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I'm having difficulties to solve this problem for a 6th grade kid:

The sum of $11$ different natural numbers is $70$. Show that their product is divisible by $720$.

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  • $\begingroup$ Naturals numbers? $\endgroup$ May 10, 2018 at 7:28
  • $\begingroup$ @ovetz13 yes. I edited the question. I managed to reach something like this: 11*x + Q = 70. Where x is natural, not 0 $\endgroup$
    – Florin M.
    May 10, 2018 at 7:30
  • $\begingroup$ @FlorinM. Just so you know, you accepted the wrong answer. $\endgroup$
    – rubik
    May 10, 2018 at 7:43
  • $\begingroup$ I undo the accepted answer $\endgroup$
    – Florin M.
    May 10, 2018 at 7:48
  • $\begingroup$ It's expected that you will show what you have tried and where you are stuck $\endgroup$ May 10, 2018 at 19:20

2 Answers 2

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The lowest eleven numbers we can pick are those from $1$ to $11$, and they sum to $66$, so we only have to add $4$ to one (or more) of them to make $70$. By keeping in mind that the numbers must be distinct, the only five possibilities are: $$\begin{align} &(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15)\tag{11 + 4}\\ &(1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14)\tag{10 + 4}\\ &(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13)\tag{9 + 4}\\ &(1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12)\tag{8 + 4}\\ &(1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13)\tag{10 + 2, 11 + 2}\\ \end{align}$$

(Why we cannot add $1$ and $3$ is left to the reader.)

Now, $720 = 6! = 2\cdot3\cdot4\cdot5\cdot6,$ and as it can be seen those factors are always present.


Note: It's not really needed to enumerate all the possibilities because we just have to prove that the product $6!$ is always present, and that follows from the fact that we cannot add $4$, $3$, $2$, $1$ to any of the numbers $1-6$ because otherwise we would have duplicates. Still, I believe that a 6th grader might understand the problem better if you let them derive all the possibilities.

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$$1+2+3+..+11=66$$ So we have to put 4 to one of the numbers so the sum is 70. There are few options. $11+4$, $10+4$, $9+4$ or $8+4$. So we have the numbers: $$1*2*3*4*5*6*7*8*9*10*14$$ $$1*2*3*4*5*6*7*8*9*14*11$$ $$1*2*3*4*5*6*7*8*13*10*11$$ $$1*2*3*4*5*6*7*12*9*10*11$$

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  • $\begingroup$ Why not also 7 + 4? The sum will also be 70. The 11th number which we want to be modified ( + 4 ) can be chosen aleatory ? $\endgroup$
    – Florin M.
    May 10, 2018 at 7:38
  • $\begingroup$ The numbers have to be different $7+4=11$ which we already have. $\endgroup$ May 10, 2018 at 7:39
  • $\begingroup$ This is incorrect because you can split the $4$ and add it to different numbers. There are five possibilities, not four. $\endgroup$
    – rubik
    May 10, 2018 at 7:40
  • $\begingroup$ @rubik You are right, I do think there are 5 cases $\endgroup$
    – Florin M.
    May 10, 2018 at 7:48
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    $\begingroup$ To elaborate: this answer is not incorrect per se, but if fails to conclude the reasoning. $\endgroup$
    – rubik
    May 10, 2018 at 7:58

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