Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | <math>1197</math> | + | We can check every possible case by considering m = 1 and n = 2 through 50, m = 2 and n = 3 through 50, etc. Note that <math>40(50)=2000</math> whereby it is obvious that for m = 1 through 40, all possible n work. This accounts for <math>49+48\dots+10=\frac{49(50)}{2}-45=1180</math> cases. We individually check the remaining cases. Note that <math>41(49)=2009</math> and <math>41(50)=2050</math> so m = 41 contributes <math>49-42+1=8</math> cases. Similarly, m = 42 contributes 5 cases, m = 43 contributes 3 cases, and m = 44 contributes 1 case. This sums to <math>\boxed{1197}</math>. |
== See also == | == See also == | ||
Latest revision as of 09:43, 6 March 2024
Problem
How many ordered pairs 
of positive integers satisfying 
have the property that their product 
is less than 
?
Solution
We can check every possible case by considering m = 1 and n = 2 through 50, m = 2 and n = 3 through 50, etc. Note that 
whereby it is obvious that for m = 1 through 40, all possible n work. This accounts for 
cases. We individually check the remaining cases. Note that 
and 
so m = 41 contributes 
cases. Similarly, m = 42 contributes 5 cases, m = 43 contributes 3 cases, and m = 44 contributes 1 case. This sums to 
.
See also
| 2015 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
