Difference between revisions of "1970 AHSME Problems/Problem 33"
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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> | ||
| − | + | Credit: Math1331Math | |
== See also == | == See also == | ||
Revision as of 17:15, 7 October 2016
Problem
Find the sum of digits of all the numbers in the sequence 
.
Solution
[b]Solution 1[/b]
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 
[b]Solution 2[/b]
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. 
