395, and 400−495. For the multiples of 5 in the interval 500−595, there are the same corresponding twelve occurrences of the digit 5 plus twenty more for the digit in hundreds place for each of the twenty multiples of 5 in 500−595, for a total of thirty-two occurrences of the digit 5. For the multiples of 5 in each of the intervals 600−695, 700−795, 800−895, 900−995, 1,000−1,095, 1,100−1,195, 1,200−1,295, 1,300−1,395, and 1,400–1,495, there are twelve occurrences of the digit 5. Finally, there is one occurrence of the digit 5 in 1,500. Therefore, the total number of occurrences of the digit 5 in the first 300 multiples of 5 is 14(12) + 32 + 1 = 201.
The correct answer is E.
The number of times the digit 5 would be written is the number of times the digit 5 will appear in the units place plus the number of times the digit 5 will appear in the tens place plus the number of times the digit 5 will appear in the hundreds place.
Number of times the digit 5 will appear in the units place: This will be the number of terms in the sequence 5, 15, 25, 35, …, 1485, 1495. Adding 5 to each member of this sequence does not change the number of terms, and doing this gives the sequence 10, 20, 30, 40, …, 1490, 1500, which clearly has 150 terms (dividing the terms by 10 gives 1, 2, 3, 4, …, 149, 150). Thus, the digit 5 appears 150 times in the units place.
Number of times the digit 5 will appear in the tens place: This will be the number of terms in the sequence 50, 55, 150, 155, 250, 255, …, 1450, 1455. The digit 5 appears in the tens place twice for each consecutive change in the hundreds digit. Thus, the digit 5 appears 2(15) = 30 times in the tens place.
Number of times the digit 5 will appear in the hundreds place: This will be the number of terms in the sequence 500, 505, 510, 515, …, 590, 595, 1500. Thus, the digit 5 appears 20 + 1 = 21 times in the hundreds place.
Therefore, the number of times the digit 5 would be written is 150 + 30 + 21 = 201.
Tip The method used above to count the number of terms in the sequence 5, 15, 25, 35, …, 1,485, 1,495 can be applied to any arithmetic sequence, and it avoids the necessity of remembering certain formulas. For example, to determine the number of terms in the sequence 13, 19, 25, 31, 37, 43, …, 301, we first observe that consecutive differences are equal to 6, so we subtract from each term a number chosen so that the first term becomes (1)(6) = 6. Thus, we subtract 7 from each term and obtain the sequence 6, 12, 18, 24, 30, 36, …, 294, which has the same number of terms as the original sequence.The number of terms in this new sequence is now easy to find—divide each term of this new sequence by 6, and it will be clear that the number of terms is 49.
Alternatively, in the 2-digit multiples of 5, namely the multiples of 5 in the interval 5−95, there are twelve occurrences of the digit 5. The same number of occurrences of the digit 5 appear in the multiples of 5 in each of the intervals 100−195, 200−295, 300−395, and 400−495. For the multiples of 5 in the interval 500−595, there are the same corresponding twelve occurrences of the digit 5 plus twenty more for the digit in hundreds place for each of the twenty multiples of 5 in 500−595, for a total of thirty-two occurrences of the digit 5. For the multiples of 5 in each of the intervals 600−695, 700−795, 800−895, 900−995, 1,000−1,095, 1,100−1,195, 1,200−1,295, 1,300−1,395, and 1,400–1,495, there are twelve occurrences of the digit 5. Finally, there is one occurrence of the digit 5 in 1,500. Therefore, the total number of occurrences of the digit 5 in the first 300 multiples of 5 is 14(12) + 32 + 1 = 201.
The correct answer is E.
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