I am not sure how to find an answer to this question. Is there a way to solve it without simply trial and error?

Do there exist ten distinct positive integers such that their arithmetic mean is

(a) six times as great as their greatest common divisor?

(b) five times as great as their greatest common divisor?


1 Answer 1


Let us call the ten positive integers $a_1 , a_2 , \dots , a_{10}$ and let us call their GCD $d$. Note that $d \mid a_i$ for each $i$. We can write $a_i = b_i d $ with $b_i$ a positive integer.

The condition for a) is: $$\frac{a_1 + a_2 + \dots + a_{10}}{10} = 6 d$$

Let us divide by $d$ to get:

$$\frac{b_1 + b_2 + \dots + b_{10}}{10} = 6 $$

So the question turns into are their $10$ distinct positive integers, $b_i$ such that their arithmetic mean is $6$, equivalently $$b_1 + b_2 + \dots + b_{10} = 60 $$ are their $10$ distinct positive integers $b_i$ such that their sum is $60$. (To be precise we should also say whose GCD is $1$ but this is a detail).

Recall that the sum of the first $10$ positive integers is already $1+ 2 + \dots + 10 = 55$. So let us start $b_1 = 1, b_2 = 2, \dots, b_9 = 9$ and then $b_10= 15$.
You can now choose $d$ to be whatever you like and multiply $b_i$ by it to get the $a_i$. If you want to make it simple just take $d=1$ and you are done.

For b) it would be that the sum must be $50$. This is not possible.

  • $\begingroup$ I don't understand the part where you say "So you are looking..." Could you please further explain. Thank you. $\endgroup$
    – anonymous
    Feb 21, 2015 at 0:04
  • $\begingroup$ The quantity can be written as $\frac{a_1}{d}+ \frac{a_2}{d} + \dots$; call $\frac{a_i}{d}= b_i$. So you need $b_1 + \dots + b_{10}= 60$. The $b_i$ are all integers as the $d$ divides the $a_i$ being the GCD. $\endgroup$
    – quid
    Feb 21, 2015 at 1:16
  • $\begingroup$ I am really sorry, but I am still really confused. Could you please try and explain how to find the answer to this question more simply. I really appreciate your support. $\endgroup$
    – anonymous
    Feb 21, 2015 at 2:29
  • $\begingroup$ I expanded the answer a bit; if it is still unclear please be specific what is unclear. Or what is the first thing that is unclear. $\endgroup$
    – quid
    Feb 21, 2015 at 12:11
  • $\begingroup$ For part (a) would 1,2,3,4,5,6,7,9,11,12 be correct? $\endgroup$
    – anonymous
    Feb 21, 2015 at 22:09

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