Difference between revisions of "2019 CIME I Problems/Problem 11"
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| − | We define a positive integer to be < | + | We define a positive integer to be <i>multiplicative</i> if it can be written as the sum of three distinct positive integers <math>x, y, z</math> such that <math>y</math> is a multiple of <math>x</math> and <math>z</math> is a multiple of <math>y</math>. Find the sum of all the positive integers which are not <i>multiplicative</i>. |
=Solution 1= | =Solution 1= | ||
| − | The positive integers which are not < | + | The positive integers which are not <i>multiplicative</i> are <math>1, 2, 3, 4, 5, 6, 8, 12, 24</math>. These sum to <math>\boxed{65}</math>. |
==See also== | ==See also== | ||
{{CIME box|year=2019|n=I|num-b=10|num-a=12}} | {{CIME box|year=2019|n=I|num-b=10|num-a=12}} | ||
{{MAC Notice}} | {{MAC Notice}} | ||
Latest revision as of 15:04, 6 October 2020
We define a positive integer to be multiplicative if it can be written as the sum of three distinct positive integers 
such that 
is a multiple of 
and 
is a multiple of . Find the sum of all the positive integers which are not multiplicative.
Solution 1
The positive integers which are not multiplicative are 
. These sum to 
.
See also
| 2019 CIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All CIME Problems and Solutions | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
