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I have this statement:

The central term of three consecutive odd ones can be determined, if it is known that the sum of these is:

i) At most $75$

ii) At least $63$

My attempt was:

Let a $2n-1$ an odd number, then $(2n-1) + (2n+1) +(2n+3) =6n+3$

Using $i) n \leq12$ Exist infinity odd numbers.

Using $ii) n \geq10$ Exist infinity odd numbers.

Using $i), ii)$ together. $n= \{10,11,12\}$

Since the sum of three consecutive numbers is in the form $6n+3$, the sum must be an odd number, so the unique possible value is $11$. But if $6n+3=11$, $2n+1$ isn't a integer number, therefore i can't assure.

But, my second solution was the same, but let $2n+1$ an odd number, then:

$(2n+1)+(2n+3)+(2n+5)=6n+9$

Using $i) n \leq 11$ Exist infinity odd numbers.

Using $ii) n \geq 9$ Exist infinity odd numbers.

Using both together, $n= \{9,10,11\}$

The sum is of the form $6n+9$, Therefore is an odd sum too.

Then the possible value are $9,11$, But with $6n+9=11$, the central term isn't an integer number, however with $6n+9=9 => n = 0$, thus the central term is $3$ that work with $1+3+5 = 9$. According to the guide, the correct solution is

Additional information is required

That is according to my first solution, but no with my second solution. What is wrong with my second solution?

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    $\begingroup$ "the sum of three consecutive numbers is in the form 6n+3, the sum must be an odd number" The sum is $6n+3$. $6n +3$ is always an odd number. "so the unique possible value is 11". $n$ isnt that sum. $6n+3$ is the sum. If $n=10,11,12$ then the sum is $6n+3= 63, 69$ or $75$. "But if 6n+3=11" $6n+3$ doesn't equal $11$. $n$ equals $11$ (or $10$ or $12$). $6n + 3= 69$ (or $63$ or $75$). $\endgroup$ – fleablood Sep 13 '19 at 3:21
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In the first case, for the consecutive odd terms $2n-1,$ $2n+1,$ $2n+3,$ you correctly deduce that the sum is $6n+3$ and that $n \in \{10,11,12\}.$

But then you start treating $n$ as if it were the sum. That is a mistake. The variable $n$ is just some integer that you put into some formulas for convenience. It is not equal to any of the numbers of interest in the question: not the middle term, not either of the other terms, and certainly not the sum.

Note that $6n+3$ is just as odd when $n$ is $10$ or $12$ as it is when $n=11.$ So you have no reason based on "the sum must be odd" to reject any of the three values of $n.$ Likewise you have no reason to take any of the elements of $\{10,11,12\}$ and say it must be equal to $6n+3.$ Again, $n$ is just an arbitrary integer parameter you introduced, and $6n+3$ is the sum, which is something completely different from any possible value of $n.$

You have the same kind of mistake in your second attempt. You come up with $1 + 3 + 5 = 9,$ but $9$ is not at least $63$ as was required for the sum of the three consecutive odd numbers. In fact $n = 9$ is a perfectly fine solution, but then the terms are $$(2n+1)+(2n+3)+(2n+5) = 19+21+23 = 63.$$ On the other hand, $n=10$ and $n=11$ are perfectly fine solutions too when you correctly write out all the formulas. So you cannot conclude that $n=9.$

It is fine to introduce an arbitrary integer variable like $n$ so that you have something to base formulas on, but you must always remember what the variables you introduce are not (in this case, not the sum of the three terms).


My approach: let $S$ be the sum, $m$ the middle term. We know $m$ is odd and the other two terms are $m - 2$ and $m + 2.$ The sum of the three terms is therefore $3m.$ That gives us $m = S/3.$ Since $63 \leq S \leq 75,$ we have $21 \leq m \leq 25.$ But by inspection, $m = 21,$ $m = 23,$ and $m=25$ all solve the given conditions. That is why more information is required.

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  • $\begingroup$ @YiFan That's right, thanks! $\endgroup$ – David K Sep 13 '19 at 2:58
  • $\begingroup$ Thanks for good explanation. $\endgroup$ – Eduardo Sebastian Sep 13 '19 at 3:33

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