1970 AHSME Problems/Problem 33

Revision as of 18:01, 7 October 2016 by Ramanan369 (talk | contribs) (Solution)

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

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 $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{\text{A)}180001}.$

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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