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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

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    $\begingroup$ Positive($> 0$) implies non-zero ($\neq 0$) $\endgroup$ – ab123 Aug 11 at 15:51
  • $\begingroup$ @ab123 you mean (0,0,4) is our solution $\endgroup$ – Abhishek Kumar Aug 11 at 16:07
  • $\begingroup$ No, we need $(a, b, c)$ such that all three of them are positive $\endgroup$ – ab123 Aug 11 at 16:09
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It is the question of combinations every variable can have the values from 1-4 except the variable c bcz 70*4=280 and 50+ 30=80 which exceeds 343.so variable c can have values from 1-3.No we will make combinations for each value of a,b,c and hence the answer will come and it will be 30

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  • $\begingroup$ $30(a-1)\le 193$ 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. $\endgroup$ – LutzL Aug 11 at 16:39
  • $\begingroup$ @LutzL He's saying that if you exhaustively test all the possibilities, $30$ are admissible, which agrees with my calculations. $\endgroup$ – saulspatz Aug 11 at 16:52
  • $\begingroup$ @saulspatz: But there is still the triple $(7,1,1)$, which is also admissible, missing from the cases stated in the first sentence. $\endgroup$ – LutzL Aug 11 at 17:08
  • $\begingroup$ @LutzL I can't really understand what he is saying in detail. $\endgroup$ – saulspatz Aug 11 at 17:42
  • $\begingroup$ @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. $\endgroup$ – LutzL Aug 11 at 18:06

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