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Find the number of ordered triplets = b, c of positive integers such that 30a + 50b + 70 C is equal to or less than 343

Find the number of ordered triplets = b, c of positive integers such that 30a + 50b + 70 C is equal to or less than 343

Grade:10

1 Answers

Aditya Gupta
2081 Points
4 years ago
obviously c cant be more than 4, as 70c would become greater than equal to 350.
so c can be 1, 2, 3, 4
but c also cannot be 4, because if c=4, then 30a + 50b would have to be less than or equal to 343 – 280= 63, but the minimum value of 30a + 50b is 30+50= 80
so c can only be 1, 2, 3.
when c=3, 30a + 50b should be less than or equal to 343 – 210= 133. in this case, when b=2, a=1 and when b=1, a can be 1, 2.
when c=2, 30a + 50b should be less than or equal to 343 – 140= 203. in this case, when b=3, a=1, when b=2, a can be 1, 2, 3 and when b=1, a can be 1,2,3,4,5
when c=1, 30a + 50b should be less than or equal to 343 – 70= 273. in this case, when b=4, a=1, 2, when b= 3, a= 1,2,3,4, when b=2, a=1,2,3,4,5 and when b=1, a=1,2,3,4,5,6,7.
these are all the possible triplets. the key insight here was to begin with the variable having the largest coeffcient (in this case c with coeff 70).
you can count all these triplets and if i have counted them right then number of ordered triplets= 30
kindly approve :)

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