Difference between revisions of "1970 AHSME Problems/Problem 33"
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We can find the sum using the following method. We break it down into cases. The first case is the numbers <math>1</math> to <math>9</math>. The second case is the numbers <math>10</math> to <math>99</math>. The third case is the numbers <math>100</math> to <math>999</math>. The fourth case is the numbers <math>1,000</math> to <math>9,999</math>. And lastly, the sum of the digits in <math>10,000</math>. The first case is just the sum of the numbers <math>1</math> to <math>9</math> which is, using <math>\frac{n(n+1)}{2}</math>, <math>45</math>. In the second case, every number <math>1</math> to <math>9</math> is used <math>19</math> times. <math>10</math> times in the tens place, and <math>9</math> times in the ones place. So the sum is just <math>19(45)</math>. Similarly, in the third case, every number <math>1</math> to <math>9</math> is used <math>100</math> times in the hundreds place, <math>90</math> times in the tens place, and <math>90</math> times in the ones place, for a total sum of <math>280(45)</math>. By the same method, every number <math>1</math> to <math>9</math> is used <math>1,000</math> times in the thousands place, <math>900</math> times in the hundreds place, <math>900</math> times in the tens place, and <math>900</math> times in the ones place, for a total of <math>3700(45)</math>. Thus, our final sum is <math>45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{\text{A)}180001}.</math> | We can find the sum using the following method. We break it down into cases. The first case is the numbers <math>1</math> to <math>9</math>. The second case is the numbers <math>10</math> to <math>99</math>. The third case is the numbers <math>100</math> to <math>999</math>. The fourth case is the numbers <math>1,000</math> to <math>9,999</math>. And lastly, the sum of the digits in <math>10,000</math>. The first case is just the sum of the numbers <math>1</math> to <math>9</math> which is, using <math>\frac{n(n+1)}{2}</math>, <math>45</math>. In the second case, every number <math>1</math> to <math>9</math> is used <math>19</math> times. <math>10</math> times in the tens place, and <math>9</math> times in the ones place. So the sum is just <math>19(45)</math>. Similarly, in the third case, every number <math>1</math> to <math>9</math> is used <math>100</math> times in the hundreds place, <math>90</math> times in the tens place, and <math>90</math> times in the ones place, for a total sum of <math>280(45)</math>. By the same method, every number <math>1</math> to <math>9</math> is used <math>1,000</math> times in the thousands place, <math>900</math> times in the hundreds place, <math>900</math> times in the tens place, and <math>900</math> times in the ones place, for a total of <math>3700(45)</math>. Thus, our final sum is <math>45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{\text{A)}180001}.</math> | ||
===Solution 2=== | ===Solution 2=== | ||
− | Consider the numbers from <math>0000-9999</math>. We have <math> | + | Consider the numbers from <math>0000-9999</math>. We have <math>4000</math> digits and each has equal an probability of being <math>0,1,2....9</math>. |
Thus, our desired sum is <math>4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{\text{A)}180001}.</math> | Thus, our desired sum is <math>4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{\text{A)}180001}.</math> | ||
Revision as of 17:00, 28 November 2018
Problem
Find the sum of digits of all the numbers in the sequence .
Solution
Solution 1
We can find the sum using the following method. We break it down into cases. The first case is the numbers to . The second case is the numbers to . The third case is the numbers to . The fourth case is the numbers to . And lastly, the sum of the digits in . The first case is just the sum of the numbers to which is, using , . In the second case, every number to is used times. times in the tens place, and times in the ones place. So the sum is just . Similarly, in the third case, every number to is used times in the hundreds place, times in the tens place, and times in the ones place, for a total sum of . By the same method, every number to is used times in the thousands place, times in the hundreds place, times in the tens place, and times in the ones place, for a total of . Thus, our final sum is
Solution 2
Consider the numbers from . We have digits and each has equal an probability of being . Thus, our desired sum is
Credit: Math1331Math
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
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