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How do you get a result of 50 from 7 odd numbers? - Answers

50 = 5x3 + 45x7/15 + 315-301. You can see from this that there is an infinite number of ways of getting the last pair of odd numbers with a difference of 14. You could use the useful trick that for 3 equally spaced numbers a, b, c, then the result of a-b+c is always b, and a+b+c is 3b. So you could have 47 + (7-9+11)/(1-3+5), or 47 + (7+9+11)/(1+3+5). There is an infinite number of ways of using that trick. (I assume that by "7 odd numbers" you mean "7 different positive integers not divisible by 2 exactly").



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How do you get a result of 50 from 7 odd numbers? - Answers

https://math.answers.com/algebra/How_do_you_get_a_result_of_50_from_7_odd_numbers

50 = 5x3 + 45x7/15 + 315-301. You can see from this that there is an infinite number of ways of getting the last pair of odd numbers with a difference of 14. You could use the useful trick that for 3 equally spaced numbers a, b, c, then the result of a-b+c is always b, and a+b+c is 3b. So you could have 47 + (7-9+11)/(1-3+5), or 47 + (7+9+11)/(1+3+5). There is an infinite number of ways of using that trick. (I assume that by "7 odd numbers" you mean "7 different positive integers not divisible by 2 exactly").



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https://math.answers.com/algebra/How_do_you_get_a_result_of_50_from_7_odd_numbers

How do you get a result of 50 from 7 odd numbers? - Answers

50 = 5x3 + 45x7/15 + 315-301. You can see from this that there is an infinite number of ways of getting the last pair of odd numbers with a difference of 14. You could use the useful trick that for 3 equally spaced numbers a, b, c, then the result of a-b+c is always b, and a+b+c is 3b. So you could have 47 + (7-9+11)/(1-3+5), or 47 + (7+9+11)/(1+3+5). There is an infinite number of ways of using that trick. (I assume that by "7 odd numbers" you mean "7 different positive integers not divisible by 2 exactly").

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      50 = 5x3 + 45x7/15 + 315-301. You can see from this that there is an infinite number of ways of getting the last pair of odd numbers with a difference of 14. You could use the useful trick that for 3 equally spaced numbers a, b, c, then the result of a-b+c is always b, and a+b+c is 3b. So you could have 47 + (7-9+11)/(1-3+5), or 47 + (7+9+11)/(1+3+5). There is an infinite number of ways of using that trick. (I assume that by "7 odd numbers" you mean "7 different positive integers not divisible by 2 exactly").
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