Difference between revisions of "1970 AHSME Problems/Problem 33"
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''Credit: Math1331Math'' | ''Credit: Math1331Math'' | ||
+ | ===Solution 3=== | ||
+ | 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. | ||
+ | |||
+ | ''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 .
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 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 davidaops.
Credit to 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 |
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