Difference between revisions of "2023 WSMO Team Round Problems/Problem 1"
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==Solution== | ==Solution== | ||
| + | From the condition, possible numbers of pencils Bob could have are <cmath>3,13,23,33,43,\boxed{53},63,\ldots</cmath> and from the second condition we have <cmath>5,17,29,41,\boxed{53},65,\ldots.</cmath> So, th emsallest number of pencils Bob could have is <math>\boxed{53}.</math> | ||
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| + | ~pinkpig | ||
Latest revision as of 13:48, 13 September 2025
Problem
Bob has a number of pencils. When Bob splits them into groups of 10, he has 3 left over. When he splits them into groups of 12, he has 5 left over. Find the smallest number of pencils Bob can have.
Solution
From the condition, possible numbers of pencils Bob could have are ![\[3,13,23,33,43,\boxed{53},63,\ldots\]](http://latex.artofproblemsolving.com/f/8/3/f8391cea57837e04a1cf6e1624681ddc07cb27ec.png)
and from the second condition we have ![\[5,17,29,41,\boxed{53},65,\ldots.\]](http://latex.artofproblemsolving.com/0/0/b/00b45ab2052187236ce1e7489dd696465bcb4d56.png)
So, th emsallest number of pencils Bob could have is 
~pinkpig
