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Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 1"

(Created page with "== Problem == Find a <math>3</math>-digit integer less than <math>200</math> where each digit is odd and the sum of the cubes of the digits is the original number. == Solution ...")
 
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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 ==
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{{UNC Math Contest box|year=2010|n=II|before=First Question|num-a=2}}
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{{UNCO Math Contest box|year=2010|n=II|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 21:05, 9 February 2015

Problem

Find a $3$-digit integer less than $200$ 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 $1bc$. We know that $b$ and $c$ are odd, and that $b^3 + c^3 + 1 = 100 + 10b + c$. This means that both $b$ and $c$ are within the set ${1,3,5,7,9}$. We also notice that since $7^3$ and $9^3$ are both far greater than $200$, they can not be our digits, so the remaining set is ${1,3,5}$. Looking at only the last digits of $1^3 = 1, 3^3 = 27, 5^3 = 125$, we see that only when $125$ and $27$ are together can we get a last digit of $5$ or $3$, which we see is $125+27+1 = 153$. Upon checking, $\boxed{\textbf{153}}$ 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
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