Difference between revisions of "1970 AHSME Problems/Problem 33"
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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{\text{(A)}180001}.</math> | 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{\text{(A)}180001}.</math> | ||
| + | |||
| + | Solution by Math1331Math and davidaops. | ||
== See also == | == See also == | ||
Revision as of 16:11, 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
Solution 3
As in Solution 2, we consider the four digit numbers from 
We see that we have 
digits with each digit appearing equally.
Thus, the digit sum will be the average of the digits multiplied by 
This is easy. The digit average comes out to be 
since all digits are consecutive.
So, our answer will be 
However, since we purposely did not include 
we add one to get our final answer as 
Solution by Math1331Math and davidaops.
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. 
