Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 1"
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== Solution == | == Solution == | ||
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| + | We have a three digit number that can be represented as <math>1bc</math>. We know that <math>b</math> and <math>c</math> are odd, and that <math>b^3 + c^3 + 1 = 100 + 10b + c</math>. This means that both <math>b</math> and <math>c</math> are within the set <math>{1,3,5,7,9}</math>. We also notice that since <math>7^3</math> and <math>9^3</math> are both far greater than <math>200</math>, they can not be our digits, so the remaining set is <math>{1,3,5}</math>. Looking at only the last digits of <math>1^3 = 1, 3^3 = 27, 5^3 = 125</math>, we see that only when <math>125</math> and <math>27</math> are together can we get a last digit of <math>5</math> or <math>3</math>, which we see is <math>125+27+1 = 153</math>. Upon checking, <math>\boxed{\textbf{153}}</math> is a working solution. | ||
== See also == | == See also == | ||
Latest revision as of 21:05, 9 February 2015
Problem
Find a 
-digit integer less than 
where each digit is odd and the sum of the cubes of the digits is
the original number.
Solution
We have a three digit number that can be represented as 
. We know that 
and 
are odd, and that 
. This means that both and are within the set 
. We also notice that since 
and 
are both far greater than , they can not be our digits, so the remaining set is 
. Looking at only the last digits of 
, we see that only when 
and 
are together can we get a last digit of 
or , which we see is 
. Upon checking, 
is a working solution.
See also
| 2010 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
