Difference between revisions of "1970 AHSME Problems/Problem 33"
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== Solution == | == Solution == | ||
− | + | ""Solution 1"" | |
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"" | |
Consider the numbers from <math>0000-9999</math>. We have <math>40000</math> digits and each has equal probability of being <math>0,1,2....9</math> | Consider the numbers from <math>0000-9999</math>. We have <math>40000</math> digits and each has equal probability of being <math>0,1,2....9</math> | ||
Our requested sum then is <math>4000(45)+1=\boxed{\text{A)}180001}.</math> | Our requested sum then is <math>4000(45)+1=\boxed{\text{A)}180001}.</math> |
Revision as of 18:18, 7 October 2016
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 probability of being
Our requested sum then 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.