Difference between revisions of "1970 AHSME Problems/Problem 33"

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== Solution ==
 
== Solution ==
""Solution 1""
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'''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""
+
'''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 18:19, 7 October 2016

Problem

Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$.

$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$

Solution

Solution 1 We can find the sum using the following method. We break it down into cases. The first case is the numbers $1$ to $9$. The second case is the numbers $10$ to $99$. The third case is the numbers $100$ to $999$. The fourth case is the numbers $1,000$ to $9,999$. And lastly, the sum of the digits in $10,000$. The first case is just the sum of the numbers $1$ to $9$ which is, using $\frac{n(n+1)}{2}$, $45$. In the second case, every number $1$ to $9$ is used $19$ times. $10$ times in the tens place, and $9$ times in the ones place. So the sum is just $19(45)$. Similarly, in the third case, every number $1$ to $9$ is used $100$ times in the hundreds place, $90$ times in the tens place, and $90$ times in the ones place, for a total sum of $280(45)$. By the same method, every number $1$ to $9$ is used $1,000$ times in the thousands place, $900$ times in the hundreds place, $900$ times in the tens place, and $900$ times in the ones place, for a total of $3700(45)$. Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{\text{A)}180001}.$ Solution 2 Consider the numbers from $0000-9999$. We have $40000$ digits and each has equal probability of being $0,1,2....9$ Our requested sum then is $4000(45)+1=\boxed{\text{A)}180001}.$ Credit: Math1331Math

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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All AHSME Problems and Solutions

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