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 == | ||
− | {{ | + | {{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 22:05, 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 |