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So my professor asked us to do an Olympiad exercise which says that the sum of five consecutive odd integers is $55$, find those integers. But I've never seen such an exercise so it is quite new and if I learn how then I will be able to solve similar exercises.

I tried to think of it in terms of geometry but that didn't help me.

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  • $\begingroup$ let the first be n, then what are the others? $\endgroup$
    – Lost1
    Dec 27, 2013 at 19:52
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    $\begingroup$ It's simpler to let the third number be n. $\endgroup$ Dec 27, 2013 at 20:31
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    $\begingroup$ Just curious - what specifically did you try with geometry? There are reasonable ways to use geometry here. $\endgroup$
    – 2'5 9'2
    Dec 28, 2013 at 2:27
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    $\begingroup$ $1$kviews, $10$ votes and $10$ answers with a lot of votes! I can't think how did this question go this far... $\endgroup$
    – chubakueno
    Dec 28, 2013 at 18:24

10 Answers 10

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Avoiding explicit algebra computation, their average would be $11$ and they would have to surround that average symmetrically.

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  • $\begingroup$ This is, in my opinion, the "right" way to solve the problem. But if someone doesn't immediately think of it, then I think the next best approach (better than setting up algebraic equations) is trial and error. $\endgroup$ Dec 28, 2013 at 1:33
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    $\begingroup$ Why do you think that? How would you justify trial and error over an algebraic approach? $\endgroup$
    – Asinomás
    Dec 28, 2013 at 2:45
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    $\begingroup$ Neat - my first gold badge. $\endgroup$
    – 2'5 9'2
    Dec 28, 2013 at 3:21
  • $\begingroup$ Cool, glad my noobness could help you. :) $\endgroup$
    – Asinomás
    Dec 28, 2013 at 3:58
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    $\begingroup$ @user4140 Aside from changing the number from $11$ to something else, I would use the exact same wording if it were an even number of numbers, an odd number of even numbers, etc. $\endgroup$
    – 2'5 9'2
    Dec 28, 2013 at 19:23
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let k be the smallest number. Then you have $k+(k+2)+(k+4)+(k+6)+(k+8)=55.$ So you get $5k+20=55\rightarrow 5k=35\rightarrow k=7$

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    $\begingroup$ Perhaps a better choice of notation would've been $k\pm1$, $k\pm3$, and $k+5$. $\endgroup$
    – Lucian
    Dec 28, 2013 at 4:04
  • $\begingroup$ But then k would not be the least element. Other answers use k as the third least element (aka the average) this one uses k as the smallest. $\endgroup$
    – Asinomás
    Dec 28, 2013 at 4:10
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Hint: Choose one of those equations and solve it: $$\begin{align} & \text{The first integer:} & \qquad x+\color{blue}{x+2+x+4+x+6+x+8}=55 \\\,\\ & \text{The second integer:} & \qquad \color{red}{x-2}+x+\color{blue}{x+2+x+4+x+6}=55 \\\,\\ & \text{The third integer:} & \qquad \color{red}{x-4+x-2}+x+\color{blue}{x+2+x+4}=55 \\\,\\ & \text{The fourth integer:} & \qquad \color{red}{x-6+x-4+x-2}+x+\color{blue}{x+2}=55 \\\,\\ & \text{The fifth integer:} & \qquad \color{red}{x-8+x-6+x-4+x-2}+x=55 \end{align}$$ Obviously the simplest choice is to solve for the third integer, since the $2$s and $4$s cancel out, and you're left with the simple equation, $5x=55$.

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Hint $\, $ The sum of an odd number $k$ of terms in arithmetic progression is $k$ times the middle term, since the differences $\,\color{#c00}{\pm\Delta}\,$ from the middle term $\,n\,$ cancel out on addition, e.g. for $\,k=7$

$$\begin{eqnarray} n &\,+\,& n\!\color{#c00}{+\!a} &\,+\,& n\!\color{#c00}{+\!2a} &\,+\,& n\!\color{#c00}{+\!3a} \\ &\,+\,& n\!\color{#c00}{-\!a} &\,+\,& n\!\color{#c00}{-\!2a} &\,+\,& n\!\color{#c00}{-\!3a} \\ \hline =\ 7n \end{eqnarray}$$

Thus, in your case we have $\ 5n = 55\ \Rightarrow\ n =\,\ldots$

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  • $\begingroup$ How could one downvote such excellent answer? $\endgroup$
    – user93957
    Dec 28, 2013 at 22:21
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    $\begingroup$ @Adobe Perhaps they thought I should have said it with words instead of algebra - as in Alex's later answer. I was tempted to do that, but I think the algebra makes more precise the key idea (innate reflection symmetry). This is but one of many beautiful proofs that work by pairing elements using reflections or involutions, e.g. see Wilson's theorem for groups. Also, be aware that this question was on the SE Hot List, so it got many visitors from other sites. So votes may not be representative of MSE. $\endgroup$ Dec 28, 2013 at 22:34
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    $\begingroup$ @Adobe See also the legendary example: Gauss's grade school method of quickly summing the first $\,100\,$ integers. $\endgroup$ Dec 28, 2013 at 22:45
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    $\begingroup$ @Adobe - by accident. On my iPad, meant to upvote, dog jumped on me, couldn't undo. My vote is locked until Bill edits the answer. I was in my usual money.SE just now and saw my history showed a Math DV. Just realized what happened. Sorry, all. $\endgroup$ Dec 29, 2013 at 21:06
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We have

$$\sum_{k=p}^{p+4} (2k+1)=55\iff2\times\frac{5(p+4+p)}{2}+5=55\iff p=3$$ so the first odd number is $7$.

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  • $\begingroup$ A variant is to use the formula for the sum of an arithmetic series, $ \ s \ = \ \frac{n}{2} \ \cdot \ [2a_1 \ + \ (n - 1) \cdot d] \ $ , with $ \ a_1 \ $ being the initial term, $ \ d \ $ being the difference between terms , and $ \ n \ $ being the number of terms. For this question, we need to solve $ \ \frac{5}{2} \ \cdot \ [2a_1 \ + \ (5 - 1) \cdot 2] \ = \ 55 \ $ for $ \ a_1 \ $ . $\endgroup$ Dec 27, 2013 at 22:11
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    $\begingroup$ ${}{}{}{}{}{}+1$ $\endgroup$
    – Mikasa
    Dec 29, 2013 at 5:53
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    $\begingroup$ $\widetilde{\color{red}{\boxed{+1}}}$ $\endgroup$
    – user93957
    Dec 29, 2013 at 15:42
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    $\begingroup$ ${}{}{}$Thanks @Adobe $\endgroup$
    – user63181
    Dec 29, 2013 at 15:44
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As consecutive odd numbers differ by $2$

and we have odd number of integer terms (namely $5$), its beneficial take the numbers to be $n,n\pm2,n\pm4$

Had the number of odd integers been even, we could take $n\pm1,n\pm3,\cdots$

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I did it this way:

We need 5 consecutive odd numbers who add up to 55, so their mid/centre or the 3rd (arithmetic mean) number must be 55/5 which is 11(as the sequence is symmetric about this number). The previous two and the next two odd numbers complete the needed sequence.

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Let's take the numbers as $x-4,x-2,x,x+2,x+4$. Now, our equality takes the form

$$(x-4) + (x-2) + x + (x+2) + (x+4) = 55,$$

so $5x = 55$, i.e., $x = 11$.

We conclude that the numbers are $7,9,11,13,15$.

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For odd numbers the answer is :$(x-4)+ (x-2)+x+(x+2)+(x+4)=55$ i.e. $5x= 55$ therefore $x=11$, and the series would be: $7+9+11+13+15=55$.

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The sum of five consecutive odd integers is $55$.

Let $n$ = any integer.

Then $2n$ = an even integer

and $2n + 1$ = an odd integer.

Thus $\color{red}{\underbrace{2n +1}},\underbrace{(2n+1)+\color{blue}{2}},\underbrace{ (2n+1)+\color{blue}{4}},\underbrace{(2n+1)+\color{blue}{6}},\underbrace{(2n+1)+\color{blue}8} $

are five consecutive odd integers. The sum is $$10n + 25 =55$$ $$10n=30$$$$n=3$$ This gives $\color{red}{\overbrace{2(3)+1}}=7$ as the $\color{red}{first}$ of the five consecutive odd integers whose sum is $55$.

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