# Find the number of ordered triples (a,b,c) of positive integers such that $30a + 50b + 70c \le 343.$

Find the number of ordered triples $$(a,b,c)$$ of positive integers such that $$30a + 50b + 70c \le 343.$$

My confusion is that while solving the question a, b,c can be zero or not

• Positive($> 0$) implies non-zero ($\neq 0$) – ab123 Aug 11 at 15:51
• @ab123 you mean (0,0,4) is our solution – Abhishek Kumar Aug 11 at 16:07
• No, we need $(a, b, c)$ such that all three of them are positive – ab123 Aug 11 at 16:09

• $30(a-1)\le 193$ gives $a\in 1,..,7$, next $50(b-1)\le 193$ gives $b\in 1,..,4$ and $70(c-1)\le 193$ gives $c\in 1,..,3$. Not every possible combination will be admissible. – LutzL Aug 11 at 16:39
• @LutzL He's saying that if you exhaustively test all the possibilities, $30$ are admissible, which agrees with my calculations. – saulspatz Aug 11 at 16:52
• @saulspatz: But there is still the triple $(7,1,1)$, which is also admissible, missing from the cases stated in the first sentence. – LutzL Aug 11 at 17:08
• @saulspatz : The first sentence states that $a,b$ can have values $1-4$ while $c$ can have values $1-3$. However, $a=7$ is also possible. If the total number 30 is correct, then it was not obtained by the reasoning given in this answer. – LutzL Aug 11 at 18:06