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

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''Credit: Math1331Math''
 
''Credit: Math1331Math''
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===Solution 3===
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As in Solution 2, we consider the four digit numbers from <math>0000-9999.</math> We see that we have <math>10000\times4=40000</math> digits with each digit  appearing equally.
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Thus, the digit sum will be the average of the digits multiplied by <math>40000.</math> The digit average comes out to be <math>\frac{0+9}{2},</math> since all digits are consecutive.
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So, our answer will be <math>\frac{9}{2} \times 40000 = 180000.</math> However, since we purposely did not include <math>10000,</math> we add one to get our final answer as <math>\boxed{\textbf{(A)}180001}.</math>
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Solution by davidaops.
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''Credit to Math1331Math''
  
 
== See also ==
 
== See also ==

Latest revision as of 15:54, 15 December 2018

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 an probability of being $0,1,2....9$. Thus, our desired sum is $4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{\text{A)}180001}.$

Credit: Math1331Math

Solution 3

As in Solution 2, we consider the four digit numbers from $0000-9999.$ We see that we have $10000\times4=40000$ digits with each digit appearing equally.

Thus, the digit sum will be the average of the digits multiplied by $40000.$ The digit average comes out to be $\frac{0+9}{2},$ since all digits are consecutive.

So, our answer will be $\frac{9}{2} \times 40000 = 180000.$ However, since we purposely did not include $10000,$ we add one to get our final answer as $\boxed{\textbf{(A)}180001}.$

Solution by davidaops.

Credit to Math1331Math

See also

1970 AHSC (ProblemsAnswer KeyResources)
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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