Difference between revisions of "1970 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | For the sums of the digits of be equal to 43, this means that the number's digits must fall under one of two categories. It either has 4 9's and a 7, or 3 9's and 2 8's. You can use casework from there to find that there are 15 5-digit numbers in which the sum of the digits is equal to 43. To find the number of ways that the number is divisible by 11, use the divisibility rule for 11, letting the number be represented by <math>abcde</math>. This gives 3 possible numbers that are divisible by 11, so the answer is <math>\frac{3}{15}=\fbox{B}</math> |
== See also == | == See also == | ||
Revision as of 19:28, 12 October 2019
Problem
If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to 43, what is the probability that this number will be divisible by 11?
Solution
For the sums of the digits of be equal to 43, this means that the number's digits must fall under one of two categories. It either has 4 9's and a 7, or 3 9's and 2 8's. You can use casework from there to find that there are 15 5-digit numbers in which the sum of the digits is equal to 43. To find the number of ways that the number is divisible by 11, use the divisibility rule for 11, letting the number be represented by 
. This gives 3 possible numbers that are divisible by 11, so the answer is 
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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. 
