How many positive integers which are less or equal with $2013$ such that $3$ or $5$ divide the number.
Problem
Source:
Tags:
05.03.2013 15:15
$LCM(3,5)=15$ Ans $=[\frac{2013}{15}]+[\frac{2013}{15^2}]+[\frac{2013}{15^3}]+...=134+8+0+0...=142 $
05.03.2013 16:21
3 or 5 means $ =[\frac{2013}{3}]+[\frac{2013}{5}]-[\frac{2013}{15}]= 671+402-134=969$ my english not Good but i understand 3 or 5 like this if im wrong can anybody explain redoxs solution?
06.03.2013 05:08
byrmc wrote: 3 or 5 means $ =[\frac{2013}{3}]+[\frac{2013}{5}]-[\frac{2013}{15}]= 671+402-134=969$ my english not Good but i understand 3 or 5 like this if im wrong can anybody explain redoxs solution? You are correct. Redox might have misread the problem...he is finding the powers of 15 but he is neglecting individual $5's$ and $3's$ which can make more $15's$.k However, this is not what the problem asked....
30.06.2013 18:21
Actually, byrmc is not completely correct $671+402-134=\boxed{939}$ But this problem is very very easy and I'd expect it in the middle school contest.